If $ - \frac{\pi }{2} < x < \frac{\pi }{2}$, then the value of $\log \,\,\sec \,x$ is

$2\,\,{\coth ^{ - 1}}\left( {{\rm{cose}}{{\rm{c}}^2}\,\frac{x}{2} - 1} \right)$

$2\,\,{\coth ^{ - 1}}\left( {{\rm{cose}}{{\rm{c}}^2}\frac{x}{2} + 1} \right)$

$2\,\,{\rm{cosec}}{{\rm{h}}^{ - 1}}\left( {{\rm{co}}{{\rm{t}}^2}\frac{x}{2} - 1} \right)$

$2\,{\rm{cosec}}{{\rm{h}}^{ - 1}}\left( {{\rm{co}}{{\rm{t}}^2}\frac{x}{2} + 1} \right)$

${\rm{sec}}{{\rm{h}}^{ - 1}}\,\,\left( {\frac{1}{2}} \right)$ is

$\log \,\,\left( {\sqrt 3 + \sqrt 2 } \right)$

$\log \,\,\left( {\sqrt 3 + 1} \right)$

${\rm{sec}}{{\rm{h}}^{ - 1}}\,\,\left( {\sin \,\,\theta } \right)$ is equal to

$\log \,\,\tan \,\,\frac{\theta }{2}$

$\log \,\,\sin \,\,\frac{\theta }{2}$

$\log \,\,\cos \,\,\frac{\theta }{2}$

The real part of $\sin \log (x+i y)$ is

$\sin \left(\log \sqrt{x^2+y^2}\right) \sinh \left(\tan ^{-1} \frac{y}{x}\right)$

Back to HYPERBOLIC FUNCTIONS

Your Progress

0 / 17 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • HYPERBOLIC FUNCTIONS

Hyperbolic functions practice · Mixed difficulty

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →

1 Easy

1 of 17

Application +4 | -1

\cos (x+h y)

is cqual to:

View answer and explanation

Correct option: C

\cos x \cosh y-i \sin x \sinh y

Option C (cos x cosh y - i sin x sinh y) is correct as it represents the expansion of cos(x + iy) using Euler's formula and hyperbolic functions.

2 Medium

2 of 17

Analyzing +4 | -1

Let

{H}: {xy}-{y}-2 {x}-2=0

. If normal drawn to

{H}

{P}(3,4)

intersects the curve again at

Q(\alpha+1, \beta+1)

, then

\frac{\alpha}{\beta}

View answer and explanation

Correct option: A

Option A (2) is correct because after transforming the hyperbola to its standard form and finding the normal at P, the ratio α/β is calculated to be 2.

3 Easy

3 of 17

Application +4 | -1

The imaginary part of

\log \sin (x+i y)

is:

View answer and explanation

Correct option: B

\tan ^{-1}\left(\frac{\cos x \sinh y}{\sin x \cosh y}\right)

Option B (

\tan ^{-1}(\frac{{\cos x \sinh y}}{{\sin x \cosh y}})

) is correct because it correctly identifies the imaginary part of the expression after applying trigonometric identities and logarithmic properties.

4 Hard

4 of 17

Application +4 | -1

{\sinh ^{ - 1}}\,\,2 + {\sinh ^{ - 1}}\,\,3 = x\; \Rightarrow \cosh \,\,x

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}\,\,\left( {12 + 2\,\sqrt {50} } \right)

Option C (

\frac{1}{2}\,\,\left( {12 + 2\,\sqrt {50} } \right)

) is correct because it is derived by applying the hyperbolic trigonometric identities to simplify the given expression.

5 Easy

5 of 17

Application +4 | -1

\frac{{1 + \mathrm{tanh\;}\,\,\frac{x}{2}}}{{1 - \mathrm{tanh\;}\,\,\frac{x}{2}}}

is equal to

View answer and explanation

Correct option: B

{e^x}

Option B (

{e^x}

) is correct because simplifying the hyperbolic tangent expression using exponential forms results in

{e^x}

6 Easy

6 of 17

Understanding +4 | -1

Which of the following functions is not defined at

x=0

View answer and explanation

Correct option: B

\operatorname{cosech} x

Option B (

\operatorname{cosech} x

) is correct because

\operatorname{cosech} x = \frac{1}{\sinh x}

and

\sinh x = 0

x=0

, making

\operatorname{cosech} x

undefined at

x=0

7 Easy

7 of 17

Application +4 | -1

The real and imaginary part of

2 \sin (\alpha+i \beta) \cos (\alpha-i \beta)

are

View answer and explanation

Correct option: C

\sin 2 \alpha, \sinh 2 \beta

Option C (

\sin 2 \alpha, \sinh 2 \beta

) is correct because it correctly identifies the real and imaginary parts after applying trigonometric identities and hyperbolic functions.

8 Easy

8 of 17

Application +4 | -1

- \frac{\pi }{2} < x < \frac{\pi }{2}

, then the value of

\log \,\,\sec \,x

View answer and explanation

Correct option: A

2\,\,{\coth ^{ - 1}}\left( {\mathrm{cose}{\mathrm{c}^2}\,\frac{x}{2} - 1} \right)

Option A (

2\,\,{\coth ^{ - 1}}\left( {\mathrm{cose}{\mathrm{c}^2}\,\frac{x}{2} - 1} \right)

) is correct because it is the result of applying componendo and dividendo rule and hyperbolic identities to the given logarithmic expression.

9 Easy

9 of 17

Understanding +4 | -1

\frac{1+\tanh x}{1-\tanh x}

is equal to:

View answer and explanation

Correct option: A

e^{2 x}

Option A (

e^{2 x}

) is correct because by converting

\tanh x

to its exponential form and simplifying the expression, it reduces to

e^{2x}

10 Easy

10 of 17

Application +4 | -1

{\cosh ^{ - 1}}\,\,x = \log \,\,\left( {2 + \sqrt 3 } \right)

, then

x

is equal to

View answer and explanation

Correct option: A

Option A is correct because it uses the inverse hyperbolic cosine definition to find the value of x.

11 Easy

11 of 17

Application +4 | -1

\mathrm{sec}{\mathrm{h}^{ - 1}}\,\,\left( {\frac{1}{2}} \right)

View answer and explanation

Correct option: C

\log \,\,\left( {2 + \sqrt 3 } \right)

Option C (

\log \,\left( {2 + \sqrt 3 } \right)

) is correct because it is the result of applying the inverse hyperbolic function formula for sech^-1(x).

12 Easy

12 of 17

Application +4 | -1

\mathrm{sec}{\mathrm{h}^{ - 1}}\,\,\left( {\sin \,\,\theta } \right)

is equal to

View answer and explanation

Correct option: D

\log \,\,\cot \,\,\frac{\theta }{2}

Option D (

\log \,\cot \,\frac{\theta }{2}

) is correct because the inverse hyperbolic secant simplifies to this logarithmic form using hyperbolic identities.

13 Easy

13 of 17

Application +4 | -1

The imaginary part of

\cosh (\alpha+1 \beta)

is:

View answer and explanation

Correct option: B

\sinh \alpha \sin \beta

Option B (

\sinh \alpha \sin \beta

) is correct because it is the imaginary part of the expansion of

\cosh (\alpha+i \beta)

using hyperbolic trigonometric identities.

14 Easy

14 of 17

Application +4 | -1

2\,\,{\tanh ^{ - 1}}\,\frac{1}{2}

is equal to

View answer and explanation

Correct option: C

\log \,\,3

Option C (

\log \,3

) is correct because using the identity for inverse hyperbolic tangent and simplifying leads to

\log \,3

15 Easy

15 of 17

Application +4 | -1

The imaginary part of

\sin ^2(x+i y)

is:

View answer and explanation

Correct option: D

\frac{1}{2} \sin 2 x \sinh 2 y

Option D (

\frac{{1}}{{2}} \sin 2 x \sinh 2 y

) is correct because it correctly identifies the imaginary part of the expression after applying trigonometric identities and hyperbolic functions.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

16 Easy

16 of 17

Application +4 | -1

\frac{\operatorname{cosech} x}{\sqrt{\operatorname{cosech}^2 x+1}}

equals:

View answer and explanation

Correct option: C

\operatorname{sech} x

Option C (sech x) is correct because it is the simplified form of the given expression using hyperbolic identities.

17 Easy

17 of 17

Application +4 | -1

The real part of

\sin \log (x+i y)

View answer and explanation

Correct option: C

\sin \left(\log \sqrt{x^2+y^2}\right) \cosh \left(\tan ^{-1} \frac{y}{x}\right)

Option C (

\sin (\log \sqrt{{x^2+y^2}}) \cosh (\tan ^{-1} \frac{{y}}{{x}})

) is correct because it correctly identifies the real part of the expression after applying trigonometric identities and logarithmic properties.

Revision notes and key takeaways

Review HYPERBOLIC FUNCTIONS in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 17 single-correct-answer MCQs on HYPERBOLIC FUNCTIONS.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

Continue Your Practice

📚 Switch Set (ALL)

Chapter Sets

ALL 17 MCQs EASY 15 MCQs HARD 1 MCQs MEDIUM 1 MCQs

0 /17

0✓ 0✗

0% Accuracy