Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Considering only the principal values of inverse functions, the set

A = { x $$ \ge $$ 0: tan^{$$-$$1}(2x) + tan^{$$-$$1}(3x) = $${\pi \over 4}$$}

A = { x $$ \ge $$ 0: tan

A

contains two elements

B

contains more than two elements

C

is an empty set

D

is a singleton

tan^{$$-$$1}(2x) + tan^{$$-$$1}(3x) = $$\pi $$/4

$$ \Rightarrow \,\,{{5x} \over {1 - 6{x^2}}}$$ = 1

$$ \Rightarrow $$ 6x^{2} + 5x $$-$$ 1 = 0

x = $$-$$1 or x = $${1 \over 6}$$

x = $${1 \over 6}$$

$$ \because $$ x > 0

$$ \Rightarrow \,\,{{5x} \over {1 - 6{x^2}}}$$ = 1

$$ \Rightarrow $$ 6x

x = $$-$$1 or x = $${1 \over 6}$$

x = $${1 \over 6}$$

$$ \because $$ x > 0

2

MCQ (Single Correct Answer)

All x satisfying the inequality (cot^{–1}
x)^{2}– 7(cot^{–1} x) + 10 > 0, lie in the interval :

A

(cot 2, $$\infty $$)

B

(–$$\infty $$, cot 5) $$ \cup $$ (cot 2, $$\infty $$)

C

(cot 5, cot 4)

D

(– $$\infty $$, cot 5) $$ \cup $$ (cot 4, cot 2)

cot^{$$-$$1} x > 5, cot^{$$-$$1} x < 2

$$ \Rightarrow $$ x < cot5, x > cot2

$$ \Rightarrow $$ x < cot5, x > cot2

3

MCQ (Single Correct Answer)

The value of $$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \right)$$ is -

A

$${{22} \over {23}}$$

B

$${{23} \over {22}}$$

C

$${{21} \over {19}}$$

D

$${{19} \over {21}}$$

$$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right.} } \right)$$

$$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} } \right)$$

$$\sum\limits_{n = 1}^{19} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)} $$

$$\cot \left( {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right) = {{\cot A\cot \beta + 1} \over {\cot \beta - \cot A}}$$

(Where tanA $$=$$ 20, tanB $$=$$ 1) $${{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}$$

$$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} } \right)$$

$$\sum\limits_{n = 1}^{19} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)} $$

$$\cot \left( {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right) = {{\cot A\cot \beta + 1} \over {\cot \beta - \cot A}}$$

(Where tanA $$=$$ 20, tanB $$=$$ 1) $${{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}$$

4

MCQ (Single Correct Answer)

English

Hindi

If x = sin^{$$-$$1}(sin10) and y = cos^{$$-$$1}(cos10), then y $$-$$ x is equal to :

A

0

B

10

C

7$$\pi $$

D

$$\pi $$

x = sin^{$$-$$1} sin 10 = 3$$\pi $$ $$-$$ 10

y = cos^{$$-$$1}cos 10 = 4$$\pi $$ $$-$$ 10

y $$-$$ x = (4$$\pi $$ $$-$$ 10) $$-$$ (3$$\pi $$ $$-$$ 10) = $$\pi $$

y = cos

y $$-$$ x = (4$$\pi $$ $$-$$ 10) $$-$$ (3$$\pi $$ $$-$$ 10) = $$\pi $$

यदि $$x=\sin ^{-1}(\sin 10)$$ तथा $$y=\cos ^{-1}(\cos 10)$$ है, तो $$y-x$$ बराबर है-

A

0

B

10

C

7$$\pi$$

D

$$\pi$$

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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

Sets and Relations

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