Inverse Trigonometric Functions(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

INVERSE TRIGONOMETRIC FUNCTIONS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

If $x = s i n (2 t a n^{- 1} 2), y = s i n (\frac{1}{2} t a n^{- 1} \frac{4}{3})$ then

x = 1 – y
$x^{2} = 1 - y$
$x^{2} = 1 + y$
$y^{2} = 1 - x$

Discuss Question

Answer: Option D. ->

y^{2} = 1 - x

:
D
Let

t a n^{- 1} 2 = α \Rightarrow x = s i n 2 α = \frac{4}{5} y = s i n (\frac{β}{2}) = \sqrt{\frac{1 - c o s β}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \frac{1}{\sqrt{5}}

Question 2.

$s i n^{- 1} | c o s x | - c o s^{- 1} | s i n x | = a$ has at least one solution if a $\in$

$0$
$[0, \frac{π}{2}]$
$[\frac{π}{2}, \frac{3 π}{2}]$
$(0, π)$

Discuss Question

Answer: Option A. ->

0

:
A

s i n^{- 1} | c o s x | - c o s^{- 1} | s i n x | = \frac{π}{2} - c o s^{- 1} | c o s x | - \frac{π}{2} + s i n^{- 1} | s i n x | = a \Rightarrow s i n^{- 1} | s i n x | - c o s^{- 1} | c o s x | = a \Rightarrow a = 0 \forall x

Question 3.

$c o s^{- 1} x = t a n^{- 1} \frac{\sqrt{1 - x^{2}}}{x}$ , then:

$X \in R$
$- \leq x \leq 1, x \neq 0$
$0 < x \leq 1$
$N o n e o f t h e s e$

Discuss Question

Answer: Option C. ->

0 < x \leq 1

:
C
Putting

θ = c o s^{- 1} x

in R.H.S., we have
R.H.S

= t a n^{- 1} \frac{\sqrt{1 - x^{2}}}{x} = t a n^{- 1} \frac{\sqrt{1 - c o s^{2} θ}}{c o s θ} (0 \leq θ \leq π) = t a n^{- 1} \frac{s i n θ}{c o s θ} = t a n^{- 1} t a n θ = θ = c o s^{- 1} x

when

- \frac{π}{2} < θ < \frac{π}{2}

i.e., when

0 \leq θ < \frac{π}{2}

i.e., when

0 < c o s θ \leq 1

i.e.,

0 < x \leq 1

Question 4.

${sin}^{- 1} (sin 10) =$

10
$3 π - 10$
$π + 6$
$2 π - 6$

Discuss Question

Answer: Option B. ->

3 π - 10

:
B

3 π < 10 < 3 π + \frac{π}{2} ∴ 10 rad \in Q_{3} ∴ - \frac{π}{2} < 3 π - 10 < 0 ∴ 3 π - 10 \in Q_{4} Also sin (3 π - 10) = sin 10 Hence {sin}^{- 1} (sin 10) = {sin}^{- 1} sin (3 π - 10) = 3 π - 10

Question 5.

The value of x for which $s i n (c o t^{- 1} (1 + x)) = c o s (t a n^{- 1} x)$ is:

$\frac{1}{2}$
1
0
$- \frac{1}{2}$

Discuss Question

Answer: Option D. ->

- \frac{1}{2}

:
D

s i n (c o t^{- 1} (1 + x)) = c o s (t a n^{- 1} x) \Rightarrow c o t^{- 1} (1 + x) = [\frac{π}{2} \pm t a n^{- 1} x] \Rightarrow \frac{π}{2} - t a n^{- 1} (1 + x) = \frac{π}{2} \pm t a n^{- 1} x \Rightarrow 1 + x = \pm x

x = - \frac{1}{2}

Which, on verification, satisfies the equation.

Question 6.

$3 c o s^{- 1} x - π x - \frac{π}{2} = 0$ has :

One solution
Infinite solutions
No solution
None of these

Discuss Question

Answer: Option A. -> One solution
:
A

3 c o s^{- 1} x - π x + \frac{π}{2}

Clearly graphs of

y = 3 c o s^{- 1} x

and

y = π x + \frac{π}{2}

in the domain of

c o s^{- 1} x

i.e., in [-1, 1] intersect only once, therefore there is only one solution of the given equation.

Question 7.

If $c o s^{- 1} x + c o s^{- 1} y + c o s^{- 1} z = 3 π$ , then $x y + y z + z x =$

0
1
3
-3

Discuss Question

Answer: Option C. -> 3
:
C
Given

c o s^{- 1} x + c o s^{- 1} y + c o s^{- 1} z = 3 π

∵ 0 \leq c o s^{- 1} x \leq π

∴ 0 \leq c o s^{- 1} y \leq π

and

0 \leq c o s^{- 1} z \leq π

Here

c o s^{- 1} x = c o s^{- 1} y = c o s^{- 1} z = π

\Rightarrow x = y = z = c o s π = - 1

∴ x y + y z + z x = (- 1) (- 1) + (- 1) (- 1) + (- 1) (- 1)

= 1 + 1 + 1 = 3

Question 8.

The value of $t a n (t a n^{- 1} \frac{1}{2} - t a n^{- 1} \frac{1}{3})$ =

$\frac{5}{6}$
$\frac{7}{6}$
$\frac{1}{6}$
$\frac{1}{7}$

Discuss Question

Answer: Option D. ->

\frac{1}{7}

:
D

t a n [t a n^{- 1} \frac{1}{2} - t a n^{- 1} \frac{1}{3}] = t a n [t a n^{- 1} \frac{\frac{1}{2} - \frac{1}{3}}{1 + \frac{1}{6}}]

= t a n t a n^{- 1} (\frac{1}{6} \times \frac{6}{7}) = \frac{1}{7}

Question 9.

$t a n [2 t a n^{- 1} (\frac{1}{5}) - \frac{π}{4}] =$

$\frac{17}{7}$
$- \frac{17}{7}$
$\frac{7}{17}$
$- \frac{7}{17}$

Discuss Question

Answer: Option D. ->

- \frac{7}{17}

:
D

t a n [2 t a n^{- 1} (\frac{1}{5}) - \frac{π}{4}] = t a n [t a n^{- 1} \frac{\frac{2}{5}}{1 - \frac{1}{25}} - t a n^{- 1} (1)]

= t a n [t a n^{- 1} \frac{5}{12} - t a n^{- 1} (1)] = t a n t a n^{- 1} (\frac{\frac{5}{12} - 1}{1 + \frac{5}{12}}) = - \frac{7}{17}

Question 10.

$\frac{1}{2} c o s^{- 1} (\frac{1 - x}{1 + x}) =$

$c o t^{- 1} \sqrt{x}$
$t a n^{- 1} \sqrt{x}$
$t a n^{- 1} x$
$c o t^{- 1} x$

Discuss Question

Answer: Option B. ->

t a n^{- 1} \sqrt{x}

:
B
Let

x = t a n^{2} θ \Rightarrow θ = t a n^{- 1} \sqrt{x}

Now,

\frac{1}{2} c o s^{- 1} (\frac{1 - x}{1 + x})

= \frac{1}{2} c o s^{- 1} (\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ})

= \frac{1}{2} c o s^{- 1} c o s 2 θ = \frac{2 θ}{2} = θ = t a n^{- 1} \sqrt{x}

Latest Videos

Chapter 1 - GLOBAL STEEL SCENARIO & INDIAN STEEL INDUSTRY Part 1 : (13-04-2024) INDUSTRY AND COMPANY AWARENESS (ICA)

Direction Sense Test Part 1 Reasoning (Hindi)

Chapter 1 - RMHP / OHP / OB & BP Part 1 : (14-02-2024) GPOE

Cube & Cuboid Part 1 Reasoning (Hindi)

Data Interpretation (DI) Basic Concept Reasoning (Hindi)

Counting Figures Part 1 Counting Of Straight Lines Reasoning (Hindi)

Sail E0 free webinar Webinars

Real Numbers Part 7 Class 10 Maths

Real Numbers Part 1 Class 10 Maths

Polynomials Part 1 Class 10 Maths

Latest Test Papers

DSP Combined 1 Free Revision Test SAIL JO E-0

GPOE / GPA Combined 1 Free Revision Test SAIL JO E-0

BSP Plant Specific Review Test SAIL JO E-0

BSL Combined 1 Free Revision Test SAIL JO E-0

GFM Combined 1 Free Revision Test SAIL JO E-0