Free Objective Test 01 Practice Test - 11th and 12th

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

A. x = 1 – y

x^{2} = 1 - y

x^{2} = 1 + y

y^{2} = 1 - x

SOLUTION

Solution : 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)

SOLUTION

Solution : 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

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

A. 10

3 π - 10

π + 6

2 π - 6

SOLUTION

Solution : 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 4

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

SOLUTION

Solution : 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 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}

B. 1

C. 0

- \frac{1}{2}

SOLUTION

Solution : 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 :

A. One solution

B. Infinite solutions

C. No solution

D. None of these

SOLUTION

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 $x y + y z + z x =$

A. 0

B. 1

C. 3

D. -3

SOLUTION

Solution : 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}

SOLUTION

Solution : 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}

SOLUTION

Solution : 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

SOLUTION

Solution : B

Let $x = t a n^{2} θ \Rightarrow θ = t a n^{- 1} \sqrt{x}$
Now, $\frac{1}{2} c o s^{- 1} (\frac{1 - x}{1 + x})$
$= \frac{1}{2} c o s^{- 1} (\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ})$
$= \frac{1}{2} c o s^{- 1} c o s 2 θ = \frac{2 θ}{2} = θ = t a n^{- 1} \sqrt{x}$ .

Question 11

$4 t a n^{- 1} \frac{1}{5} - t a n^{- 1} \frac{1}{239}$ is equal to

π

\frac{π}{2}

\frac{π}{3}

\frac{π}{4}

SOLUTION

Solution : D

Since $2 t a n^{- 1} x = t a n^{- 1} \frac{2 x}{1 - x^{2}}$
$∴ 4 t a n^{- 1} \frac{1}{5} = 2 [2 t a n^{- 1} \frac{1}{5}] = 2 t a n^{- 1} \frac{\frac{2}{5}}{1 - \frac{1}{25}}$
$= 2 t a n^{- 1} \frac{10}{24} = t a n^{- 1} \frac{\frac{20}{24}}{1 - \frac{100}{576}} = t a n^{- 1} \frac{120}{119}$
So, $4 t a n^{- 1} \frac{1}{5} - t a n^{- 1} \frac{1}{239} = t a n^{- 1} \frac{120}{119} - t a n^{- 1} \frac{1}{239}$
$= t a n^{- 1} \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} . \frac{1}{239}} = t a n^{- 1} \frac{(120 \times 239) - 119}{(119 \times 239) + 120}$
$\Rightarrow t a n^{- 1} 1 = \frac{π}{4}$ .

Question 12

If $3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}$ then x=

\sqrt{3}

\frac{1}{\sqrt{3}}

C. 1

D. None of these

SOLUTION

Solution : B

$3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}$
Putting $x = t a n θ$
$3 s i n^{- 1} (\frac{2 t a n θ}{1 + t a n^{2} θ}) - 4 c o s^{- 1} (\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ})$
$+ 2 t a n^{- 1} (\frac{2 t a n θ}{1 - t a n^{2} θ}) = \frac{π}{3}$
$\Rightarrow 3 s i n^{- 1} (s i n 2 θ) - 4 c o s^{- 1} (c o s 2 θ)$
$+ 2 t a n^{- 1} (t a n 2 θ) = \frac{π}{3}$
$\Rightarrow 3 (2 θ) - 4 (2 θ) + 2 (2 θ) = \frac{π}{3} \Rightarrow 6 θ - 8 θ + 4 θ = \frac{π}{3}$
$\Rightarrow θ = \frac{π}{6} \Rightarrow t a n^{- 1} x = \frac{π}{6} \Rightarrow x = t a n \frac{π}{6} = \frac{1}{\sqrt{3}}$

Question 13

The value of $s i n [2 t a n^{- 1} (\frac{1}{3})] + c o s [t a n^{- 1} (2 \sqrt{2})] =$

\frac{16}{15}

\frac{14}{15}

\frac{12}{15}

\frac{11}{15}

SOLUTION

Solution : B

$s i n [2 t a n^{- 1} (\frac{1}{3})] + c o s [t a n^{- 1} (2 \sqrt{2})]$
$= s i n [t a n^{- 1} \frac{\frac{2}{3}}{1 - \frac{1}{9}}] + c o s [t a n^{- 1} (2 \sqrt{2})]$
$= s i n [t a n^{- 1} \frac{3}{4}] + c o s [t a n^{- 1} 2 \sqrt{2}]$
$= t a n^{- 1} 2 \sqrt{2} = c o s^{- 1} \frac{1}{3}$
Also $t a n^{- 1} \frac{3}{4} = s i n^{- 1} \frac{3}{5}$
$= \frac{3}{5} + \frac{1}{3} = \frac{14}{15}$ .

Question 14

$s i n (\frac{1}{2} c o s^{- 1} \frac{4}{5}) =$

\frac{1}{\sqrt{10}}

- \frac{1}{\sqrt{10}}

\frac{1}{10}

- \frac{1}{10}

SOLUTION

Solution : A

Let $0 \leq x \leq π$
So, $c o s^{- 1} \frac{4}{5} = x \Rightarrow c o s x = \frac{4}{5}$ ....(i)
Now $s i n (\frac{1}{2} c o s^{- 1} \frac{4}{5}) = s i n (\frac{x}{2})$ ....(ii)
From (i), $c o s x = \frac{4}{5} \Rightarrow 1 - 2 s i n^{2} \frac{x}{2} = \frac{4}{5}$
$\Rightarrow 2 s i n^{2} \frac{x}{2} = 1 - \frac{4}{5} = \frac{1}{5} \Rightarrow s i n \frac{x}{2} = \sqrt{\frac{1}{10}}$ .

Question 15

The value of $c o s^{- 1} (c o s 12) - s i n^{- 1} (s i n 14)$ is

- 2

8 π - 26

4 π + 2

D. None of these

SOLUTION

Solution : D

$c o s^{- 1} (c o s 12) - s i n^{- 1} (s i n 14) \Rightarrow 4 π - 12 + 5 π - 14 = 9 π - 26$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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