Joint Entrance Examination

Graduate Aptitude Test in Engineering

NEW

New Website Launch

Experience the best way to solve previous year questions with **mock
tests** (very detailed analysis), **bookmark your favourite questions**, **practice** etc...

1

MCQ (Single Correct Answer)

The Rolle's theorem is applicable in the interval $$-$$1 $$\le$$ x $$\le$$ 1 for the function

A

f(x) = x

B

f(x) = x^{2}

C

f(x) = 2x^{3} + 3

D

f(x) = |x|

(a) f(x) = x

$$f'(x) = {{df(x)} \over {dx}} = 1$$ which is greater than zero

$$\therefore$$ f (x) is strictly increasing in [$$-$$1, 1].

So Rolle's theorem is not applicable.

(b) $$\because$$ f($$-$$1) = f(1) = 1

Also f(x) = x^{2} is continuous in [$$-$$1, 1] and differentiable in ($$-$$1, 1)

$$\therefore$$ Rolle's theorem is applicable.

(c) f(x) = 2x^{3} + 3 $$\Rightarrow$$ f'(x) = 6x^{2} > 0

$$\therefore$$ f(x) is strictly increasing in [$$-$$1, 1].

So, Rolle's theorem is not applicable.

(d) f(x) = |x| = x, x $$\ge$$ 0 and $$-$$x, x < 0

f(1) = f($$-$$1) = 1, also f(x) is continuous but f(x) is not differentiable at x = 0 $$\in$$ ($$-$$1, 1). So all conditions of Rolle's theorem is not satisfied.

2

MCQ (Single Correct Answer)

A particle is moving in a straight line. At time t, the distance between the particle from its starting point is given by x = t $$-$$ 6t^{2} + t^{3}. Its acceleration will be zero at

A

t = 1 unit time

B

t = 2 unit time

C

t = 3 unit time

D

t = 4 unit time

$$x = t - 6{t^2} + {t^3}$$

velocity $$ = {{dx} \over {dt}} = 1 - 12t + 3{t^2}$$

acceleration $$ = {d \over {dt}}\left( {{{dx} \over {dt}}} \right) = - 12 + 6t$$

If acceleration is zero, then

$$ - 12 + 6t = 0 \Rightarrow t = 2$$.

3

MCQ (Single Correct Answer)

The equation of the tangent to the conic $${x^2} - {y^2} - 8x + 2y + 11 = 0$$ at (2, 1) is

A

x + 2 = 0

B

2x + 1 = 0

C

x + y + 1 = 0

D

x $$-$$ 2 = 0

Equation of conic section $${x^2} - {y^2} - 8x + 2y + 11 = 0$$

Differentiating w.r.t. x, we get

$$2x - 2yy' - 8 + 2y' = 0 \Rightarrow y' = {{4 - x} \over {1 - y}}$$

Slope of tangent at (2, 1) is $${(y')_{(2,1)}} = {{4 - 2} \over {1 - 1}} = {2 \over 0}$$

$$\therefore$$ Equation of tangent at (2, 1) is

$$y - 1 = {2 \over 0}(x - 2) \Rightarrow x - 2 = 0$$

4

MCQ (Single Correct Answer)

A particle is projected vertically upwards and is at a height h after t_{1} seconds and again after t_{2} seconds then

A

$$h = g{t_1}{t_2}$$

B

$$h = {1 \over 2}g{t_1}{t_2}$$

C

$$h = {2 \over g}{t_1}{t_2}$$

D

$$h = \sqrt {g{t_1}{t_2}} $$

Let the initial velocity is u

$$\therefore$$ $$h = ut - {1 \over 2}g{t^2}$$

$$ \Rightarrow g{t^2} - 2ut + 2h = 0$$

This is a quadratic in t,

let it has two roots t_{1} and t_{2}

$$\therefore$$ $${t_1}{t_2} = {{2h} \over g}$$ ( Product of roots )

$$ \Rightarrow h = {1 \over 2}g{t_1}{t_2}$$.

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

WB JEE 2022 (1)

WB JEE 2021 (2)

WB JEE 2020 (4)

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