Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If $$a,\,b,\,c$$ are distinct $$ + ve$$ real numbers and $${a^2} + {b^2} + {c^2} = 1$$ then $$ab + bc + ca$$ is

A

less than 1

B

equal to 1

C

greater than 1

D

any real no.

As $$\,\,\,\,\,{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} > 0$$

$$ \Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0$$

$$ \Rightarrow 2 > 2\left( {ab + bc + ca} \right)$$

$$ \Rightarrow ab + bc + ca < 1$$

$$ \Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0$$

$$ \Rightarrow 2 > 2\left( {ab + bc + ca} \right)$$

$$ \Rightarrow ab + bc + ca < 1$$

2

MCQ (Single Correct Answer)

If $$p$$ and $$q$$ are the roots of the equation $${x^2} + px + q = 0,$$ then

A

$$p = 1,\,\,q = - 2$$

B

$$p = 0,\,\,q = 1$$

C

$$p = - 2,\,\,q = 0$$

D

$$p = - 2,\,\,q = 1$$

$$p + q = - p$$ and $$pq = q \Rightarrow q\left( {p - 1} \right) = 0$$

$$ \Rightarrow q = 0$$ or $$p=1.$$

If $$q = 0,$$ then $$p=0.$$ i.e.$$p=q$$

$$\therefore$$ $$p=1$$ and $$q=-2.$$

$$ \Rightarrow q = 0$$ or $$p=1.$$

If $$q = 0,$$ then $$p=0.$$ i.e.$$p=q$$

$$\therefore$$ $$p=1$$ and $$q=-2.$$

3

MCQ (Single Correct Answer)

Difference between the corresponding roots of $${x^2} + ax + b = 0$$ and $${x^2} + bx + a = 0$$ is same and $$a \ne b,$$ then

A

$$a + b + 4 = 0$$

B

$$a + b - 4 = 0$$

C

$$a - b - 4 = 0$$

D

$$a - b + 4 = 0$$

Let $$\alpha ,\beta $$ and $$\gamma ,\delta $$ be the roots of the equations $${x^2} + ax + b = 0$$

and $${x^2} + bx + a = 0$$ respectively.

$$\therefore$$ $$\alpha + \beta = - a,\alpha \beta = b$$

and $$\gamma + \delta = - b,\gamma \delta = a.$$

Given $$\left| {\alpha - \beta } \right| = \left| {\gamma - \delta } \right|$$

$$ \Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( {\gamma - \delta } \right)^2}$$

$$ \Rightarrow {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta = {\left( {\gamma + \delta } \right)^2} - 4\gamma \delta $$

$$ \Rightarrow {a^2} - 4b = {b^2} - 4a$$

$$ \Rightarrow \left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0$$

$$ \Rightarrow a + b + 4 = 0$$

( as $$a \ne b$$ )

and $${x^2} + bx + a = 0$$ respectively.

$$\therefore$$ $$\alpha + \beta = - a,\alpha \beta = b$$

and $$\gamma + \delta = - b,\gamma \delta = a.$$

Given $$\left| {\alpha - \beta } \right| = \left| {\gamma - \delta } \right|$$

$$ \Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( {\gamma - \delta } \right)^2}$$

$$ \Rightarrow {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta = {\left( {\gamma + \delta } \right)^2} - 4\gamma \delta $$

$$ \Rightarrow {a^2} - 4b = {b^2} - 4a$$

$$ \Rightarrow \left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0$$

$$ \Rightarrow a + b + 4 = 0$$

( as $$a \ne b$$ )

4

MCQ (Single Correct Answer)

Product of real roots of equation $${t^2}{x^2} + \left| x \right| + 9 = 0$$

A

is always positive

B

is always negative

C

does not exist

D

none of these

Product of real roots $$ = {9 \over {{t^2}}} > 0,\forall \,t\, \in R$$

$$\therefore$$ Product of real roots is always positive.

$$\therefore$$ Product of real roots is always positive.

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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

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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