Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

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)

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

2

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 $$=4$$

3

MCQ (Single Correct Answer)

The value of '$$a$$' for which one root of the quadratic equation
$$$\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0$$$

is twice as large as the other is

is twice as large as the other is

A

$$ - {1 \over 3}$$

B

$$ {2 \over 3}$$

C

$$ - {2 \over 3}$$

D

$$ {1 \over 3}$$

Let the roots of given equation be $$\alpha $$ and $$2$$$$\alpha $$ then

$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$

and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$

$$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$$

$$\therefore$$ $$2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]$$

$$ = {2 \over {{a^2} - 5a + 3}}$$

$${{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9$$

or $$9{a^2} - 6a + 1$$

$$ = 9{a^2} - 45a + 27$$

or $$39a = 26$$ or $$a = {2 \over 3}$$

$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$

and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$

$$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$$

$$\therefore$$ $$2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]$$

$$ = {2 \over {{a^2} - 5a + 3}}$$

$${{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9$$

or $$9{a^2} - 6a + 1$$

$$ = 9{a^2} - 45a + 27$$

or $$39a = 26$$ or $$a = {2 \over 3}$$

4

MCQ (Single Correct Answer)

If the sum of the roots of the quadratic equation $$a{x^2} + bx + c = 0$$ is equal to the sum of the squares of their reciprocals, then $${a \over c},\,{b \over a}$$ and $${c \over b}$$ are in

A

Arithmetic - Geometric Progression

B

Arithmetic Progression

C

Geometric Progression

D

Harmonic Progression

$$a{x^2} + bx + c = 0,$$ $$\alpha + \beta = {{ - b} \over a},\alpha \beta = {c \over a}$$

As for given condition, $$\alpha + \beta = {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}}$$

$$\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}$$

$$ = {{{{{b^2}} \over {{a^2}}} - {{2c} \over a}} \over {{{{c^2}} \over {{a^2}}}}}$$

On simplification $$2{a^2}c = a{b^2} + b{c^2}$$

$$ \Rightarrow {{2a} \over b} = {c \over a} + {b \over c}$$

$$ \Rightarrow {c \over a},{a \over b},{b \over c}$$ are in $$A.P.$$

$$\therefore$$ $${a \over c},{b \over a},\,\,\& \,\,$$ are in $$H.P.$$

As for given condition, $$\alpha + \beta = {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}}$$

$$\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}$$

$$ = {{{{{b^2}} \over {{a^2}}} - {{2c} \over a}} \over {{{{c^2}} \over {{a^2}}}}}$$

On simplification $$2{a^2}c = a{b^2} + b{c^2}$$

$$ \Rightarrow {{2a} \over b} = {c \over a} + {b \over c}$$

$$ \Rightarrow {c \over a},{a \over b},{b \over c}$$ are in $$A.P.$$

$$\therefore$$ $${a \over c},{b \over a},\,\,\& \,\,$$ are in $$H.P.$$

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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