Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If $$\left( {1 - p} \right)$$ is a root of quadratic equation $${x^2} + px + \left( {1 - p} \right) = 0$$ then its root are

A

$$ - 1,2$$

B

$$ - 1,1$$

C

$$ 0,-1$$

D

$$0,1$$

Let the second root be $$\alpha .$$

Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$

Also $$\alpha .\left( {1 - p} \right) = 1 - p$$

$$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$

$$ \Rightarrow p = 1$$ [as $$\alpha = - 1$$]

$$\therefore$$ Roots are $$\alpha = - 1$$ and $$p-1=0$$

Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$

Also $$\alpha .\left( {1 - p} \right) = 1 - p$$

$$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$

$$ \Rightarrow p = 1$$ [as $$\alpha = - 1$$]

$$\therefore$$ Roots are $$\alpha = - 1$$ and $$p-1=0$$

2

MCQ (Single Correct Answer)

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation

A

$${x^2} - 18x - 16 = 0$$

B

$${x^2} - 18x + 16 = 0$$

C

$${x^2} + 18x - 16 = 0$$

D

$${x^2} + 18x + 16 = 0$$

Let two numbers be a and b then $${{a + b} \over 2} = 9$$

and $$\sqrt {ab} = 4$$

$$\therefore$$ Equation with roots $$a$$ and $$b$$ is

$${x^2} - \left( {a + b} \right)x + ab = 0$$

$$ \Rightarrow {x^2} - 18x + 16 = 0$$

and $$\sqrt {ab} = 4$$

$$\therefore$$ Equation with roots $$a$$ and $$b$$ is

$${x^2} - \left( {a + b} \right)x + ab = 0$$

$$ \Rightarrow {x^2} - 18x + 16 = 0$$

3

MCQ (Single Correct Answer)

The real number $$x$$ when added to its inverse gives the minimum value of the sum at $$x$$ equal to

A

-2

B

2

C

1

D

-1

$$y = x + {1 \over x}$$ or $${{dy} \over {dx}} = 1 - {1 \over {{x^2}}}$$

For max. or min, $$1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1$$

$${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} \Rightarrow {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 2}} = 2$$ ($$+ve$$ minima)

$$\therefore$$ $$x=1$$

For max. or min, $$1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1$$

$${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} \Rightarrow {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 2}} = 2$$ ($$+ve$$ minima)

$$\therefore$$ $$x=1$$

4

MCQ (Single Correct Answer)

The number of real solutions of the equation $${x^2} - 3\left| x \right| + 2 = 0$$ is

A

3

B

2

C

4

D

1

$${x^2} - 3\left| x \right| + 2 = 0$$

$$ \Rightarrow {\left| x \right|^2} - 3\left| x \right| + 2 = 0$$

$$\left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 1} \right) = 0$$

$$\left| x \right| = 1,2$$ or $$x = \pm 1, \pm 2$$

$$\therefore$$ No. of solution $$=4$$

$$ \Rightarrow {\left| x \right|^2} - 3\left| x \right| + 2 = 0$$

$$\left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 1} \right) = 0$$

$$\left| x \right| = 1,2$$ or $$x = \pm 1, \pm 2$$

$$\therefore$$ No. of solution $$=4$$

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Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations