Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The area included between the parabolas y^{2} = 4x and x^{2} = 4y is

A

$${8 \over 3}$$ sq. units

B

8 sq. units

C

$${16 \over 3}$$ sq. units

D

12 sq. units

Equation of curves are

y^{2} = 4x ..... (i)

and x^{2} = 4y .... (ii)

From (i) and (ii)

$${\left( {{{{x^2}} \over 4}} \right)^2} = 4x \Rightarrow {x^4} - 64x = 0$$

$$ \Rightarrow x({x^3} - 64) = 0 \Rightarrow x = 0$$ or $$x = 4$$

$$\therefore$$ Required area $$ = \int\limits_0^4 {\left( {\sqrt {4x} - {{{x^2}} \over 4}} \right)dx = \left[ {2.{{2{x^{3/2}}} \over 3} - {{{x^3}} \over {12}}} \right]_0^4} $$

$$ = {{32} \over 3} - {{64} \over {12}} = {{32} \over 3} - {{16} \over 3} = {{16} \over 3}$$ sq. unit.

2

MCQ (Single Correct Answer)

The line which is parallel to x-axis and crosses the curve y = $$\sqrt x$$ at an angle 45$$^\circ$$ is

A

y = $${1 \over 4}$$

B

y = $${1 \over 2}$$

C

y = 1

D

y = 4

Since equation of any line parallel to y-axis is y = k where k is any real constant. We draw a tangent which makes an angle 45$$^\circ$$ with the line parallel to x-axis.

Given equation of curve is

$$y = \sqrt x $$ ..... (1)

Taking derivative

$${{dy} \over {dx}} = {1 \over {2\sqrt x }}$$

Since $${{dy} \over {dx}} = \tan 45^\circ = 1$$

$$\therefore$$ $${1 \over {2\sqrt x }} = 1$$ or $$x = {1 \over 4}$$

Substituting $$x = {1 \over 4}$$ in (1)

$$y = \sqrt {{1 \over 4}} = {1 \over 2}$$

$$\therefore$$ Point $$P\left( {{1 \over 4},{1 \over 2}} \right)$$, it also passes through the line parallel to x-axis.

$$\therefore$$ $$y = {1 \over 2}$$ is required equation of line.

3

MCQ (Single Correct Answer)

The area enclosed between the curve y = 1 + x^{2}, the y-axis, and the straight line y = 5 is given by

A

$${{14} \over 3}$$ square units

B

$${{7} \over 3}$$ square units

C

5 square units

D

$${{16} \over 3}$$ square units

Given equation of curve is y = 1 + x^{2} or x^{2} = (y $$-$$ 1)

This is equation of parabola having vertex (0, 1)

$$\therefore$$ Required area $$ = \int\limits_1^5 {\sqrt {y - 1} \,dy} $$

$$ = {2 \over 3}\left[ {{{(y - 1)}^{3/2}}} \right]_1^5$$

$$ = {2 \over 3}\left[ {{{(4)}^{3/2}}} \right] = {2 \over 3} \times 8 = {{16} \over 3}$$ sq. units

4

MCQ (Single Correct Answer)

The function f(x) which satisfies $$f(x) = f( - x) = {{f'(x)} \over x}$$ is given by

A

$$f(x) = {1 \over 2}{e^{{x^2}}}$$

B

$$f(x) = {1 \over 2}{e^{ - {x^2}}}$$

C

$$f(x) = {x^2}{e^{{x^2}/2}}$$

D

$$f(x) = {e^{{x^2}/2}}$$

$$f(x) = {{f'(x)} \over x} \Rightarrow {{f'(x)} \over {f(x)}} = x$$

Integrating both sides, we get

$$\int {{{f'(x)} \over {f(x)}}dx = \int {x\,dx \Rightarrow \log |f(x)| = {{{x^2}} \over 2} + k} } $$

$$ \Rightarrow |f(x)| = {e^{({x^2}/2) + k}}{e^k}\,.\,{e^{{{{x^2}} \over 2}}}$$

$$ \Rightarrow |f(x)| = c{e^{{x^2}/2}}$$ ($$c = {e^k}$$ is an arbitrary constant)

$$ \Rightarrow f(x) = c{e^{{x^2}/2}}$$

keeping c = 1, $$f(x) = {e^{{x^2}/2}}$$.

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

WB JEE 2022 (3)

WB JEE 2021 (3)

WB JEE 2020 (3)

WB JEE 2019 (1)

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Logarithms

Sequence and Series

Quadratic Equations

Complex Numbers

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Matrices and Determinants

Vector Algebra

Three Dimensional Geometry

Probability

Statistics

Sets and Relations

Functions

Definite Integration

Application of Integration

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Differential Equations

Straight Lines and Pair of Straight Lines

Circle

Parabola

Ellipse and Hyperbola