LAC CAT-1: Partial Derivatives MCQ Question Bank
Course: Linear Algebra and Calculus
Course Code: 25APS0101
University: Chitkara University, Punjab • Semester 01
Scope: Complete Unit 1 Curriculum (CAT-1 Scope) • 163 Solved Examination Problems with Step-by-Step Derivations
Question Bank Quick Navigator
| Category | Thematic Focus | Question Range | Topics Covered |
|---|---|---|---|
| Category A | First & Higher-Order Partial Derivatives & Point Evaluations | Q1 – Q23 | Partial derivative fundamentals, higher-order derivatives ($f_{xx}, f_{yy}, f_{xy}$), mixed partials, and evaluation at specific coordinate points. |
| Category B | Homogeneous Functions & Euler's Theorem | Q24 – Q48 | Degree calculation, first-order Euler relation ($x u_x + y u_y = nu$), second-order Euler corollary ($n(n-1)u$), and composite functions ($g(u) = n \frac{f}{f'}$). |
| Category C | Tangent Planes & Normal Lines to Surfaces | Q49 – Q58 | Surface gradient vectors $\nabla F$, direction ratios, tangent plane equations, and symmetric Cartesian normal lines in three dimensions. |
| Category D | Multivariable Chain Rule & Composite Functions | Q59 – Q81 | Chain rule for single-parameter and two-parameter dependencies, polar coordinates ($r, \theta$), total rates $dz/dt$, and cyclic coordinate identities. |
| Category E | Implicit Function Differentiation & Total Derivatives | Q82 – Q93 | Implicit derivative formula $dy/dx = -F_x / F_y$, total derivative $dz/dx = z_x + z_y(dy/dx)$, and total differential equations $dz = z_x dx + z_y dy$. |
| Category F | Approximations, Differentials & Engineering Error Analysis | Q94 – Q104 | Absolute error approximation $du$, relative error $\delta u / u$, and percentage error propagation in physical laws (pendulum, kinetic energy, area, volume). |
| Category G | Jacobians & Coordinate Transformations | Q105 – Q121 | Jacobian determinants $\frac{\partial(u,v)}{\partial(x,y)}$, reciprocal property ($J \cdot J' = 1$), chain rule, functional dependence, polar, cylindrical, and spherical systems. |
| Category H | Taylor's & Maclaurin's Series Expansions | Q122 – Q141 | Single-variable and multivariable Taylor/Maclaurin series, evaluation of origin constant terms, linear/quadratic differential terms, and power series coefficients. |
| Category I | Differential-Equation Identities | Q142 – Q144 | Verification of partial differential equations, harmonic functions, wave and heat equations, and multivariable functional identities. |
| Category J | Maxima, Minima, Saddle Points & Critical Points | Q145 – Q160 | Critical and stationary points, second-derivative discriminant test ($D = rt - s^2$), local maxima, local minima, and saddle point classifications. |
| Category K | Constrained Extrema & Lagrange Multipliers | Q161 – Q163 | Constrained optimization, Lagrange multiplier method ($\nabla f = \lambda \nabla g$), geometric extrema on curves and surfaces, and stationary point evaluations. |
Category A: First & Higher-Order Partial Derivatives & Point Evaluations
Partial derivative fundamentals, higher-order derivatives ($f_{xx}, f_{yy}, f_{xy}$), mixed partials, and evaluation at specific coordinate points.
Ques 1. If $z=x^{3}y^{2}+4xy,$ then $\frac{\partial z}{\partial x}$ is
- (a) $3x^{2}y^{2}+4y$
- (b) $3xy^{2}+4x$
- (c) $x^{2}y^{2}+4y$
- (d) $3x^{2}y+4$
Step-by-Step Solution:
$z = x^{3}y^{2}+4xy$
$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^{3}y^{2}) + \frac{\partial}{\partial x}(4xy)$
$\frac{\partial z}{\partial x} = 3x^{2}y^{2} + 4y(1)$
$\frac{\partial z}{\partial x} = 3x^{2}y^{2} + 4y$Correct Option: (a)
Ques 2. If $z=x^{2}+y^{2}+xy$, then $\partial z/\partial y$ equals
- (a) $2y+x$
- (b) $2x+y$
- (c) $2y-x$
- (d) $x+y$
Step-by-Step Solution:
$z = x^{2}+y^{2}+xy$
$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^{2}) + \frac{\partial}{\partial y}(y^{2}) + \frac{\partial}{\partial y}(xy)$
$\frac{\partial z}{\partial y} = 0 + 2y + x(1)$
$\frac{\partial z}{\partial y} = 2y + x$Correct Option: (a)
Ques 3. If $u=e^{xy}$, then $\partial u/\partial x$ is
- (a) $xy e^{xy}$
- (b) $x e^{xy}$
- (c) $y e^{xy}$
- (d) $e^{xy}$
Step-by-Step Solution:
$u = e^{xy}$
$\frac{\partial u}{\partial x} = e^{xy} \cdot \frac{\partial}{\partial x}(xy)$
$\frac{\partial u}{\partial x} = e^{xy} \cdot (y \cdot 1)$
$\frac{\partial u}{\partial x} = y e^{xy}$Correct Option: (c)
Ques 4. If $u=\ln(x^{2}+y^{2})$, then $\partial u/\partial x$ equals
- (a) $2x/(x^{2}+y^{2})$
- (b) $x/(x^{2}+y^{2})$
- (c) $2y/(x^{2}+y^{2})$
- (d) $1/(x^{2}+y^{2})$
Step-by-Step Solution:
$u = \ln(x^{2}+y^{2})$
$\frac{\partial u}{\partial x} = \frac{1}{x^{2}+y^{2}} \cdot \frac{\partial}{\partial x}(x^{2}+y^{2})$
$\frac{\partial u}{\partial x} = \frac{1}{x^{2}+y^{2}} \cdot (2x + 0)$
$\frac{\partial u}{\partial x} = \frac{2x}{x^{2}+y^{2}}$Correct Option: (a)
Ques 5. If $z=x\sin y+y\cos x$, then $\partial z/\partial x$ equals:
- (a) $\sin y-y\sin x$
- (b) $\sin y+y\cos x$
- (c) $\cos y-y\sin x$
- (d) $\sin x+y\cos x$
Step-by-Step Solution:
$z = x \sin y + y \cos x$
$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x \sin y) + \frac{\partial}{\partial x}(y \cos x)$
$\frac{\partial z}{\partial x} = \sin y \cdot (1) + y \cdot (-\sin x)$
$\frac{\partial z}{\partial x} = \sin y - y \sin x$Correct Option: (a)
Ques 6. If $u=\tan^{-1}(xy)$, then $\partial u/\partial x$ is
- (a) $y/(1+x^{2}y^{2})$
- (b) $x/(1+x^{2}y^{2})$
- (c) $xy/(1+x^{2}y^{2})$
- (d) $1/(1+x^{2}y^{2})$
Step-by-Step Solution:
$u = \tan^{-1}(xy)$
$\frac{\partial u}{\partial x} = \frac{1}{1 + (xy)^{2}} \cdot \frac{\partial}{\partial x}(xy)$
$\frac{\partial u}{\partial x} = \frac{1}{1 + x^{2}y^{2}} \cdot (y \cdot 1)$
$\frac{\partial u}{\partial x} = \frac{y}{1 + x^{2}y^{2}}$Correct Option: (a)
Ques 7. If $u=x^{2}y+3y^{2},$ then $u_{yy}$ equals
- (a) 6
- (b) 3
- (c) 2x
- (d) 0
Step-by-Step Solution:
$u = x^{2}y + 3y^{2}$
$u_{y} = \frac{\partial}{\partial y}(x^{2}y) + \frac{\partial}{\partial y}(3y^{2})$
$u_{y} = x^{2}(1) + 6y$
$u_{y} = x^{2} + 6y$
$u_{yy} = \frac{\partial}{\partial y}(x^{2} + 6y)$
$u_{yy} = 0 + 6(1)$
$u_{yy} = 6$Correct Option: (a)
Ques 8. If $z=x^{2}y+xy^{2}$, then $\partial z/\partial x$ at (1, 2) is
- (a) 8
- (b) 6
- (c) 10
- (d) 12
Step-by-Step Solution:
$z = x^{2}y + xy^{2}$
$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^{2}y) + \frac{\partial}{\partial x}(xy^{2})$
$\frac{\partial z}{\partial x} = 2xy + y^{2}(1)$
$\frac{\partial z}{\partial x} = 2xy + y^{2}$
$\left. \frac{\partial z}{\partial x} \right|_{(1, 2)} = 2(1)(2) + (2)^{2}$
$\left. \frac{\partial z}{\partial x} \right|_{(1, 2)} = 4 + 4$
$\left. \frac{\partial z}{\partial x} \right|_{(1, 2)} = 8$Correct Option: (a)
Ques 9. If $z=x^{3}+y^{3}-3xy$, then $\partial z/\partial y$ at (1, 2) equals
- (a) 9
- (b) 10
- (c) 8
- (d) 6
Step-by-Step Solution:
$z = x^{3} + y^{3} - 3xy$
$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^{3}) + \frac{\partial}{\partial y}(y^{3}) - \frac{\partial}{\partial y}(3xy)$
$\frac{\partial z}{\partial y} = 0 + 3y^{2} - 3x(1)$
$\frac{\partial z}{\partial y} = 3y^{2} - 3x$
$\left. \frac{\partial z}{\partial y} \right|_{(1, 2)} = 3(2)^{2} - 3(1)$
$\left. \frac{\partial z}{\partial y} \right|_{(1, 2)} = 12 - 3$
$\left. \frac{\partial z}{\partial y} \right|_{(1, 2)} = 9$Correct Option: (a)
Ques 10. If $u=e^{xy}$, find $\partial u/\partial y$ at (1, 0).
- (a) 1
- (b) 0
- (c) e
- (d) 2
Step-by-Step Solution:
$u = e^{xy}$
$\frac{\partial u}{\partial y} = e^{xy} \cdot \frac{\partial}{\partial y}(xy)$
$\frac{\partial u}{\partial y} = e^{xy} \cdot (x)$
$\left. \frac{\partial u}{\partial y} \right|_{(1, 0)} = e^{(1)(0)} \cdot (1)$
$\left. \frac{\partial u}{\partial y} \right|_{(1, 0)} = e^{0} \cdot 1$
$\left. \frac{\partial u}{\partial y} \right|_{(1, 0)} = 1 \cdot 1 = 1$Correct Option: (a)
Ques 11. If $z=x^{2}y^{2}$, then $z_{xy}$ at (2, 3) equals
- (a) 24
- (b) 12
- (c) 36
- (d) 18
Step-by-Step Solution:
$z = x^{2}y^{2}$
$z_{x} = \frac{\partial}{\partial x}(x^{2}y^{2}) = 2xy^{2}$
$z_{xy} = \frac{\partial}{\partial y}(2xy^{2}) = 2x(2y) = 4xy$
$\left. z_{xy} \right|_{(2, 3)} = 4(2)(3)$
$\left. z_{xy} \right|_{(2, 3)} = 24$Correct Option: (a)
Ques 12. If $u=\ln(x+y)$, then $\partial u/\partial y$ at (1, 1) equals
- (a) $1/2$
- (b) 1
- (c) $1/4$
- (d) 2
Step-by-Step Solution:
$u = \ln(x+y)$
$\frac{\partial u}{\partial y} = \frac{1}{x+y} \cdot \frac{\partial}{\partial y}(x+y)$
$\frac{\partial u}{\partial y} = \frac{1}{x+y} \cdot (0 + 1)$
$\frac{\partial u}{\partial y} = \frac{1}{x+y}$
$\left. \frac{\partial u}{\partial y} \right|_{(1, 1)} = \frac{1}{1+1} = \frac{1}{2}$Correct Option: (a)
Ques 13. If $z=\sqrt{x^{2}+y^{2}}$, then $\partial z/\partial x$ at (3, 4) equals
- (a) $3/5$
- (b) $4/5$
- (c) $1/5$
- (d) 1
Step-by-Step Solution:
$z = (x^{2}+y^{2})^{1/2}$
$\frac{\partial z}{\partial x} = \frac{1}{2}(x^{2}+y^{2})^{-1/2} \cdot \frac{\partial}{\partial x}(x^{2}+y^{2})$
$\frac{\partial z}{\partial x} = \frac{1}{2\sqrt{x^{2}+y^{2}}} \cdot (2x)$
$\frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}}}$
$\left. \frac{\partial z}{\partial x} \right|_{(3, 4)} = \frac{3}{\sqrt{3^{2}+4^{2}}}$
$\left. \frac{\partial z}{\partial x} \right|_{(3, 4)} = \frac{3}{\sqrt{9+16}} = \frac{3}{\sqrt{25}} = \frac{3}{5}$Correct Option: (a)
Ques 14. If $f=\log(x\tan^{-1}y)$, then $f_{xy}$ is equal to
- (a) $-\dfrac{1}{x^2}$
- (b) $0$
- (c) $\dfrac{1}{x^2}$
- (d) none of these
Step-by-Step Solution:
$ f=\log x+\log(\tan^{-1}y) $
$ f_x=\frac1x\implies f_{xy}=0 $
Correct Option: (b)
Ques 15. If $f(x,y,z)=0$, then the value of $\dfrac{\partial x}{\partial y}\cdot\dfrac{\partial y}{\partial z}\cdot\dfrac{\partial z}{\partial x}$ is
- (a) $1$
- (b) $-1$
- (c) $0$
- (d) none of these
Step-by-Step Solution:
By the cyclic partial derivative rule,
$ \frac{\partial x}{\partial y}\cdot\frac{\partial y}{\partial z}\cdot\frac{\partial z}{\partial x}=-1 $
Correct Option: (b)
Ques 16. If $f(x,y)=e^{xy^2}$, the total differential of the function at the point $(1,2)$ is
- (a) $e(dx+dy)$
- (b) $e^4(dx+dy)$
- (c) $e^4(4dx+dy)$
- (d) $4e^4(dx+dy)$
Step-by-Step Solution:
$ f_x=y^2e^{xy^2},\qquad f_y=2xye^{xy^2} $
At $(1,2)$, $f_x=f_y=4e^4$. Hence,
$ df=f_xdx+f_ydy=4e^4(dx+dy) $
Correct Option: (d)
Ques 17. If $f(x,y)=0$, then $\dfrac{dy}{dx}$ is equal to
- (a) $\dfrac{f_x}{f_y}$
- (b) $-\dfrac{f_x}{f_y}$
- (c) $\dfrac{f_y}{f_x}$
- (d) $-\dfrac{f_y}{f_x}$
Step-by-Step Solution:
$ f_x+f_y\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{f_x}{f_y} $
Correct Option: (b)
Ques 18. If $z=f(x,y)$, then $dz$ is equal to
- (a) $f_x\,dx+f_y\,dy$
- (b) $f_y\,dx+f_x\,dy$
- (c) $f_x\,dx-f_y\,dy$
- (d) $f_y\,dx-f_x\,dy$
Step-by-Step Solution:
$ dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy=f_x,dx+f_y,dy $
Correct Option: (a)
Ques 19. If $f(x,y)$ is such that $f_x=e^x\cos y$ and $f_y=e^x\sin y$, then which of the following is true?
- (a) $f(x,y)=e^{x+y}\sin(x+y)$
- (b) $f(x,y)=e^x\sin(x+y)$
- (c) $f(x,y)$ does not exist
- (d) none of above
Step-by-Step Solution:
$ f_{xy}=-e^x\sin y,\qquad f_{yx}=e^x\sin y $
Since $f_{xy}\ne f_{yx}$ in general, such an $f$ does not exist.
Correct Option: (c)
Ques 20. If $u=x^y$, the values of $\dfrac{\partial u}{\partial x},\dfrac{\partial u}{\partial y}$ are:
- (a) $xy^{x-1},y^x\log y$
- (b) $x^y\log x,yx^{y-1}$
- (c) $yx^{y-1},y^x\log y$
- (d) $yx^{y-1},x^y\log x$
Step-by-Step Solution:
$ u_x=yx^{y-1} $
Also, $u=e^{y\log x}$, so
$ u_y=x^y\log x $
Correct Option: (d)
Ques 21. If $x+y+z=\log z$, then $z_x$ is:
- (a) $\dfrac{z}{1-z}$
- (b) $\dfrac{z}{1+z}$
- (c) $\dfrac{1-z}{z}$
- (d) $\dfrac{1+z}{z}$
Step-by-Step Solution:
Differentiating with respect to $x$:
$ 1+z_x=\frac{z_x}{z} \implies z_x=\frac{z}{1-z} $
Correct Option: (a)
Ques 22. The plane $x=1$ intersects the paraboloid $z=x^2+y^2$, is a parabola. The slope of the tangent to the parabola at $(1,2,5)$ is:
- (a) $4$
- (b) $2$
- (c) $8$
- (d) $16$
Step-by-Step Solution:
For $x=1$,
$ z=1+y^2\implies \frac{dz}{dy}=2y $
At $y=2$, the slope is $4$.
Correct Option: (a)
Ques 23. If $u=\sin^{-1}(x-y)$, where $x=3t$, $y=4t^3$, then $\dfrac{du}{dt}$ is:
- (a) $\sqrt{1-t^2}$
- (b) $\dfrac{3}{\sqrt[3]{1-t^2}}$
- (c) $\dfrac{3}{\sqrt{1-t^2}}$
- (d) $3\sqrt{1-t^2}$
Step-by-Step Solution:
$ x-y=3t-4t^3=\sin(3\sin^{-1}t) $
Thus $u=3\sin^{-1}t$, so
$ \frac{du}{dt}=\frac{3}{\sqrt{1-t^2}} $
Correct Option: (c)
Category B: Homogeneous Functions & Euler's Theorem
Degree calculation, first-order Euler relation ($x u_x + y u_y = nu$), second-order Euler corollary ($n(n-1)u$), and composite functions ($g(u) = n \frac{f}{f'}$).
Ques 24. If $u=x^{3}+3x^{2}y+3xy^{2}+y^{3}$, then $x u_{x}+y u_{y}$ equals
- (a) 2u
- (b) 3u
- (c) 4u
- (d) u
Step-by-Step Solution:
$u = x^{3}+3x^{2}y+3xy^{2}+y^{3}$
$u(tx, ty) = (tx)^{3}+3(tx)^{2}(ty)+3(tx)(ty)^{2}+(ty)^{3}$
$u(tx, ty) = t^{3}x^{3}+3t^{3}x^{2}y+3t^{3}xy^{2}+t^{3}y^{3}$
$u(tx, ty) = t^{3}(x^{3}+3x^{2}y+3xy^{2}+y^{3})$
$u(tx, ty) = t^{3}u(x, y)$
$n = 3$
$x u_{x} + y u_{y} = n u$
$x u_{x} + y u_{y} = 3u$Correct Option: (b)
Ques 25. If $u=\frac{x^{2}+y^{2}}{\sqrt{(x+y)}},$ the degree of the homogeneous function is
- (a) $1/2$
- (b) $3/2$
- (c) $5/2$
- (d) 3
Step-by-Step Solution:
$u(x, y) = \frac{x^{2}+y^{2}}{(x+y)^{1/2}}$
$u(tx, ty) = \frac{(tx)^{2}+(ty)^{2}}{((tx)+(ty))^{1/2}}$
$u(tx, ty) = \frac{t^{2}(x^{2}+y^{2})}{t^{1/2}(x+y)^{1/2}}$
$u(tx, ty) = t^{2 - 1/2} \frac{x^{2}+y^{2}}{(x+y)^{1/2}}$
$u(tx, ty) = t^{3/2} u(x, y)$
$n = 3/2$Correct Option: (b)
Ques 26. If $u$ is homogeneous of degree 6 and $u=64$ at a point, then $x u_{x}+y u_{y}=$
- (a) 64
- (b) 128
- (c) 256
- (d) 384
Step-by-Step Solution:
$n = 6$
$u = 64$
$x u_{x} + y u_{y} = n u$
$x u_{x} + y u_{y} = 6(64)$
$x u_{x} + y u_{y} = 384$Correct Option: (d)
Ques 27. If $u=x^{3}y^{2}/(x+y),$ then the degree of u is
- (a) 2
- (b) 3
- (c) 4
- (d) 5
Step-by-Step Solution:
$u(tx, ty) = \frac{(tx)^{3}(ty)^{2}}{(tx)+(ty)}$
$u(tx, ty) = \frac{t^{3}x^{3}t^{2}y^{2}}{t(x+y)}$
$u(tx, ty) = \frac{t^{5}(x^{3}y^{2})}{t(x+y)}$
$u(tx, ty) = t^{5-1} \frac{x^{3}y^{2}}{x+y}$
$u(tx, ty) = t^{4} u(x, y)$
$n = 4$Correct Option: (c)
Ques 28. If $u=\sqrt{x^{2}+xy+y^{2}}$, then $u$ is homogeneous of degree:
- (a) 0
- (b) $1/2$
- (c) 1
- (d) 2
Step-by-Step Solution:
$u(tx, ty) = \sqrt{(tx)^{2}+(tx)(ty)+(ty)^{2}}$
$u(tx, ty) = \sqrt{t^{2}x^{2}+t^{2}xy+t^{2}y^{2}}$
$u(tx, ty) = \sqrt{t^{2}(x^{2}+xy+y^{2})}$
$u(tx, ty) = (t^{2})^{1/2} \sqrt{x^{2}+xy+y^{2}}$
$u(tx, ty) = t^{1} u(x, y)$
$n = 1$Correct Option: (c)
Ques 29. If $u=x^{3}\sin(y/x)$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$ equals:
- (a) u
- (b) 2u
- (c) 3u
- (d) 4u
Step-by-Step Solution:
$u(tx, ty) = (tx)^{3} \sin \left(\frac{ty}{tx}\right)$
$u(tx, ty) = t^{3}x^{3} \sin \left(\frac{y}{x}\right)$
$u(tx, ty) = t^{3} u(x, y)$
$n = 3$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3u$Correct Option: (c)
Ques 30. If $u$ is homogeneous of degree 6, then $x^{2}u_{xx}+2xyu_{xy}+y^{2}u_{yy}$ is:
- (a) 24u
- (b) 30u
- (c) 36u
- (d) 42u
Step-by-Step Solution:
$n = 6$
$x^{2}u_{xx} + 2xyu_{xy} + y^{2}u_{yy} = n(n-1)u$
$x^{2}u_{xx} + 2xyu_{xy} + y^{2}u_{yy} = 6(6-1)u$
$x^{2}u_{xx} + 2xyu_{xy} + y^{2}u_{yy} = 6(5)u$
$x^{2}u_{xx} + 2xyu_{xy} + y^{2}u_{yy} = 30u$Correct Option: (b)
Ques 31. If $u=(x^{4}+2x^{2}y^{2}+y^{4})/(x^{2}+y^{2})$, then the value of $x\cdot\partial u/\partial x+y\cdot\partial u/\partial y$ is:
- (a) u
- (b) 2u
- (c) 3u
- (d) 4u
Step-by-Step Solution:
$u = \frac{(x^{2}+y^{2})^{2}}{x^{2}+y^{2}}$
$u = (x^{2}+y^{2})^{1}$
$u = x^{2}+y^{2}$
$u(tx, ty) = (tx)^{2}+(ty)^{2} = t^{2}(x^{2}+y^{2}) = t^{2}u$
$n = 2$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 2u$Correct Option: (b)
Ques 32. If $u=x^{5}\cos(y/x)$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$ is equal to:
- (a) 3u
- (b) 4u
- (c) 5u
- (d) 6u
Step-by-Step Solution:
$u(tx, ty) = (tx)^{5} \cos \left(\frac{ty}{tx}\right)$
$u(tx, ty) = t^{5}x^{5} \cos \left(\frac{y}{x}\right)$
$u(tx, ty) = t^{5} u(x, y)$
$n = 5$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 5u$Correct Option: (c)
Ques 33. If u is a homogeneous function of degree 4 and $u(2,1)=8$, then the value of $2\cdot\partial u/\partial x+\partial u/\partial y$ at (2, 1) is:
- (a) 8
- (b) 16
- (c) 24
- (d) 32
Step-by-Step Solution:
$n = 4$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 4u$
$x=2, y=1 \implies 2 \frac{\partial u}{\partial x} + 1 \frac{\partial u}{\partial y} = 4 \cdot u(2,1)$
$2 \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 4(8)$
$2 \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 32$Correct Option: (d)
Ques 34. If $u=(x+y)^{4}/(x^{2}+y^{2})$, then the degree of homogeneity of u is
- (a) 0
- (b) 1
- (c) 2
- (d) 4
Step-by-Step Solution:
$u(tx, ty) = \frac{(tx+ty)^{4}}{(tx)^{2}+(ty)^{2}}$
$u(tx, ty) = \frac{t^{4}(x+y)^{4}}{t^{2}(x^{2}+y^{2})}$
$u(tx, ty) = t^{4-2} \frac{(x+y)^{4}}{x^{2}+y^{2}}$
$u(tx, ty) = t^{2} u(x, y)$
$n = 2$Correct Option: (c)
Ques 35. If $u=(x^{2}+y^{2})^{\frac{3}{2}},$ then $x^{2}u_{xx}+2xyu_{xy}+y^{2}u_{yy}$ is equal to
- (a) 3u
- (b) 6u
- (c) $(15/4)u$
- (d) $(3/4)u$
Step-by-Step Solution:
$u(tx, ty) = ((tx)^{2}+(ty)^{2})^{3/2}$
$u(tx, ty) = (t^{2}(x^{2}+y^{2}))^{3/2}$
$u(tx, ty) = (t^{2})^{3/2} (x^{2}+y^{2})^{3/2}$
$u(tx, ty) = t^{3} u(x, y)$
$n = 3$
$x^{2}u_{xx} + 2xyu_{xy} + y^{2}u_{yy} = n(n-1)u$
$x^{2}u_{xx} + 2xyu_{xy} + y^{2}u_{yy} = 3(3-1)u$
$x^{2}u_{xx} + 2xyu_{xy} + y^{2}u_{yy} = 3(2)u = 6u$Correct Option: (b)
Ques 36. If $u=x^{4}f(y/x)$ where f is a differentiable function, then $x\cdot\partial u/\partial x+y\cdot\partial u/\partial y$ is
- (a) 2u
- (b) 3u
- (c) 4u
- (d) 5u
Step-by-Step Solution:
$u(tx, ty) = (tx)^{4} f\left(\frac{ty}{tx}\right)$
$u(tx, ty) = t^{4}x^{4} f\left(\frac{y}{x}\right)$
$u(tx, ty) = t^{4} u(x, y)$
$n = 4$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 4u$Correct Option: (c)
Ques 37. If $u=\log(x^{3}+y^{3}+z^{3}-3xyz)$, then $x u_{x}+y u_{y}+z u_{z}$ is equal to:
- (a) 3u
- (b) 2u
- (c) 0
- (d) 3
Step-by-Step Solution:
$f(u) = e^{u} = x^{3}+y^{3}+z^{3}-3xyz$
$f(tx, ty, tz) = (tx)^{3}+(ty)^{3}+(tz)^{3}-3(tx)(ty)(tz)$
$f(tx, ty, tz) = t^{3}(x^{3}+y^{3}+z^{3}-3xyz)$
$n = 3$
$x u_{x} + y u_{y} + z u_{z} = n \frac{f(u)}{f'(u)}$
$x u_{x} + y u_{y} + z u_{z} = 3 \frac{e^{u}}{e^{u}}$
$x u_{x} + y u_{y} + z u_{z} = 3(1) = 3$Correct Option: (d)
Ques 38. If $u=\sin^{-1}[(x+y)/(\sqrt{x}+\sqrt{y})]$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$ is:
- (a) $\frac{1}{2}\tan u$
- (b) $-\frac{1}{2}\tan u$
- (c) $\frac{1}{4}\tan u$
- (d) $-\frac{1}{4}\tan u$
Step-by-Step Solution:
$f(u) = \sin(u) = \frac{x+y}{x^{1/2}+y^{1/2}}$
$f(tx, ty) = \frac{tx+ty}{(tx)^{1/2}+(ty)^{1/2}}$
$f(tx, ty) = \frac{t(x+y)}{t^{1/2}(x^{1/2}+y^{1/2})}$
$f(tx, ty) = t^{1 - 1/2} f(u)$
$f(tx, ty) = t^{1/2} f(u)$
$n = 1/2$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n \frac{f(u)}{f'(u)}$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{1}{2} \frac{\sin(u)}{\cos(u)}$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{1}{2} \tan(u)$Correct Option: (a)
Ques 39. If $z=\cos\left(\dfrac{x}{y}\right)+\sin\left(\dfrac{x}{y}\right)$, then $xz_x+yz_y$ is equal to
- (a) $z$
- (b) $2z$
- (c) $0$
- (d) $4z$
Step-by-Step Solution:
$z$ is homogeneous of degree $0$. By Euler's theorem,
$ xz_x+yz_y=0z=0 $
Correct Option: (c)
Ques 40. If $u=\log(x^3+y^3+z^3-3xyz)$, then $xu_x+yu_y+zu_z$ is equal to
- (a) $3u$
- (b) $2u$
- (c) $0$
- (d) $3$
Step-by-Step Solution:
Let $v=x^3+y^3+z^3-3xyz$. Since $v$ is homogeneous of degree $3$,
$ xv_x+yv_y+zv_z=3v $
Since $u=\log v$,
$ xu_x+yu_y+zu_z=\frac{3v}{v}=3 $
Correct Option: (d)
Ques 41. If $u=x^2+y^2+z^2$ be such that $xu_x+yu_y+zu_z=\lambda u$, then $\lambda$ is equal to
- (a) $1$
- (b) $2$
- (c) $0$
- (d) none of above
Step-by-Step Solution:
$u$ is homogeneous of degree $2$. Therefore,
$ xu_x+yu_y+zu_z=2u $
Hence $\lambda=2$.
Correct Option: (b)
Ques 42. If $u(x,y)=x^2\tan^{-1}\left(\dfrac{y}{x}\right)-y^2\tan^{-1}\left(\dfrac{x}{y}\right)$, $x>0,y>0$, then
$ x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2} $
is equal to
- (a) $0$
- (b) $u$
- (c) $2u$
- (d) $3u$
Step-by-Step Solution:
$u$ is homogeneous of degree $2$. By Euler's second theorem,
$ x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=n(n-1)u=2(1)u=2u $
Correct Option: (c)
Ques 43. If $u=f(r,s,t)$, $r=\dfrac{x}{y}$, $s=\dfrac{y}{z}$, $t=\dfrac{z}{x}$, then $xu_x+yu_y+zu_z$ is:
- (a) $0$
- (b) $1$
- (c) $2$
- (d) $3$
Step-by-Step Solution:
The ratios $r,s,t$ are unchanged when $(x,y,z)\mapsto(kx,ky,kz)$. Hence $u$ is homogeneous of degree $0$:
$ xu_x+yu_y+zu_z=0 $
Correct Option: (a)
Ques 44. If $u=x^2f\left(\dfrac{y}{x}\right)$, then:
- (a) $x\dfrac{\partial u}{\partial x}-y\dfrac{\partial u}{\partial y}=0$
- (b) $x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=0$
- (c) $x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=1$
- (d) $x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=2u$
Step-by-Step Solution:
$u$ is homogeneous of degree $2$. By Euler's theorem,
$ xu_x+yu_y=2u $
Correct Option: (d)
Ques 45. If $u=\log(x^2+y^2+z^2)$, then the value of $xu_x+yu_y+zu_z$ is equal to
- (a) $0$
- (b) $2u$
- (c) $2$
- (d) $2e^u$
Step-by-Step Solution:
$ u_x=\frac{2x}{x^2+y^2+z^2} $
Similarly for $u_y,u_z$. Therefore,
$ xu_x+yu_y+zu_z=\frac{2(x^2+y^2+z^2)}{x^2+y^2+z^2}=2 $
Correct Option: (c)
Ques 46. If $u=\log x+\log y$, then the value of $xu_x+yu_y$ is
- (a) $u$
- (b) $2u$
- (c) $2$
- (d) $0$
Step-by-Step Solution:
$ u_x=\frac1x,\qquad u_y=\frac1y $
$ xu_x+yu_y=1+1=2 $
Correct Option: (c)
Ques 47. If $u$ is a homogeneous function of order $n$, then $\dfrac{\partial u}{\partial x}$ and $\dfrac{\partial u}{\partial y}$, both are homogeneous functions of the order:
- (a) $n$
- (b) $n-1$
- (c) $n+1$
- (d) $n\pm1$
Step-by-Step Solution:
Differentiation lowers the degree of a homogeneous function by $1$. Hence $u_x$ and $u_y$ have degree $n-1$.
Correct Option: (b)
Ques 48. If $u=\sin^{-1}\left(\dfrac{x+y}{\sqrt{x}+\sqrt{y}}\right)$, then $x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}$ is
- (a) $\dfrac12\tan u$
- (b) $-\dfrac12\tan u$
- (c) $\dfrac14\tan u$
- (d) $-\dfrac14\tan u$
Step-by-Step Solution:
Let
$ v=\sin u=\frac{x+y}{\sqrt{x}+\sqrt{y}} $
$v$ is homogeneous of degree $\tfrac12$. Therefore,
$ xv_x+yv_y=\frac12v $
Since $v_x=\cos u\,u_x$ and $v_y=\cos u\,u_y$,
$ \cos u(xu_x+yu_y)=\frac12\sin u $
$ xu_x+yu_y=\frac12\tan u $
Correct Option: (a)
Category C: Tangent Planes & Normal Lines to Surfaces
Surface gradient vectors $\nabla F$, direction ratios, tangent plane equations, and symmetric Cartesian normal lines in three dimensions.
Ques 49. The normal line to $z=x^{2}+y^{2}$ at (1, 2, 5) is
- (a) $(x-1)/2=(y-2)/4=(z-5)/(-1)$
- (b) $(x-1)/1=(y-2)/2=(z-5)/1$
- (c) $(x-1)/4=(y-2)/2=(z-5)/1$
- (d) $(x-1)/2=(y-2)/2=(z-5)/2$
Step-by-Step Solution:
$F(x,y,z) = x^{2} + y^{2} - z = 0$
$\frac{\partial F}{\partial x} = 2x$
$\frac{\partial F}{\partial y} = 2y$
$\frac{\partial F}{\partial z} = -1$
$\left. \frac{\partial F}{\partial x} \right|_{(1, 2, 5)} = 2(1) = 2$
$\left. \frac{\partial F}{\partial y} \right|_{(1, 2, 5)} = 2(2) = 4$
$\left. \frac{\partial F}{\partial z} \right|_{(1, 2, 5)} = -1$
$\frac{x-1}{2} = \frac{y-2}{4} = \frac{z-5}{-1}$Correct Option: (a)
Ques 50. The equation of the normal line to the surface $xyz=a^{3}$ at the point $(x_1, y_1, z_1)$ is:
- (a) $(x-x_{1})/(y_{1}z_{1})=(y-y_{1})/(z_{1}x_{1})=(z-z_{1})/(x_{1}y_{1})$
- (b) $-(x-x_{1})/(y_{1}z_{1})=(y-y_{1})/(z_{1}x_{1})=-(z-z_{1})/(x_{1}y_{1}).$
- (c) $(x-x_{1})/(y_{1}z_{1})=-(y-y_{1})/(z_{1}x_{1})=-(z-z_{1})/(x_{1}y_{1})$
- (d) $-(x-x_{1})/(y_{1}z_{1})=-(y-y_{1})/(z_{1}x_{1})=(z-z_{1})/(x_{1}y_{1})$
Step-by-Step Solution:
$F(x,y,z) = xyz - a^{3} = 0$
$\frac{\partial F}{\partial x} = yz$
$\frac{\partial F}{\partial y} = xz$
$\frac{\partial F}{\partial z} = xy$
$\left. \frac{\partial F}{\partial x} \right|_{(x_{1}, y_{1}, z_{1})} = y_{1}z_{1}$
$\left. \frac{\partial F}{\partial y} \right|_{(x_{1}, y_{1}, z_{1})} = x_{1}z_{1}$
$\left. \frac{\partial F}{\partial z} \right|_{(x_{1}, y_{1}, z_{1})} = x_{1}y_{1}$
$\frac{x-x_{1}}{y_{1}z_{1}} = \frac{y-y_{1}}{x_{1}z_{1}} = \frac{z-z_{1}}{x_{1}y_{1}}$Correct Option: (a)
Ques 51. The equation of the tangent plane to the surface $x^{2}+y^{2}+z^{2}=14$ at the point (1, 2, 3) is
- (a) $x+2y+3z=14$
- (b) $2x+4y+6z=14$
- (c) $x+2y+3z=7$
- (d) $2x+2y+2z=14$
Step-by-Step Solution:
$F(x,y,z) = x^{2}+y^{2}+z^{2} - 14 = 0$
$\frac{\partial F}{\partial x} = 2x \implies 2(1) = 2$
$\frac{\partial F}{\partial y} = 2y \implies 2(2) = 4$
$\frac{\partial F}{\partial z} = 2z \implies 2(3) = 6$
$2(x-1) + 4(y-2) + 6(z-3) = 0$
$2x - 2 + 4y - 8 + 6z - 18 = 0$
$2x + 4y + 6z - 28 = 0$
$x + 2y + 3z - 14 = 0$
$x + 2y + 3z = 14$Correct Option: (a)
Ques 52. The equation of the normal line to the surface $x^{2}+y^{2}+z^{2}=9$ at the point (2, 1, 2) is
- (a) $(x-2)/2=(y-1)/1=(z-2)/2$
- (b) $(x-2)/1=(y-1)/2=(z-2)/2$
- (c) $(x-2)/4=(y-1)/2=(z-2)/4$
- (d) $(x-2)/2=(y-1)/2=(z-2)/1$
Step-by-Step Solution:
$F(x,y,z) = x^{2}+y^{2}+z^{2}-9 = 0$
$\frac{\partial F}{\partial x} = 2x \implies 2(2) = 4$
$\frac{\partial F}{\partial y} = 2y \implies 2(1) = 2$
$\frac{\partial F}{\partial z} = 2z \implies 2(2) = 4$
$\frac{x-2}{4} = \frac{y-1}{2} = \frac{z-2}{4}$
$\frac{x-2}{2} = \frac{y-1}{1} = \frac{z-2}{2}$Correct Option: (a)
Ques 53. The normal line to the surface $x^{2}+2y^{2}+3z^{2}=18$ at (2, 1, 2) is
- (a) $(x-2)/4=(y-1)/4=(z-2)/12$
- (b) $(x-2)/2=(y-1)/2=(z-2)/6$
- (c) $(x-2)/4=(y-1)/2=(z-2)/6$
- (d) $(x-2)/1=(y-1)/2=(z-2)/3$
Step-by-Step Solution:
$F(x,y,z) = x^{2}+2y^{2}+3z^{2}-18 = 0$
$\frac{\partial F}{\partial x} = 2x \implies 2(2) = 4$
$\frac{\partial F}{\partial y} = 4y \implies 4(1) = 4$
$\frac{\partial F}{\partial z} = 6z \implies 6(2) = 12$
$\frac{x-2}{4} = \frac{y-1}{4} = \frac{z-2}{12}$Correct Option: (a)
Ques 54. The tangent plane to the surface $e^{x}+y+z=5$ at the point (0, 2, 2) is
- (a) $x+y+z=4$
- (b) $x+y-z=0$
- (c) $x-y+z=0$
- (d) $2x+y+z=5$
Step-by-Step Solution:
$F(x,y,z) = e^{x}+y+z-5 = 0$
$\frac{\partial F}{\partial x} = e^{x} \implies e^{0} = 1$
$\frac{\partial F}{\partial y} = 1 \implies 1$
$\frac{\partial F}{\partial z} = 1 \implies 1$
$1(x-0) + 1(y-2) + 1(z-2) = 0$
$x + y - 2 + z - 2 = 0$
$x + y + z = 4$Correct Option: (a)
Ques 55. The tangent plane to the surface $\ln x+\ln y+\ln z=0$ at (1, 1, 1) is:
- (a) $x+y-z=1$
- (b) $x+y+z=3$
- (c) $2x+y+z=3$
- (d) $x+2y+z=3$
Step-by-Step Solution:
$F(x,y,z) = \ln(x)+\ln(y)+\ln(z) = 0$
$\frac{\partial F}{\partial x} = \frac{1}{x} \implies \frac{1}{1} = 1$
$\frac{\partial F}{\partial y} = \frac{1}{y} \implies \frac{1}{1} = 1$
$\frac{\partial F}{\partial z} = \frac{1}{z} \implies \frac{1}{1} = 1$
$1(x-1) + 1(y-1) + 1(z-1) = 0$
$x - 1 + y - 1 + z - 1 = 0$
$x + y + z = 3$Correct Option: (b)
Ques 56. The normal line to the surface $xy=z$ at (1, 2, 2) is
- (a) $(x-1)/2=(y-2)/1=(z-2)/(-1)$
- (b) $(x-1)/1=(y-2)/2=(z-2)/1$
- (c) $(x-1)/2=(y-2)/2=(z-2)/1$
- (d) $(x-1)/1=(y-2)/1=(z-2)/2$
Step-by-Step Solution:
$F(x,y,z) = xy - z = 0$
$\frac{\partial F}{\partial x} = y \implies 2$
$\frac{\partial F}{\partial y} = x \implies 1$
$\frac{\partial F}{\partial z} = -1 \implies -1$
$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-2}{-1}$Correct Option: (a)
Ques 57. The tangent plane to the surface $xz+y=5$ at (1, 2, 3) is
- (a) $3x+y+z=8$
- (b) $x+3y+z=8$
- (c) $3x-y+z=8$
- (d) $x+y+3z=8$
Step-by-Step Solution:
$F(x,y,z) = xz + y - 5 = 0$
$\frac{\partial F}{\partial x} = z \implies 3$
$\frac{\partial F}{\partial y} = 1 \implies 1$
$\frac{\partial F}{\partial z} = x \implies 1$
$3(x-1) + 1(y-2) + 1(z-3) = 0$
$3x - 3 + y - 2 + z - 3 = 0$
$3x + y + z = 8$Correct Option: (a)
Ques 58. The equation of normal line to the surface $xyz=a^3$ at the point $(x_1,y_1,z_1)$ is
- (a) $\dfrac{x-x_1}{y_1z_1}=\dfrac{y-y_1}{z_1x_1}=\dfrac{z-z_1}{x_1y_1}$
- (b) $-\dfrac{x-x_1}{y_1z_1}=\dfrac{y-y_1}{z_1x_1}=-\dfrac{z-z_1}{x_1y_1}$
- (c) $\dfrac{x-x_1}{y_1z_1}=-\dfrac{y-y_1}{z_1x_1}=-\dfrac{z-z_1}{x_1y_1}$
- (d) $-\dfrac{x-x_1}{y_1z_1}=-\dfrac{y-y_1}{z_1x_1}=\dfrac{z-z_1}{x_1y_1}$
Step-by-Step Solution:
For $F(x,y,z)=xyz-a^3$,
$ \nabla F=(yz,zx,xy) $
At $(x_1,y_1,z_1)$, the normal direction is $(y_1z_1,z_1x_1,x_1y_1)$. Hence the normal line is
$ \frac{x-x_1}{y_1z_1}=\frac{y-y_1}{z_1x_1}=\frac{z-z_1}{x_1y_1} $
Correct Option: (a)
Category D: Multivariable Chain Rule & Composite Functions
Chain rule for single-parameter and two-parameter dependencies, polar coordinates ($r, \theta$), total rates $dz/dt$, and cyclic coordinate identities.
Ques 59. If $z=f(x,y)$ where $x=r\cos\theta, y=r\sin\theta$, then $\partial z/\partial r$ is:
- (a) $f_{x}\cos\theta + f_{y}\sin\theta$
- (b) $f_{x}\sin\theta + f_{y}\cos\theta$
- (c) $f_{x}\cos\theta - f_{y}\sin\theta$
- (d) $f_{x} + f_{y}$
Step-by-Step Solution:
$\frac{\partial z}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r}$
$\frac{\partial x}{\partial r} = \frac{\partial}{\partial r}(r \cos\theta) = \cos\theta$
$\frac{\partial y}{\partial r} = \frac{\partial}{\partial r}(r \sin\theta) = \sin\theta$
$\frac{\partial z}{\partial r} = f_{x} \cos\theta + f_{y} \sin\theta$Correct Option: (a)
Ques 60. If $z=x^{2}+y^{2}$ where $x=t^{2}$, $y=e^{t}$, then $dz/dt$ equals:
- (a) $4t^{3}+2e^{2t}$
- (b) $2t^{2}+2e^{t}$
- (c) $\underline{4t+2e^{t}}$
- (d) $2t^{3}+e^{2t}$
Step-by-Step Solution:
$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$
$\frac{\partial z}{\partial x} = 2x, \quad \frac{\partial z}{\partial y} = 2y$
$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = e^{t}$
$\frac{dz}{dt} = (2x)(2t) + (2y)(e^{t})$
$\frac{dz}{dt} = 2(t^{2})(2t) + 2(e^{t})(e^{t})$
$\frac{dz}{dt} = 4t^{3} + 2e^{2t}$Correct Option: (a)
Ques 61. Let $u=f(x,y)$ where $x=rs, y=r/s$. Then $\partial u/\partial r$ is:
- (a) $s\cdot f_{x} + (1/s)\cdot f_{y}$
- (b) $r\cdot f_{x} + s\cdot f_{y}$
- (c) $s\cdot f_{x} - (1/s)\cdot f_{y}$
- (d) $f_{x} + f_{y}$
Step-by-Step Solution:
$\frac{\partial u}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r}$
$\frac{\partial x}{\partial r} = \frac{\partial}{\partial r}(rs) = s$
$\frac{\partial y}{\partial r} = \frac{\partial}{\partial r}(r/s) = 1/s$
$\frac{\partial u}{\partial r} = f_{x}(s) + f_{y}(1/s)$
$\frac{\partial u}{\partial r} = s \cdot f_{x} + (1/s) \cdot f_{y}$Correct Option: (a)
Ques 62. If $z=f(x,y)$, where $x=x(t)$ and $y=y(t)$, then $dz/dt$ is given by:
- (a) $\partial z/\partial x+\partial z/\partial y$
- (b) $(\partial z/\partial x)(dx/dt)+(\partial z/\partial y)(dy/dt)$
- (c) $(\partial z/\partial x)(dy/dt)+(\partial z/\partial y)(dx/dt)$
- (d) $(\partial z/\partial x)\cdot(\partial z/\partial y)$
Step-by-Step Solution:
$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$
Correct Option: (b)
Ques 63. If $z=x+y$ where $x=t^{2}$, $y=t^{3}$, then $dz/dt$ at $t=1$ is:
- (a) 5
- (b) 6
- (c) 4
- (d) 3
Step-by-Step Solution:
$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$
$\frac{\partial z}{\partial x} = 1, \quad \frac{\partial z}{\partial y} = 1$
$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^{2}$
$\frac{dz}{dt} = 1(2t) + 1(3t^{2}) = 2t + 3t^{2}$
$\left. \frac{dz}{dt} \right|_{t=1} = 2(1) + 3(1)^{2} = 5$Correct Option: (a)
Ques 64. If $u=f(x,y)$ and x, y are both functions of two variables r and s, then $\partial u/\partial r$ equals:
- (a) $(\partial u/\partial x)(\partial x/\partial r)+(\partial u/\partial y)(\partial y/\partial r)$
- (b) $(\partial u/\partial x)(\partial x/\partial s)+(\partial u/\partial y)(\partial y/\partial s)$
- (c) $\partial u/\partial x+\partial u/\partial y$
- (d) $(\partial u/\partial r)(\partial x/\partial u)$
Step-by-Step Solution:
$\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial r}$
Correct Option: (a)
Ques 65. If $z=x^{2}+y^{2}$ where $x=t$, $y=t^{2}$, then $dz/dt$ at $t=1$ is:
- (a) 2
- (b) 6
- (c) 4
- (d) 10
Step-by-Step Solution:
$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$
$\frac{\partial z}{\partial x} = 2x, \quad \frac{\partial z}{\partial y} = 2y$
$\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = 2t$
$\frac{dz}{dt} = 2x(1) + 2y(2t) = 2x + 4yt$
$\left. \frac{dz}{dt} \right|_{t=1} \implies x=1, y=1^{2}=1$
$\frac{dz}{dt} = 2(1) + 4(1)(1) = 6$Correct Option: (b)
Ques 66. For $z=f(x,y)$, $x=r\cos\theta$, $y=r\sin\theta$, the expression $\partial z/\partial\theta$ involves:
- (a) only $\partial z/\partial x$
- (b) only $\partial z/\partial y$
- (c) both $\partial z/\partial x$ and $\partial z/\partial y$
- (d) neither
Step-by-Step Solution:
$\frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}$
Correct Option: (c)
Ques 67. A function $z=f(x,y)$ is called a composite function when:
- (a) x and y are independent variables
- (b) x and y are themselves functions of other variables
- (c) z is constant
- (d) f is linear
Step-by-Step Solution:
$x = g(t, u, \dots), y = h(t, u, \dots)$
Correct Option: (b)
Ques 68. If $w=f(x,y,z)$ and x, y, z are functions of t, then $dw/dt$ has how many terms (by chain rule)?
- (a) 1
- (b) 2
- (c) 3
- (d) 4
Step-by-Step Solution:
$\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}$
$1 + 1 + 1 = 3$Correct Option: (c)
Ques 69. If $z=f(x,y)$ and $x=e^{t}$, $y=e^{t}$, the chain rule requires:
- (a) $dx/dt$ and $dy/dt$
- (b) only $dx/dt$
- (c) only $\partial z/\partial x$
- (d) second derivatives
Step-by-Step Solution:
$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$
Correct Option: (a)
Ques 70. If $z=f(x,y)$, where $x=2r+s$, $y=r-2s$, then $\partial z/\partial r$ equals
- (a) $2\partial z/\partial x+\partial z/\partial y$
- (b) $2\partial z/\partial x-\partial z/\partial y$
- (c) $\partial z/\partial x+2\partial z/\partial y$
- (d) $\partial z/\partial x-\partial z/\partial y$
Step-by-Step Solution:
$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$
$\frac{\partial x}{\partial r} = \frac{\partial}{\partial r}(2r+s) = 2$
$\frac{\partial y}{\partial r} = \frac{\partial}{\partial r}(r-2s) = 1$
$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}(2) + \frac{\partial z}{\partial y}(1) = 2\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}$Correct Option: (a)
Ques 71. The chain rule for composite functions extends the concept of:
- (a) Implicit differentiation
- (b) Ordinary derivative chain rule to multivariable functions
- (c) Integration by parts
- (d) Partial integration
Step-by-Step Solution:
$\frac{dz}{dt} = \sum_{i} \frac{\partial z}{\partial x_{i}} \frac{dx_{i}}{dt}$
Correct Option: (b)
Ques 72. If $z=x^{2}y$, $x=t^{2}$, $y=t^{3}$, then $dz/dt$ is:
- (a) $2t^{5}+3t^{5}$
- (b) $4t^{6}+3t^{6}$
- (c) $7t^{6}$
- (d) $5t^{5}$
Step-by-Step Solution:
$z = (t^{2})^{2}(t^{3}) = t^{4} \cdot t^{3} = t^{7}$
$\frac{dz}{dt} = \frac{d}{dt}(t^{7}) = 7t^{6}$Correct Option: (c)
Ques 73. If $u=x^{2}+y^{2}$, $x=r\cos\theta$, $y=r\sin\theta$, then $\partial u/\partial r$ equals:
- (a) 2r
- (b) r
- (c) $2r^{2}$
- (d) $r^{2}$
Step-by-Step Solution:
$u = (r \cos\theta)^{2} + (r \sin\theta)^{2}$
$u = r^{2}\cos^{2}\theta + r^{2}\sin^{2}\theta = r^{2}(\cos^{2}\theta + \sin^{2}\theta) = r^{2}$
$\frac{\partial u}{\partial r} = \frac{\partial}{\partial r}(r^{2}) = 2r$Correct Option: (a)
Ques 74. If $z=f(x,y)$, $x=uv,$ $y=u/v,$ then $\partial z/\partial u$ is:
- (a) $v\cdot\partial z/\partial x+(1/v)\cdot\partial z/\partial y$
- (b) $v\cdot\partial z/\partial x-(1/v)\cdot\partial z/\partial y$
- (c) $u\cdot\partial z/\partial x+(1/u)\cdot\partial z/\partial y$
- (d) $v\cdot\partial z/\partial x+v\cdot\partial z/\partial y$
Step-by-Step Solution:
$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$
$\frac{\partial x}{\partial u} = \frac{\partial}{\partial u}(uv) = v$
$\frac{\partial y}{\partial u} = \frac{\partial}{\partial u}(u/v) = 1/v$
$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}(v) + \frac{\partial z}{\partial y}(1/v)$
$\frac{\partial z}{\partial u} = v \cdot \frac{\partial z}{\partial x} + (1/v) \cdot \frac{\partial z}{\partial y}$Correct Option: (a)
Ques 75. If $w=x^{2}+y^{2}+z^{2}$, where $x=t$, $y=t^{2}$, $z=t^{3}$, then $dw/dt$ at $t=1$ is:
- (a) 14
- (b) 12
- (c) 16
- (d) 18
Step-by-Step Solution:
$w = (t)^{2} + (t^{2})^{2} + (t^{3})^{2}$
$w = t^{2} + t^{4} + t^{6}$
$\frac{dw}{dt} = 2t + 4t^{3} + 6t^{5}$
$\left. \frac{dw}{dt} \right|_{t=1} = 2(1) + 4(1)^{3} + 6(1)^{5}$
$\left. \frac{dw}{dt} \right|_{t=1} = 2 + 4 + 6 = 12$Correct Option: (b)
Ques 76. If $z=xy$, $x=e^{t}\cos t$, $y=e^{t}\sin t$, then $dz/dt$ equals:
- (a) $2e^{2t}\sin t\cos t$
- (b) $e^{2t}(\cos 2t + \sin 2t)$
- (c) $e^{2t}$
- (d) $e^{t}$
Step-by-Step Solution:
$z = (e^{t}\cos t)(e^{t}\sin t) = e^{2t}\sin t \cos t$
$z = \frac{1}{2} e^{2t} \sin(2t)$
$\frac{dz}{dt} = \frac{1}{2} \left[ \frac{d}{dt}(e^{2t}) \sin(2t) + e^{2t} \frac{d}{dt}(\sin(2t)) \right]$
$\frac{dz}{dt} = \frac{1}{2} [2e^{2t}\sin(2t) + e^{2t}(2\cos(2t))]$
$\frac{dz}{dt} = e^{2t}\sin(2t) + e^{2t}\cos(2t)$
$\frac{dz}{dt} = e^{2t}(\cos 2t + \sin 2t)$Correct Option: (b)
Ques 77. If $z=f(x,y)$ is homogeneous of degree $n$, and $x=r\cos\theta$, $y=r\sin\theta$, then by Euler's theorem, $x(\partial z/\partial x)+y(\partial z/\partial y)$ equals:
- (a) nz
- (b) $z/n$
- (c) $n^{2}z$
- (d) z
Step-by-Step Solution:
$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = nz$
Correct Option: (a)
Ques 78. If $u=f(x-y,y-x)$, then $\partial u/\partial x+\partial u/\partial y$ is:
- (a) 0
- (b) 1
- (c) 2u
- (d) u
Step-by-Step Solution:
$r = x - y, \quad s = y - x$
$u = f(r, s)$
$\frac{\partial u}{\partial x} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial f}{\partial s}\frac{\partial s}{\partial x} = f_{r}(1) + f_{s}(-1)$
$\frac{\partial u}{\partial y} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial f}{\partial s}\frac{\partial s}{\partial y} = f_{r}(-1) + f_{s}(1)$
$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = f_{r} - f_{s} - f_{r} + f_{s} = 0$Correct Option: (a)
Ques 79. If $z=f(x,y)$ with $x=a+h$, $y=b+k$ (a small change), the total differential dz used in composite differentiation is:
- (a) $(\partial z/\partial x)\,dx+(\partial z/\partial y)\,dy$
- (b) $(\partial z/\partial x)\,dx-(\partial z/\partial y)\,dy$
- (c) $(\partial z/\partial x)(\partial z/\partial y)$
- (d) $dx+dy$
Step-by-Step Solution:
$dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy$
Correct Option: (a)
Ques 80. If $z=x^{2}-y^{2}$, $x=r\cos\theta$, $y=r\sin\theta$, then $\partial z/\partial\theta$ equals:
- (a) $-2r^{2}\sin\theta\cos\theta - 2r^{2}\sin\theta\cos\theta$
- (b) $-4r^{2}\sin\theta\cos\theta$
- (c) $2r^{2}\cos(2\theta)$
- (d) $-2r^{2}\sin(2\theta)$
Step-by-Step Solution:
$z = (r \cos\theta)^{2} - (r \sin\theta)^{2} = r^{2}\cos^{2}\theta - r^{2}\sin^{2}\theta$
$z = r^{2}(\cos^{2}\theta - \sin^{2}\theta) = r^{2}\cos(2\theta)$
$\frac{\partial z}{\partial\theta} = \frac{\partial}{\partial\theta}(r^{2}\cos(2\theta))$
$\frac{\partial z}{\partial\theta} = r^{2}(-\sin(2\theta) \cdot 2) = -2r^{2}\sin(2\theta)$Correct Option: (d)
Ques 81. If $u=x^{2}y+y^{2}z+z^{2}x$, and $x=t$, $y=t^{2}$, $z=t^{3}$, the number of chain-rule terms needed to compute $du/dt$ is:
- (a) 2
- (b) 3
- (c) 4
- (d) 6
Step-by-Step Solution:
$\frac{du}{dt} = \frac{\partial u}{\partial x}\frac{dx}{dt} + \frac{\partial u}{\partial y}\frac{dy}{dt} + \frac{\partial u}{\partial z}\frac{dz}{dt}$
$1 + 1 + 1 = 3$Correct Option: (b)
Category E: Implicit Function Differentiation & Total Derivatives
Implicit derivative formula $dy/dx = -F_x / F_y$, total derivative $dz/dx = z_x + z_y(dy/dx)$, and total differential equations $dz = z_x dx + z_y dy$.
Ques 82. If $F(x,y)=0$ defines y implicitly as a function of x, then $dy/dx$ is given by:
- (a) $-(\partial F/\partial x)/(\partial F/\partial y)$
- (b) $-(\partial F/\partial y)/(\partial F/\partial x)$
- (c) $(\partial F/\partial x)/(\partial F/\partial y)$
- (d) $(\partial F/\partial x)\times(\partial F/\partial y)$
Step-by-Step Solution:
$F(x,y) = 0$
$\frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy = 0$
$\frac{\partial F}{\partial y} dy = -\frac{\partial F}{\partial x} dx$
$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}$Correct Option: (a)
Ques 83. If $x^{2}+y^{2}=25$, then $dy/dx$ at the point (3, 4) is:
- (a) $-3/4$
- (b) $3/4$
- (c) $-4/3$
- (d) $4/3$
Step-by-Step Solution:
$F = x^{2}+y^{2}-25 = 0$
$\frac{\partial F}{\partial x} = 2x$
$\frac{\partial F}{\partial y} = 2y$
$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y} = -\frac{2x}{2y} = -\frac{x}{y}$
$\left. \frac{dy}{dx} \right|_{(3, 4)} = -\frac{3}{4}$Correct Option: (a)
Ques 84. If $z=f(x,y)$ and y is itself a function of x, the total derivative $dz/dx$ is given by:
- (a) $\partial z/\partial x$
- (b) $\partial z/\partial x+(\partial z/\partial y)(dy/dx)$
- (c) $\partial z/\partial x-(\partial z/\partial y)(dy/dx)$
- (d) $(\partial z/\partial y)(dy/dx)$
Step-by-Step Solution:
$\frac{dz}{dx} = \frac{\partial z}{\partial x}\frac{dx}{dx} + \frac{\partial z}{\partial y}\frac{dy}{dx}$
$\frac{dz}{dx} = \frac{\partial z}{\partial x}(1) + \frac{\partial z}{\partial y}\frac{dy}{dx}$Correct Option: (b)
Ques 85. If $xy=c$ (a constant), then $dy/dx$ by implicit differentiation is:
- (a) $-y/x$
- (b) $y/x$
- (c) $-x/y$
- (d) $x/y$
Step-by-Step Solution:
$xy = c$
$\frac{d}{dx}(xy) = \frac{d}{dx}(c)$
$x \frac{dy}{dx} + y(1) = 0$
$x \frac{dy}{dx} = -y$
$\frac{dy}{dx} = -\frac{y}{x}$Correct Option: (a)
Ques 86. The total differential of $z=f(x,y)$ is defined as:
- (a) $(\partial z/\partial x)\,dx+(\partial z/\partial y)\,dy$
- (b) $(\partial z/\partial x\partial y+\partial z/\partial y)\,dx$
- (c) $\partial z/\partial x+\partial z/\partial y$
- (d) $dx+dy$
Step-by-Step Solution:
$dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy$
Correct Option: (a)
Ques 87. For $z=f(x,y)$ with $y=y(x)$ the total derivative $dz/dx$ (by chain rule) has how many terms?
- (a) 1
- (b) 2
- (c) 3
- (d) 4
Step-by-Step Solution:
$\frac{dz}{dx} = \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}\frac{dy}{dx}$
$1 + 1 = 2$Correct Option: (b)
Ques 88. If $F(x,y)=0$ and $\partial F/\partial y=0$ at a point, then $dy/dx$ at that point is:
- (a) 0
- (b) 1
- (c) undefined
- (d) equal to $\partial F/\partial x$
Step-by-Step Solution:
$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}$
$\frac{dy}{dx} = -\frac{\partial F/\partial x}{0} = \text{undefined}$Correct Option: (c)
Ques 89. If $x^{3}+y^{3}=$ 3axy then $dy/dx$ equals:
- (a) $(ay-x^{2})/(y^{2}-ax)$
- (b) $(x^{2}-ay)/(y^{2}-ax)$
- (c) $(ay-x^{2})/(ax-y^{2})$
- (d) $(x^{2}+ay)/(y^{2}+ax)$
Step-by-Step Solution:
$F(x,y) = x^{3}+y^{3}-3axy = 0$
$\frac{\partial F}{\partial x} = 3x^{2} - 3ay$
$\frac{\partial F}{\partial y} = 3y^{2} - 3ax$
$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y} = -\frac{3x^{2} - 3ay}{3y^{2} - 3ax}$
$\frac{dy}{dx} = -\frac{3(x^{2} - ay)}{3(y^{2} - ax)} = -\frac{x^{2} - ay}{y^{2} - ax} = \frac{ay - x^{2}}{y^{2} - ax}$Correct Option: (a)
Ques 90. If $x^{2}+xy+y^{2}=7,$ then $dy/dx$ equals:
- (a) $-(2x+y)/(x+2y)$
- (b) $(2x+y)/(x+2y)$
- (c) $-(x+2y)/(2x+y)$
- (d) $(2x-y)/(x-2y)$
Step-by-Step Solution:
$F(x,y) = x^{2}+xy+y^{2}-7 = 0$
$\frac{\partial F}{\partial x} = 2x + y$
$\frac{\partial F}{\partial y} = x + 2y$
$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y} = -\frac{2x+y}{x+2y}$Correct Option: (a)
Ques 91. If $z=x^{2}+y^{2}$ and $y=x^{2}$ (so y depends directly on x), the total derivative $dz/dx$ at $x=1$ is:
- (a) 2
- (b) 4
- (c) 6
- (d) 8
Step-by-Step Solution:
$z = x^{2} + (x^{2})^{2} = x^{2} + x^{4}$
$\frac{dz}{dx} = \frac{d}{dx}(x^{2} + x^{4})$
$\frac{dz}{dx} = 2x + 4x^{3}$
$\left. \frac{dz}{dx} \right|_{x=1} = 2(1) + 4(1)^{3} = 2 + 4 = 6$Correct Option: (c)
Ques 92. If $e^{x}+e^{y}=e^{x+y}$ then $dy/dx$ equals:
- (a) $-(e^{x}-e^{x+y})/(e^{y}-e^{x+y})$
- (b) $(e^{x}-e^{y})/e^{x+y}$
- (c) $e^{x-y}$
- (d) $-e^{y-x}$
Step-by-Step Solution:
$F(x,y) = e^{x}+e^{y}-e^{x+y} = 0$
$\frac{\partial F}{\partial x} = e^{x} - e^{x+y}$
$\frac{\partial F}{\partial y} = e^{y} - e^{x+y}$
$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y} = -\frac{e^{x}-e^{x+y}}{e^{y}-e^{x+y}}$Correct Option: (a)
Ques 93. If $z=x^{2}y^{3}$ the total differential dz equals:
- (a) $2xy^{3}dx+3x^{2}y^{2}dy$
- (b) $2xy^{3}dx-3x^{2}y^{2}dy$
- (c) $x^{2}y^{3}(dx+dy)$
- (d) $2x\,dx+3y\,dy$
Step-by-Step Solution:
$dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy$
$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^{2}y^{3}) = 2xy^{3}$
$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^{2}y^{3}) = 3x^{2}y^{2}$
$dz = 2xy^{3}dx + 3x^{2}y^{2}dy$Correct Option: (a)
Category F: Approximations, Differentials & Engineering Error Analysis
Absolute error approximation $du$, relative error $\delta u / u$, and percentage error propagation in physical laws (pendulum, kinetic energy, area, volume).
Ques 94. If $u=f(x,y)$, the total differential du representing the absolute error in u is given by:
- (a) $\partial u/\partial x+\partial u/\partial y$
- (b) $(\partial u/\partial x)\,dx+(\partial u/\partial y)\,dy$
- (c) $(\partial u/\partial x)\,dx-(\partial u/\partial y)\,dy$
- (d) $(\partial^{2}u/\partial x\partial y)\,dx\,dy$
Step-by-Step Solution:
$du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy$
Correct Option: (b)
Ques 95. If $x$ is measured with an absolute error $\delta x$, then the percentage error in $x$ is defined as:
- (a) $\delta x/x$
- (b) $(x/\delta x)\times100$
- (c) $(\delta x/x)\times100$
- (d) $\delta x\times100$
Step-by-Step Solution:
$\text{Percentage error} = (\text{Relative error}) \times 100 = \left(\frac{\delta x}{x}\right) \times 100$
Correct Option: (c)
Ques 96. If $u=x^{n},$ the relative error $\delta u/u$ in terms of the relative error in x is:
- (a) $n(\delta x/x)$
- (b) $(1/n)(\delta x/x)$
- (c) $n^{2}(\delta x/x)$
- (d) $(\delta x/x)^{n}$
Step-by-Step Solution:
$u = x^{n}$
$\ln u = n \ln x$
$d(\ln u) = d(n \ln x)$
$\frac{du}{u} = n \frac{dx}{x}$
$\frac{\delta u}{u} \approx n \frac{\delta x}{x}$Correct Option: (a)
Ques 97. If $u=xy$ and the percentage errors in x and y are 1% and 2% respectively, the percentage error in u is approximately:
- (a) 2%
- (b) 1%
- (c) 3%
- (d) 0.5%
Step-by-Step Solution:
$u = xy$
$\ln u = \ln x + \ln y$
$\frac{\delta u}{u} = \frac{\delta x}{x} + \frac{\delta y}{y}$
$\frac{\delta u}{u} \times 100 = \left(\frac{\delta x}{x} \times 100\right) + \left(\frac{\delta y}{y} \times 100\right)$
$\frac{\delta u}{u} \times 100 = 1\% + 2\% = 3\%$Correct Option: (c)
Ques 98. If $u=x/y$ the percentage error in u when x and y have percentage errors $e_{1}$ and $e_{2}$ respectively is:
- (a) $e_{1}-e_{2}$
- (b) $e_{1}+e_{2}$
- (c) $e_{1}/e_{2}$
- (d) $e_{1}\cdot e_{2}$
Step-by-Step Solution:
$u = x/y$
$\ln u = \ln x - \ln y$
$du/u = dx/x - dy/y$
$\text{Maximum error: } \left| \frac{\delta u}{u} \right| = \left| \frac{\delta x}{x} \right| + \left| \frac{\delta y}{y} \right|$
$\left| \frac{\delta u}{u} \right| \times 100 = e_{1} + e_{2}$Correct Option: (b)
Ques 99. The period of a simple pendulum is $T=2\pi\sqrt{l/g}$. If the error in length $l$ is 2%, the percentage error in $T$ is:
- (a) 2%
- (b) 1%
- (c) 4%
- (d) 0.5%
Step-by-Step Solution:
$T = 2\pi (l/g)^{1/2}$
$\ln T = \ln(2\pi) + \frac{1}{2}\ln l - \frac{1}{2}\ln g$
$\frac{\delta T}{T} = \frac{1}{2} \frac{\delta l}{l} - 0$
$\frac{\delta T}{T} \times 100 = \frac{1}{2} \left( \frac{\delta l}{l} \times 100 \right)$
$\frac{\delta T}{T} \times 100 = \frac{1}{2} (2\%) = 1\%$Correct Option: (b)
Ques 100. The area of a circle of radius r is $A=\pi r^{2}$. If the radius is measured with an error of 1.5%, the percentage error in the calculated area is:
- (a) 1.5%
- (b) 3.0%
- (c) 2.25%
- (d) 0.75%
Step-by-Step Solution:
$A = \pi r^{2}$
$\ln A = \ln \pi + 2 \ln r$
$\frac{\delta A}{A} = 2 \frac{\delta r}{r}$
$\frac{\delta A}{A} \times 100 = 2(1.5\%) = 3.0\%$Correct Option: (b)
Ques 101. In calculating the volume of a sphere $V=(4/3)\pi r^{3}$, if r increases by 1%, the percentage increase in V is:
- (a) 1%
- (b) 3%
- (c) $4/3$%
- (d) $1/3$%
Step-by-Step Solution:
$V = \frac{4}{3}\pi r^{3}$
$\ln V = \ln(4\pi/3) + 3\ln r$
$\frac{\delta V}{V} = 3 \frac{\delta r}{r}$
$\frac{\delta V}{V} \times 100 = 3(1\%) = 3\%$Correct Option: (b)
Ques 102. If the kinetic energy of a body is given by $K=(1/2)mv^{2}$ and the errors in measuring mass m and velocity v are 1% and 2% respectively, the maximum error in K is:
- (a) 3%
- (b) 5%
- (c) 4%
- (d) 2%
Step-by-Step Solution:
$K = \frac{1}{2}mv^{2}$
$\ln K = \ln(1/2) + \ln m + 2\ln v$
$\frac{\delta K}{K} = \frac{\delta m}{m} + 2\frac{\delta v}{v}$
$\left| \frac{\delta K}{K} \right| \times 100 = 1\% + 2(2\%)$
$\left| \frac{\delta K}{K} \right| \times 100 = 1\% + 4\% = 5\%$Correct Option: (b)
Ques 103. If $g=2s/t^{2}$ and the relative errors in s and t are 1% and 2% respectively, then the relative error in g is:
- (a) 3%
- (b) 5%
- (c) 1%
- (d) 4%
Step-by-Step Solution:
$g = 2st^{-2}$
$\ln g = \ln 2 + \ln s - 2\ln t$
$dg/g = ds/s - 2dt/t$
$\left| \frac{\delta g}{g} \right| = \left| \frac{\delta s}{s} \right| + 2\left| \frac{\delta t}{t} \right|$
$\left| \frac{\delta g}{g} \right| \times 100 = 1\% + 2(2\%) = 5\%$Correct Option: (b)
Ques 104. The percentage error in the area of a rectangle when an error of +1% is made in its length and breadth is:
- (a) 2%
- (b) 1%
- (c) 3%
- (d) 4%
Step-by-Step Solution:
$A = l \cdot b$
$\ln A = \ln l + \ln b$
$\frac{\delta A}{A} = \frac{\delta l}{l} + \frac{\delta b}{b}$
$\frac{\delta A}{A} \times 100 = 1\% + 1\% = 2\%$Correct Option: (a)
Category G: Jacobians & Coordinate Transformations
Jacobian determinants $\frac{\partial(u,v)}{\partial(x,y)}$, reciprocal property ($J \cdot J' = 1$), chain rule, functional dependence, polar, cylindrical, and spherical systems.
Ques 105. According to the Reciprocal Property of Jacobians, if $J=\partial(u,v)/\partial(x,y)$ and $J^{\prime}=\partial(x,y)/\partial(u,v)$ then:
- (a) $J\cdot J^{\prime}=0$
- (b) $J\cdot J^{\prime}=1$
- (c) $J+J^{\prime}=1$
- (d) $J=J^{\prime}$
Step-by-Step Solution:
$J = \frac{\partial(u,v)}{\partial(x,y)}$
$J' = \frac{\partial(x,y)}{\partial(u,v)}$
$J \cdot J' = \frac{\partial(u,v)}{\partial(x,y)} \cdot \frac{\partial(x,y)}{\partial(u,v)} = \frac{\partial(u,v)}{\partial(u,v)} = 1$Correct Option: (b)
Ques 106. If two functions $u(x,y)$ and $v(x,y)$ are functionally dependent then their Jacobian $J=\partial(u,v)/\partial(x,y)$ is:
- (a) 1
- (b) ∞
- (c) 0
- (d) -1
Step-by-Step Solution:
$f(u, v) = 0$
$\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} u_{x} & u_{y} \\ v_{x} & v_{y} \end{vmatrix} = 0$Correct Option: (c)
Ques 107. If $x=r\cos\theta$ and $y=r\sin\theta$, then the Jacobian $\partial(x,y)/\partial(r,\theta)$ is:
- (a) $r$
- (b) $1/r$
- (c) $r^{2}$
- (d) $1$
Step-by-Step Solution:
$J = \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix}$
$J = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix}$
$J = (\cos\theta)(r\cos\theta) - (-r\sin\theta)(\sin\theta)$
$J = r\cos^{2}\theta + r\sin^{2}\theta = r(\cos^{2}\theta + \sin^{2}\theta) = r$Correct Option: (a)
Ques 108. Using the Chain Rule for Jacobians, if u, v are functions of r, s and r, s are functions of x, y, then $\partial(u,v)/\partial(x,y)$ equals:
- (a) $\partial(u,v)/\partial(r,s)+\partial(r,s)/\partial(x,y)$
- (b) $\partial(u,v)/\partial(r,s)\cdot\partial(r,s)/\partial(x,y)$
- (c) $[\partial(u,v)/\partial(r,s)]/[\partial(r,s)/\partial(x,y)]$
- (d) $\partial(r,s)/\partial(u,v)\cdot\partial(x,y)/\partial(r,s)$
Step-by-Step Solution:
$\frac{\partial(u,v)}{\partial(x,y)} = \frac{\partial(u,v)}{\partial(r,s)} \cdot \frac{\partial(r,s)}{\partial(x,y)}$
Correct Option: (b)
Ques 109. If $u=x+y$ and $v=x-y,$ then the value of $\partial(u,v)/\partial(x,y)$ is:
- (a) 0
- (b) 2
- (c) -2
- (d) 1
Step-by-Step Solution:
$J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}$
$J = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix}$
$J = 1(-1) - 1(1) = -1 - 1 = -2$Correct Option: (c)
Ques 110. For spherical polar coordinates $x=r\sin\theta\cos\phi$, $y=r\sin\theta\sin\phi$, $z=r\cos\theta$, the Jacobian $\partial(x,y,z)/\partial(r,\theta,\phi)$ is:
- (a) $r\sin\theta$
- (b) $r^{2}\sin\theta$
- (c) $r\cos\theta$
- (d) $r\sin^{2}\theta$
Step-by-Step Solution:
$x = r\sin\theta\cos\phi, \quad y = r\sin\theta\sin\phi, \quad z = r\cos\theta$
$J = \frac{\partial(x,y,z)}{\partial(r,\theta,\phi)} = \begin{vmatrix} \sin\theta\cos\phi & r\cos\theta\cos\phi & -r\sin\theta\sin\phi \\ \sin\theta\sin\phi & r\cos\theta\sin\phi & r\sin\theta\cos\phi \\ \cos\theta & -r\sin\theta & 0 \end{vmatrix}$
$J = r^{2}\sin\theta$Correct Option: (b)
Ques 111. If $u=x^{2}$ and $v=y^{2}$ then $\partial(u,v)/\partial(x,y)$ is:
- (a) 4xy
- (b) 2xy
- (c) $x+y$
- (d) 0
Step-by-Step Solution:
$J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}$
$J = \begin{vmatrix} 2x & 0 \\ 0 & 2y \end{vmatrix}$
$J = (2x)(2y) - (0)(0) = 4xy$Correct Option: (a)
Ques 112. If $\partial(u,v)/\partial(x,y)=4,$ then the value of $\partial(x,y)/\partial(u,v)$ is:
- (a) -4
- (b) 4
- (c) $1/4$
- (d) 0
Step-by-Step Solution:
$J \cdot J' = 1$
$J' = \frac{1}{J}$
$J' = \frac{1}{4}$Correct Option: (c)
Ques 113. For cylindrical coordinates $x=\rho\cos\phi$, $y=\rho\sin\phi$, $z=z$, the Jacobian $\partial(x,y,z)/\partial(\rho,\phi,z)$ is:
- (a) $\rho$
- (b) $\rho^{2}$
- (c) $1$
- (d) $\rho z$
Step-by-Step Solution:
$x = \rho\cos\phi, \quad y = \rho\sin\phi, \quad z = z$
$J = \begin{vmatrix} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial \rho} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial z} \end{vmatrix}$
$J = \begin{vmatrix} \cos\phi & -\rho\sin\phi & 0 \\ \sin\phi & \rho\cos\phi & 0 \\ 0 & 0 & 1 \end{vmatrix}$
$J = 1 \cdot (\rho\cos^{2}\phi + \rho\sin^{2}\phi) = \rho(\cos^{2}\phi + \sin^{2}\phi) = \rho$Correct Option: (a)
Ques 114. If $u=x(1-y)$ and $v=xy,$ then the Jacobian $\partial(u,v)/\partial(x,y)$ is:
- (a) x
- (b) y
- (c) -x
- (d) 1
Step-by-Step Solution:
$u = x - xy, \quad v = xy$
$J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}$
$J = \begin{vmatrix} 1-y & -x \\ y & x \end{vmatrix}$
$J = (1-y)(x) - (-x)(y)$
$J = x - xy + xy = x$Correct Option: (a)
Ques 115. If $u=x+y+z$, $v=y+z$, $w=z$, then $\partial(u,v,w)/\partial(x,y,z)$ is:
- (a) 0
- (b) 1
- (c) 2
- (d) $x+y+z$
Step-by-Step Solution:
$J = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{vmatrix}$
$J = 1(1 \cdot 1 - 0 \cdot 1) - 1(0) + 1(0)$
$J = 1$Correct Option: (b)
Ques 116. If $u=x^{2}-y^{2}$ and $v=2xy$ where $x=r\cos\theta$ and $y=r\sin\theta$, then $\partial(u,v)/\partial(r,\theta)$ is:
- (a) $4r^{3}$
- (b) $2r^{2}$
- (c) $r^{3}$
- (d) 4r
Step-by-Step Solution:
$u = r^{2}\cos^{2}\theta - r^{2}\sin^{2}\theta = r^{2}\cos(2\theta)$
$v = 2(r\cos\theta)(r\sin\theta) = r^{2}\sin(2\theta)$
$J = \begin{vmatrix} \frac{\partial u}{\partial r} & \frac{\partial u}{\partial \theta} \\ \frac{\partial v}{\partial r} & \frac{\partial v}{\partial \theta} \end{vmatrix}$
$J = \begin{vmatrix} 2r\cos(2\theta) & -2r^{2}\sin(2\theta) \\ 2r\sin(2\theta) & 2r^{2}\cos(2\theta) \end{vmatrix}$
$J = (2r\cos(2\theta))(2r^{2}\cos(2\theta)) - (-2r^{2}\sin(2\theta))(2r\sin(2\theta))$
$J = 4r^{3}\cos^{2}(2\theta) + 4r^{3}\sin^{2}(2\theta) = 4r^{3}(\cos^{2}(2\theta) + \sin^{2}(2\theta)) = 4r^{3}$Correct Option: (a)
Ques 117. If $x=au$, $y=bv$, $z=cw$, then Jacobian of $(x,y,z)$ w.r.t. $(u,v,w)$
- (a) $abc$
- (b) $a^2b^2c^2$
- (c) $abcuvw$
- (d) $3abc$
Step-by-Step Solution:
$ \frac{\partial(x,y,z)}{\partial(u,v,w)}= \begin{vmatrix} a&0&0\ 0&b&0\ 0&0&c \end{vmatrix}=abc $
Correct Option: (a)
Ques 118. If $x=r\cos\varphi\sin\theta$, $y=r\sin\varphi\sin\theta$, $z=r\cos\theta$, then the value of $\dfrac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}$ is:
- (a) $0$
- (b) $r$
- (c) $r^2\sin\theta$
- (d) $r^2\cos\theta$
Step-by-Step Solution:
This is the standard spherical-coordinate Jacobian:
$ \frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}=r^2\sin\theta $
Correct Option: (c)
Ques 119. The Jacobian $\dfrac{\partial(u,v)}{\partial(x,y)}$ for the function $u=e^x\sin y$, $v=x+\log\sin y$ is
- (a) $1$
- (b) $\sin x\sin y-xy\cos x\cos y$
- (c) $0$
- (d) $\dfrac{e^x}{x}$
Step-by-Step Solution:
$ u_x=e^x\sin y,\quad u_y=e^x\cos y,\quad v_x=1,\quad v_y=\cot y $
$ \frac{\partial(u,v)}{\partial(x,y)}=u_xv_y-u_yv_x=e^x\cos y-e^x\cos y=0 $
Correct Option: (c)
Ques 120. If the function $u,v,w$ of three variables $x,y,z$ are not independent, then Jacobian of $u,v,w$ w.r.t. $x,y,z$ is always equal to
- (a) $1$
- (b) $0$
- (c) $\infty$
- (d) Jacobian of $x,y,z$ w.r.t. $u,v,w$.
Step-by-Step Solution:
Functional dependence makes the Jacobian rows or columns linearly dependent. Therefore,
$ \frac{\partial(u,v,w)}{\partial(x,y,z)}=0 $
Correct Option: (b)
Ques 121. If $x=r\cos\theta$, $y=r\sin\theta$, $z=z$, then the value of $\dfrac{\partial(x,y,z)}{\partial(r,\theta,z)}$ is
- (a) $\theta$
- (b) $r$
- (c) $0$
- (d) $r^2\sin\theta$
Step-by-Step Solution:
$ \frac{\partial(x,y,z)}{\partial(r,\theta,z)}= \begin{vmatrix} \cos\theta&-r\sin\theta&0\ \sin\theta&r\cos\theta&0\ 0&0&1 \end{vmatrix}=r $
Correct Option: (b)
Category H: Taylor's & Maclaurin's Series Expansions
Single-variable and multivariable Taylor/Maclaurin series, evaluation of origin constant terms, linear/quadratic differential terms, and power series coefficients.
Ques 122. What is the constant term (first term) in the Maclaurin series expansion of $f(x)=e^{x}$?
- (a) 0
- (b) 1
- (c) e
- (d) x
Step-by-Step Solution:
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^{2} + \dots$
$f(0) = e^{0} = 1$Correct Option: (b)
Ques 123. The Maclaurin series expansion of $\sin x$ contains only:
- (a) Even powers of x
- (b) Odd powers of x
- (c) Positive integers only
- (d) Negative powers of x
Step-by-Step Solution:
$\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots$
Correct Option: (b)
Ques 124. In the Taylor series expansion of $f(x)$ about $x=a$, the coefficient of $(x-a)$ is given by:
- (a) $f(a)$
- (b) $f^{\prime}(a)$
- (c) $f^{\prime\prime}(a)/2!$
- (d) $f^{\prime\prime}(a)$
Step-by-Step Solution:
$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^{2} + \dots$
Correct Option: (b)
Ques 125. The Maclaurin series expansion of $\cos x$ begins as:
- (a) $x-x^{3}/6$
- (b) $1-x^{2}/2$
- (c) $1+x+x^{2}/2$
- (d) $x+x^{2}/2$
Step-by-Step Solution:
$f(x) = \cos x \implies f(0) = 1$
$f'(x) = -\sin x \implies f'(0) = 0$
$f''(x) = -\cos x \implies f''(0) = -1$
$f(x) = 1 + 0x + \frac{-1}{2!}x^{2} = 1 - \frac{x^{2}}{2}$Correct Option: (b)
Ques 126. For a two-variable function $f(x,y),$ the first term in its Maclaurin series expansion is:
- (a) $f(1,1)$
- (b) $f(x,y)$
- (c) $f(0,0)$
- (d) 0
Step-by-Step Solution:
$f(x,y) = f(0,0) + (x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y})f(0,0) + \dots$
Correct Option: (c)
Ques 127. The value of $f_{x}(0,0)$ for $f(x,y)=x^{2}+3xy+2y$ is:
- (a) 0
- (b) 2
- (c) 3
- (d) 1
Step-by-Step Solution:
$f(x,y) = x^{2}+3xy+2y$
$f_{x}(x,y) = 2x + 3y$
$f_{x}(0,0) = 2(0) + 3(0) = 0$Correct Option: (a)
Ques 128. The coefficient of $x^{2}$ in the Maclaurin series of $(1+x)^{-1}$ is:
- (a) 1
- (b) -1
- (c) 2
- (d) -2
Step-by-Step Solution:
$f(x) = (1+x)^{-1} \implies f(0) = 1$
$f'(x) = -(1+x)^{-2} \implies f'(0) = -1$
$f''(x) = 2(1+x)^{-3} \implies f''(0) = 2$
$\frac{f''(0)}{2!} x^{2} = \frac{2}{2} x^{2} = 1x^{2}$Correct Option: (a)
Ques 129. The first-order linear terms in the Taylor expansion of $f(x,y)$ about (a, b) are:
- (a) $(x-a)f_{x}(a,b)+(y-b)f_{y}(a,b)$
- (b) $(x-a)^{2}f_{x}(a,b)+(y-b)^{2}f_{y}(a,b)$
- (c) $f_{x}(a,b)+f_{y}(a,b)$
- (d) $(x-a)(y-b)f_{xy}(a,b)$
Step-by-Step Solution:
$f(x,y) = f(a,b) + [(x-a)f_{x}(a,b) + (y-b)f_{y}(a,b)] + \dots$
Correct Option: (a)
Ques 130. What is the second derivative $f^{\prime\prime}(0)$ for $f(x)=e^{2x}$ at $x=0$?
- (a) 1
- (b) 2
- (c) 4
- (d) 8
Step-by-Step Solution:
$f(x) = e^{2x}$
$f'(x) = 2e^{2x}$
$f''(x) = 4e^{2x}$
$f''(0) = 4e^{2(0)} = 4(1) = 4$Correct Option: (c)
Ques 131. In the Maclaurin series of $f(x,y)=e^{x}e^{y}$ the constant term is:
- (a) 0
- (b) 1
- (c) e
- (d) 2
Step-by-Step Solution:
$f(x,y) = e^{x}e^{y}$
$f(0,0) = e^{0}e^{0} = 1 \cdot 1 = 1$Correct Option: (b)
Ques 132. The expansion of $f(x)=x^{2}+2x+1$ in powers of $(x-1)$ is:
- (a) $4+4(x-1)+(x-1)^{2}$
- (b) $1+2(x-1)+(x-1)^{2}$
- (c) $4+2(x-1)+(x-1)^{2}$
- (d) $(x-1)^{2}$
Step-by-Step Solution:
$f(1) = (1)^{2} + 2(1) + 1 = 4$
$f'(x) = 2x + 2 \implies f'(1) = 2(1) + 2 = 4$
$f''(x) = 2 \implies f''(1) = 2$
$f(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^{2}$
$f(x) = 4 + 4(x-1) + \frac{2}{2}(x-1)^{2} = 4 + 4(x-1) + (x-1)^{2}$Correct Option: (a)
Ques 133. The coefficient of $x^{3}$ in the Maclaurin expansion of $f(x)=e^{3x}$ is:
- (a) 3
- (b) $9/2$
- (c) 27
- (d) 1
Step-by-Step Solution:
$f(x) = e^{3x}$
$f'(x) = 3e^{3x} \implies f'(0) = 3$
$f''(x) = 9e^{3x} \implies f''(0) = 9$
$f'''(x) = 27e^{3x} \implies f'''(0) = 27$
$\frac{f'''(0)}{3!} x^{3} = \frac{27}{6} x^{3} = \frac{9}{2} x^{3}$Correct Option: (b)
Ques 134. Expand $f(x,y)=xy$ about the point (1, 2) up to first-degree terms.
- (a) $2+2(x-1)+(y-2)$
- (b) $2+(x-1)+2(y-2)$
- (c) $2+(x-1)+(y-2)$
- (d) $2(x-1)+2(y-2)$
Step-by-Step Solution:
$f(x,y) = xy$
$f(1,2) = 1(2) = 2$
$f_{x}(x,y) = y \implies f_{x}(1,2) = 2$
$f_{y}(x,y) = x \implies f_{y}(1,2) = 1$
$f(x,y) \approx f(1,2) + f_{x}(1,2)(x-1) + f_{y}(1,2)(y-2)$
$f(x,y) \approx 2 + 2(x-1) + 1(y-2)$Correct Option: (a)
Ques 135. The first two non-zero terms in the Maclaurin series of $f(x)=\tan x$ are:
- (a) $x+x^{2}/2$
- (b) $x+x^{3}/3$
- (c) $x-x^{3}/3$
- (d) $1+x$
Step-by-Step Solution:
$f(x) = \tan x \implies f(0) = 0$
$f'(x) = \sec^{2}x \implies f'(0) = 1$
$f''(x) = 2\sec^{2}x \tan x \implies f''(0) = 0$
$f'''(x) = 4\sec^{2}x \tan^{2}x + 2\sec^{4}x \implies f'''(0) = 2$
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + \dots$
$f(x) = x + \frac{2}{6}x^{3} = x + \frac{1}{3}x^{3}$Correct Option: (b)
Ques 136. What is the coefficient of xy in the Maclaurin expansion of $f(x,y)=\sin x\sin y$?
- (a) 0
- (b) 1
- (c) -1
- (d) $1/2$
Step-by-Step Solution:
$f(x,y) = \sin x \sin y$
$f_{xy}(x,y) = \frac{\partial}{\partial y}(\cos x \sin y) = \cos x \cos y$
$f_{xy}(0,0) = \cos(0)\cos(0) = 1 \cdot 1 = 1$
$\text{Term: } \frac{f_{xy}(0,0)}{1!1!} xy = (1)xy = xy$Correct Option: (b)
Ques 137. The Taylor series expansion of $f(x)=1/x$ about $a=1$ up to the term $(x-1)^{2}$ is:
- (a) $1-(x-1)+(x-1)^{2}$
- (b) $1+(x-1)+(x-1)^{2}$
- (c) $1-(x-1)-(x-1)^{2}$
- (d) $1-2(x-1)+(x-1)^{2}$
Step-by-Step Solution:
$f(x) = x^{-1} \implies f(1) = 1$
$f'(x) = -x^{-2} \implies f'(1) = -1$
$f''(x) = 2x^{-3} \implies f''(1) = 2$
$f(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^{2}$
$f(x) = 1 - 1(x-1) + \frac{2}{2}(x-1)^{2}$
$f(x) = 1 - (x-1) + (x-1)^{2}$Correct Option: (a)
Ques 138. For $f(x,y)=x^{2}+y^{2},$ the second-order terms in the Maclaurin expansion are:
- (a) $x+y$
- (b) $x^{2}+y^{2}$
- (c) $2x+2y$
- (d) 2xy
Step-by-Step Solution:
$f(x,y) = x^{2}+y^{2}$
$f_{xx} = 2 \implies f_{xx}(0,0) = 2$
$f_{yy} = 2 \implies f_{yy}(0,0) = 2$
$f_{xy} = 0 \implies f_{xy}(0,0) = 0$
$\frac{1}{2!}(f_{xx}x^{2} + 2f_{xy}xy + f_{yy}y^{2}) = \frac{1}{2}(2x^{2} + 0 + 2y^{2}) = x^{2} + y^{2}$Correct Option: (b)
Ques 139. The Maclaurin expansion of $f(x)=\log(1-x)$ up to the $x^{2}$ term is:
- (a) $-x-x^{2}/2$
- (b) $x-x^{2}/2$
- (c) $-x+x^{2}/2$
- (d) $x+x^{2}/2$
Step-by-Step Solution:
$f(x) = \ln(1-x) \implies f(0) = \ln(1) = 0$
$f'(x) = \frac{-1}{1-x} = -(1-x)^{-1} \implies f'(0) = -1$
$f''(x) = (1-x)^{-2}(-1) = -(1-x)^{-2} \implies f''(0) = -1$
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^{2}$
$f(x) = 0 - 1x - \frac{1}{2}x^{2} = -x - \frac{x^{2}}{2}$Correct Option: (a)
Ques 140. Linear Taylor series polynomial approximation to function $f(x,y)=x^2y-2$ about point $(1,-2)$ is:
- (a) $y-4x+2$
- (b) $-4x+y$
- (c) $y+4x+2$
- (d) $y-4x-5$
Step-by-Step Solution:
At $(1,-2)$,
$ f=-4,\qquad f_x=2xy=-4,\qquad f_y=x^2=1 $
$ L=-4-4(x-1)+(y+2)=y-4x+2 $
Correct Option: (a)
Ques 141. If $f(x,y)=(50-x^2-y^2)^{\frac12}$, the approximate value of $f(3,4)-f(3.1,3.9)$ is:
- (a) $-0.02$
- (b) $0.02$
- (c) $0.04$
- (d) $-0.04$
Step-by-Step Solution:
At $(3,4)$, $f=5$, $f_x=-\dfrac35$, $f_y=-\dfrac45$, $dx=0.1$, $dy=-0.1$.
$ df\approx f_xdx+f_ydy=-\frac35(0.1)-\frac45(-0.1)=0.02 $
Therefore,
$ f(3,4)-f(3.1,3.9)\approx-0.02 $
Correct Option: (a)
Category I: Differential-Equation Identities
Verification of partial differential equations, harmonic functions, wave and heat equations, and multivariable functional identities.
Ques 142. If $z=f(x+ay)+g(x-ay)$, then
- (a) $z_{xx}=z_{yy}$
- (b) $z_{xx}=a^2z_{yy}$
- (c) $z_{yy}=a^2z_{xx}$
- (d) $z_{xx}+a^2z_{yy}=0$
Step-by-Step Solution:
$ z_{xx}=f''+g'',\qquad z_{yy}=a^2(f''+g'') $
Hence,
$ z_{yy}=a^2z_{xx} $
Correct Option: (c)
Ques 143. If $u=f(y+ax)+g(y-ax)$, then $u_{xx}-a^2u_{yy}$ is equal to
- (a) $0$
- (b) $a^2$
- (c) $a^2(f''-g'')$
- (d) $a^2(f''+g'')$
Step-by-Step Solution:
$ u_{xx}=a^2(f''+g''),\qquad u_{yy}=f''+g'' $
Therefore,
$ u_{xx}-a^2u_{yy}=0 $
Correct Option: (a)
Ques 144. If $\theta=t^ne^{-\frac{r^2}{4t}}$ such that
$ \frac1{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial\theta}{\partial r}\right)=\frac{\partial\theta}{\partial t}, $
the value of $n$ is
- (a) $\dfrac32$
- (b) $-\dfrac32$
- (c) $\dfrac12$
- (d) $1$
Step-by-Step Solution:
$ \frac{\partial\theta}{\partial t}=\theta\left(\frac nt+\frac{r^2}{4t^2}\right) $
and
$ \frac1{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial\theta}{\partial r}\right) =\theta\left(-\frac{3}{2t}+\frac{r^2}{4t^2}\right) $
Comparing gives $n=-\dfrac32$.
Correct Option: (b)
Category J: Maxima, Minima, Saddle Points & Critical Points
Critical and stationary points, second-derivative discriminant test ($D = rt - s^2$), local maxima, local minima, and saddle point classifications.
Ques 145. The function $f(x,y)=2x^2+2xy-y^3$ has
- (a) only one stationary point at $(0,0)$.
- (b) two stationary points at $(0,0)$ and $\left(\dfrac16,-\dfrac13\right)$.
- (c) two stationary points at $(0,0)$ and $(1,-1)$.
- (d) no stationary points.
Step-by-Step Solution:
$ f_x=4x+2y=0,\qquad f_y=2x-3y^2=0 $
Thus $y=-2x$, giving $y=0$ or $y=-\dfrac13$. Therefore the points are
$ (0,0),\qquad \left(\frac16,-\frac13\right) $
Correct Option: (b)
Ques 146. The function $z=5xy-4x^2+y^2-2x+y+5$ has at $x=-\dfrac9{41}$ and $y=\dfrac2{41}$
- (a) maxima
- (b) saddle point
- (c) minima
- (d) none of above
Step-by-Step Solution:
$ r=z_{xx}=-8,\qquad s=z_{xy}=5,\qquad t=z_{yy}=2 $
$ rt-s^2=(-8)(2)-25=-41<0 $
Hence the stationary point is a saddle point.
Correct Option: (b)
Ques 147. With usual notations, the properties of maximum and minima under various conditions are
| I | II |
|---|---|
| (P) Maxima | (i) $rt-s^2=0$ |
| (Q) Minima | (ii) $rt-s^2<0$ |
| (R) Saddle Point | (iii) $rt-s^2>0, r>0$ |
| (S) Case of failure | (iv) $rt-s^2>0, r<0$ |
- (a) P-i, Q-iii, R-iv, S-ii
- (b) P-ii, Q-i, R-iii, S-iv
- (c) P-iii, Q-iv, R-ii, S-i
- (d) P-iv, Q-iii, R-ii, S-i
Step-by-Step Solution:
$ \begin{aligned} rt-s^2>0,\ r<0 &\Rightarrow \text{maximum},\ rt-s^2>0,\ r>0 &\Rightarrow \text{minimum},\ rt-s^2<0 &\Rightarrow \text{saddle point},\ rt-s^2=0 &\Rightarrow \text{test fails}. \end{aligned} $
Correct Option: (d)
Ques 148. The critical points of the function $x^3+y^3-3axy=0$, $a>0$, are
- (a) $(0,0),(a,a)$
- (b) $(0,0),\left(\dfrac a2,\dfrac a2\right)$
- (c) $(0,0),\left(\dfrac a3,\dfrac a3\right)$
- (d) $(0,0),\left(\dfrac{3a}{2},\dfrac{3a}{2}\right)$
Step-by-Step Solution:
For $f=x^3+y^3-3axy$,
$ f_x=3x^2-3ay=0,\qquad f_y=3y^2-3ax=0 $
Solving gives $(0,0)$ and $(a,a)$.
Correct Option: (a)
Ques 149. The extreme value of the function $f(x,y)=xy-x^2-y^2-2x-2y+4$ is:
- (a) $0$
- (b) $8$
- (c) $-8$
- (d) $4$
Step-by-Step Solution:
$ f_x=y-2x-2=0,\qquad f_y=x-2y-2=0 $
This gives $(x,y)=(-2,-2)$. Since
$ f_{xx}f_{yy}-f_{xy}^2=4-1=3>0,\qquad f_{xx}=-2<0, $
the point is a maximum, and
$ f(-2,-2)=8 $
Correct Option: (b)
Ques 150. The critical points of the function $xy(a-x-y)$, $a>0$, are:
- (a) $(0,0),(a,a)$
- (b) $(0,0),(a,0),(0,a),\left(\dfrac a2,\dfrac a2\right)$
- (c) $(0,0),(a,0),(0,a),\left(\dfrac a3,\dfrac a3\right)$
- (d) $(0,0),\left(\dfrac{3a}{2},\dfrac{3a}{2}\right)$
Step-by-Step Solution:
$ f_x=y(a-2x-y),\qquad f_y=x(a-x-2y) $
Solving $f_x=f_y=0$ gives
$ (0,0),(a,0),(0,a),\left(\frac a3,\frac a3\right) $
Correct Option: (c)
Ques 151. Which of the following point/points behave as saddle point for the function
$ f(x,y)=x^3+3xy^2-15x^2-15y^2+72x-90 $
- (a) $(4,0),(6,0)$
- (b) $(0,4),(0,6)$
- (c) $(1,5),(-1,5)$
- (d) $(5,1),(5,-1)$
Step-by-Step Solution:
$ f_{xx}=6x-30,\qquad f_{yy}=6x-30,\qquad f_{xy}=6y $
At $(5,\pm1)$,
$ f_{xx}f_{yy}-f_{xy}^2=0-36=-36<0 $
Hence both points are saddle points.
Correct Option: (d)
Ques 152. The function $f(x,y)$ has a relative maximum at point $(a,b)$ if for small positive or negative values of $h$ and $k$ we have
- (a) $f(a,b)<f(a+h,b+k)$
- (b) $f(a,b)>f(a+h,b+k)$
- (c) $f(a,b)=f(a+h,b+k)$
- (d) none of above
Step-by-Step Solution:
At a relative maximum, the function value at $(a,b)$ is greater than the values at all sufficiently nearby points.
Correct Option: (b)
Ques 153. Which of the following statement is not true:
- (a) Extremum is a point which is either a maximum or minimum.
- (b) Saddle point is a point, where the function is neither maximum nor minimum.
- (c) Every stationary point is an extrema.
- (d) Extrema occurs only at stationary points.
Step-by-Step Solution:
A stationary point may be a maximum, minimum, saddle point, or a case where the second-derivative test fails.
Correct Option: (c)
Ques 154. For the function $f(x,y)=1-x^2-y^2$, the point $(0,0)$ is a
- (a) saddle point
- (b) maximum point
- (c) minimum point
- (d) none of above
Step-by-Step Solution:
$ f(x,y)=1-(x^2+y^2)\le 1=f(0,0) $
Hence $(0,0)$ is a maximum point.
Correct Option: (b)
Ques 155. The critical point of the function $f(x,y)=x^2+y^2-9a$, $a>0$, is
- (a) $(0,0)$
- (b) $(a,0)$
- (c) $(0,a)$
- (d) $(a,a)$
Step-by-Step Solution:
$ f_x=2x=0,\qquad f_y=2y=0 $
Therefore, the critical point is $(0,0)$.
Correct Option: (a)
Ques 156. For the function $f(x,y)=x^3+y^3-3axy$,
- (a) the point $(a,a)$ is a point of maximum, if $a>0$.
- (b) the point $(a,a)$ is a point of maximum, if $a<0$.
- (c) the point $(a,a)$ is a point of minimum, if $a<0$.
- (d) the point $(a,a)$ is a saddle point.
Step-by-Step Solution:
At $(a,a)$,
$ f_{xx}=6a,\qquad f_{yy}=6a,\qquad f_{xy}=-3a $
$ D=f_{xx}f_{yy}-f_{xy}^2=27a^2>0 $
For $a<0$, $f_{xx}<0$, so the point is a maximum.
Correct Option: (b)
Ques 157. For the function $f(x,y)=xy(a-x-y)$,
- (a) the point $\left(\dfrac a3,\dfrac a3\right)$ is a point of maximum if $a>0$.
- (b) the point $\left(\dfrac a3,\dfrac a3\right)$ is a point of maximum if $a<0$.
- (c) the point $\left(\dfrac a3,\dfrac a3\right)$ is a point of minimum if $a>0$.
- (d) the point $\left(\dfrac a3,\dfrac a3\right)$ is a saddle point.
Step-by-Step Solution:
At $\left(\dfrac a3,\dfrac a3\right)$,
$ f_{xx}=-\frac{2a}{3},\qquad f_{yy}=-\frac{2a}{3},\qquad f_{xy}=-\frac a3 $
$ D=\frac{a^2}{3}>0 $
For $a>0$, $f_{xx}<0$, so the point is a maximum.
Correct Option: (a)
Ques 158. Which of the following statement is not true:
- (a) The tangent planes to the surface at the maximum or minimum point is parallel to $xy$-plane and perpendicular to $z$-axis.
- (b) The hyperbolic paraboloid, $z-xy=0$, has a saddle point at the point $(1,1)$.
- (c) At a saddle point the function is maximum in one direction, while minimum in other direction.
- (d) Extrema occurs only at stationary points.
Step-by-Step Solution:
For $z=xy$, the stationary point satisfies $z_x=y=0$ and $z_y=x=0$. The saddle point is $(0,0)$, not $(1,1)$.
Correct Option: (b)
Ques 159. Critical points of the function $f(x,y)=x^3y^2(a-x-y)$, $a<0$, are
- (a) $(0,0),\left(\dfrac a3,\dfrac a3\right)$
- (b) $(0,0),\left(\dfrac a2,\dfrac a3\right)$
- (c) $(0,0),\left(\dfrac a3,\dfrac a2\right)$
- (d) $(0,0),\left(\dfrac a2,\dfrac a2\right)$
Step-by-Step Solution:
For $x\ne0$ and $y\ne0$,
$ f_x=x^2y^2(3a-4x-3y)=0 $
$ f_y=x^3y(2a-2x-3y)=0 $
Solving
$ 4x+3y=3a,\qquad 2x+3y=2a $
gives $x=\dfrac a2$, $y=\dfrac a3$.
Accuracy note: Strictly, every point on $x=0$ or $y=0$ is also stationary. Option B is the intended answer because it lists the nonzero isolated critical point along with the origin.
Correct Option: (b)
Ques 160. The stationary points of $x^4+y^4-2x^2+4xy-2y^2$ are:
- (a) $(0,0),(\sqrt2,-\sqrt2),(-\sqrt2,\sqrt2)$
- (b) $(\sqrt2,-\sqrt2),(-\sqrt2,\sqrt2)$
- (c) $(0,0),(-\sqrt2,\sqrt2)$
- (d) $(0,0),(\sqrt2,-\sqrt2)$
Step-by-Step Solution:
$ f_x=4(x^3-x+y)=0,\qquad f_y=4(y^3+x-y)=0 $
Solving gives
$ (0,0),(\sqrt2,-\sqrt2),(-\sqrt2,\sqrt2) $
Correct Option: (a)
Category K: Constrained Extrema & Lagrange Multipliers
Constrained optimization, Lagrange multiplier method ($\nabla f = \lambda \nabla g$), geometric extrema on curves and surfaces, and stationary point evaluations.
Ques 161. The minimum value of the function $f(x,y,z)=x^2+y^2+z^2$, where $lx+my+nz=p$, is:
- (a) $\dfrac{l^2+m^2+n^2}{p^2}$
- (b) $\dfrac{p^2}{l^2+m^2+n^2}$
- (c) $\dfrac{3p^2}{l^2+m^2+n^2}$
- (d) $\dfrac{l^2+m^2+n^2}{3p^2}$
Step-by-Step Solution:
By Cauchy-Schwarz,
$ (lx+my+nz)^2\le(l^2+m^2+n^2)(x^2+y^2+z^2) $
Thus
$ x^2+y^2+z^2\ge\frac{p^2}{l^2+m^2+n^2} $
Correct Option: (b)
Ques 162. The minimum value of the function $f(x,y,z)=x^2+y^2+z^2$ subjected to the condition $\dfrac1x+\dfrac1y+\dfrac1z=1$ is
- (a) $18$
- (b) $81$
- (c) $243$
- (d) $27$
Step-by-Step Solution:
Using Lagrange multipliers gives $x^3=y^3=z^3$, so $x=y=z$. Then
$ \frac3x=1\implies x=y=z=3 $
$ f_{min}=3^2+3^2+3^2=27 $
Correct Option: (d)
Ques 163. The shortest distance from the origin to the surface $xyz^2=2$ is:
- (a) $2$
- (b) $1$
- (c) $4$
- (d) $8$
Step-by-Step Solution:
Minimize $x^2+y^2+z^2$ subject to $xyz^2=2$. Lagrange multipliers give
$ x^2=y^2,\qquad z^2=2x^2 $
For the nearest point, take $x=y=1$, $z=\sqrt2$. Hence
$ d=\sqrt{1+1+2}=2 $
Correct Option: (a)