Free Objective Test 02 Practice Test - 11th and 12th

Question 1

The value of $t a n^{2} (s e c^{- 1} 3) + c o t^{2} (c o s e c^{- 1} 4)$ is

A. 20

B. 21

C. 23

D. 25

SOLUTION

Solution : C

Let $S e c^{- 1} 3 = α, c o s e c^{- 1} 4 = β \Rightarrow t a n^{2} α + c o t^{2} β = 9 - 1 + 16 - 1 = 23$

Question 2

The value of $c o s^{- 1} (c o s 10)$ =

10

4 π - 10

2 π + 10

2 π - 10

SOLUTION

Solution : B

$c o s^{- 1} (c o s 10) = c o s^{- 1} (c o s (4 π - 10)) = 4 π - 10$

Question 3

Find maximum value of x for which $2 t a n^{- 1} x + c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$ is independent of x.

A. 0

B. 2

C. 3

D. 1

SOLUTION

Solution : A

Let $x = t a n θ, - \frac{π}{2} < θ < \frac{π}{2} 2 θ + c o s^{- 1} c o s 2 θ$
$(i) 0 \leq 2 θ < \frac{π}{2} .2 c o s^{- 1} c o s 2 θ = 2 θ$
So given expression = $4 θ = 4 t a n^{- 1} x$
$(i i) - \frac{π}{2} < θ \leq 0 \Rightarrow - π < 2 θ \leq 0 c o s^{- 1} c o s 2 θ = - 2 θ$
So, given expression becomes independent of $θ$
For $- \frac{π}{2} < θ \leq 0 \Rightarrow - \infty < x \leq 0$

Question 4

If $2 t a n^{- 1} x = s i n^{- 1} \frac{2 x}{1 + x^{2}}$ , then:

x \in R

x \geq 1

x \leq - 1

- 1 \leq x \leq 1

SOLUTION

Solution : D

Putting $θ = t a n^{- 1} x, - \frac{π}{2} < θ < \frac{π}{2}$ , we have
R.H.S. $s i n^{- 1} \frac{2 t a n θ}{1 + t a n^{2} θ} = s i n^{- 1} s i n 2 θ = 2 θ = 2 t a n^{- 1} x$
When $- \frac{π}{2} \leq 2 θ \leq \frac{π}{2}$ i.e., $- \frac{π}{4} \leq θ \leq \frac{π}{4}$
i.e., $- 1 \leq x = t a n θ \leq 1$ .

Question 5

The value of ${cos}^{- 1} (cos \frac{5 π}{3}) + {sin}^{- 1} (sin \frac{5 π}{3})$ is

0

\frac{π}{2}

\frac{2 π}{3}

\frac{10 π}{3}

SOLUTION

Solution : A

${cos}^{- 1} (cos \frac{5 π}{3}) + {sin}^{- 1} (sin \frac{5 π}{3}) = {cos}^{- 1} [cos (2 π - \frac{π}{3})] + {sin}^{- 1} [sin (2 π - \frac{π}{3})] = \frac{π}{3} - \frac{π}{3} = 0.$

Question 6

If $s i n^{- 1} \frac{3}{5} + c o s^{- 1} (\frac{12}{13}) = s i n^{- 1} C, t h e n C =$

\frac{65}{56}

\frac{24}{65}

\frac{16}{65}

\frac{56}{65}

SOLUTION

Solution : D

$G i v e n s i n^{- 1} C = s i n^{- 1} \frac{3}{5} + c o s^{- 1} \frac{12}{13} \Rightarrow C = s i n (s i n^{- 1} \frac{3}{5} + c o s^{- 1} \frac{12}{13}) U s i n g s i n^{(} A + B) = s i n A c o s B + c o s A s i n B \Rightarrow C = \frac{3}{5} \times \frac{12}{13} + \sqrt{1 - \frac{9}{25}} \sqrt{1 - \frac{144}{169}} \Rightarrow C = \frac{56}{65} .$

Question 7

The set of values of p for which $x^{2} - p x + s i n^{- 1} (s i n 4) > 0$ for all real x is given by :

(- 4, 4)

(- \infty, - 4) \cup (4, \infty)

ϕ

D. None of these

SOLUTION

Solution : C

$x^{2} - p x + s i n^{- 1} (s i n 4) > 0$ for all real x.
$\Rightarrow x^{2} - p x + s i n^{- 1} (s i n 4) > 0$
$\Rightarrow x^{2} - p x + (π - 4) > 0 \forall x ϵ R$
$\Rightarrow D = p^{2} - 4 (π - 4) < 0$ $\Rightarrow p^{2} + 16 - 4 π < 0$
Since $16 - 4 π > 0, p^{2} + 16 - 4 π$ cannot be negative for any value of $p ϵ R$ .
$∴$ Set of values of $p = ϕ$

Question 8

Complete solution set of $t a n^{2} (s i n^{- 1} x) > 1$ is :

(- 1 - \frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, 1)

(- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) - {0}

(- 1, 1) - {0}

D. None of these

SOLUTION

Solution : A

$t a n^{2} (s i n^{- 1} x) > 1$
$\Rightarrow t a n (s i n^{- 1} x) < - 1$ or $t a n (s i n^{- 1} x) > 1$
$\Rightarrow - \frac{π}{2} < s i n^{- 1} x < - \frac{π}{4}$ or $\frac{π}{4} < s i n^{- 1} x < \frac{π}{2}$
$\Rightarrow - 1 < x < \frac{- 1}{\sqrt{2}}$ or $\frac{1}{\sqrt{2}} < x < 1$
$\Rightarrow x ϵ (- 1 - \frac{- 1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, 1)$

Question 9

Total number of positive integral value of `n' such that the equations $c o s^{- 1} x + (s i n^{- 1} y)^{2} = \frac{n π^{2}}{4}$ and $(s i n^{- 1} y)^{2} - c o s^{- 1} x = \frac{π^{2}}{16}$ are consistent, is equal to :

A. 1

B. 4

C. 3

D. 2

SOLUTION

Solution : A

$(s i n^{- 1} y)^{2} + c o s^{- 1} x = \frac{n π^{2}}{4}$
$(s i n^{- 1} y)^{2} - c o s^{- 1} x = \frac{π^{2}}{16}$
$\Rightarrow (s i n^{- 1} y)^{2} = \frac{(4 n + 1) π^{2}}{32}, c o s^{- 1} x = \frac{π^{2} (4 n - 1)}{32}$
$\Rightarrow 0 \leq \frac{(4 n + 1)}{32} π^{2} \leq \frac{π^{2}}{4}, 0 \leq \frac{(4 n - 1)}{32} π^{2} \leq π$
$\Rightarrow - \frac{1}{4} \leq n \leq \frac{7}{4}, \frac{1}{4}, \leq n \leq \frac{8}{π} + \frac{1}{4}$

Question 10

The value of $s i n^{- 1} {c o t (s i n^{- 1} \sqrt{(\frac{2 - \sqrt{3}}{4})} + c o s^{- 1} \frac{\sqrt{12}}{4} + s e c^{- 1} \sqrt{2})}$ is:

0

\frac{π}{4}

\frac{π}{6}

\frac{π}{2}

SOLUTION

Solution : A

$c o s^{- 1} \frac{\sqrt{12}}{4} = c o s^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{6}, s e c^{- 1} \sqrt{2} = \frac{π}{4}$
$s i n^{- 1} \sqrt{\frac{2 - \sqrt{3}}{4}} = s i n^{- 1} \sqrt{\frac{4 - 2 \sqrt{3}}{8}} = s i n^{- 1} \frac{\sqrt{3} - 1}{2 \sqrt{2}} = \frac{π}{12}$
$∴$ $s i n^{- 1} \sqrt{\frac{2 - \sqrt{3}}{4}} + c o s^{- 1} \frac{\sqrt{12}}{4} + s e c^{- 1} \sqrt{2} = \frac{π}{2}$
and $s i n^{- 1} (c o t \frac{π}{2}) = s i n^{- 1} 0 = 0$

Question 11

If $θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}$ , where $a_{1}, a_{2}, a_{3}, \dots a_{n}$ are in A.P. with common difference d, then $t a n θ =$

\frac{(n - 1) d}{1 + a_{1} a_{n}}

\frac{(n - 1) d}{a_{1} + a_{n}}

\frac{a_{n} - a_{1}}{a_{n} + a_{1}}

\frac{n d}{1 + a_{1} a_{n}}

SOLUTION

Solution : A

$θ = t a n^{- 1} \frac{a_{2} - a_{1}}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{a_{3} - a_{2}}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{a_{n} - a_{n - 1}}{1 + a_{n - 1} a_{n}}$
$= (t a n^{- 1} a_{2} - t a n^{- 1} a_{1}) + (t a n^{- 1} a_{3} - t a n^{- 1} a_{2}) + \dots + t a n^{- 1} \frac{a_{n} - a_{n - 1}}{1 + a_{n - 1} a_{n}} + \dots + (t a n^{- 1} a_{n} - t a n^{- 1} a_{n - 1})$
$= t a n^{- 1} a_{n} - t a n^{- 1} a_{1}$
$= t a n^{- 1} \frac{a_{n} - a_{1}}{1 + a_{1} a_{n}} = t a n^{- 1} \frac{(n - 1) d}{1 + a_{1} a_{n}}$
$∴ t a n θ = \frac{(n - 1) d}{1 + a_{1} a_{n}}$

Question 12

$c o s [c o s^{- 1} (\frac{- 1}{7}) + s i n^{- 1} (\frac{- 1}{7})] =$

\frac{- 1}{3}

0

\frac{1}{3}

\frac{4}{9}

SOLUTION

Solution : B

$c o s {c o s^{- 1} (\frac{- 1}{7}) + s i n^{- 1} (\frac{- 1}{7})} = c o s \frac{π}{2} = 0$

Question 13

If $c o s^{- 1} \sqrt{p} + c o s^{- 1} \sqrt{1 - p} + c o s^{- 1} \sqrt{1 - q} = \frac{3 π}{4}$ , then the value of q is

1

\frac{1}{\sqrt{2}}

\frac{1}{3}

\frac{1}{2}

SOLUTION

Solution : D

Let $α = c o s^{- 1} \sqrt{p} : β = c o s^{- 1} \sqrt{1 - p}$
and $γ = c o s^{- 1} \sqrt{1 - q}$ or $c o s α = \sqrt{p} : c o s β = \sqrt{1 - p}$
and $c o s γ = \sqrt{1 - q} .$
Therefore $s i n α = \sqrt{1 - p}, s i n β = \sqrt{p}$ and $s i n γ = \sqrt{q} .$
The given equation may be written as $α + β + γ = \frac{3 π}{4} o r α + β = \frac{3 π}{4} - γ$ or $c o s (α + β) = c o s (\frac{3 π}{4} - γ)$
$\Rightarrow c o s α c o s β - s i n α s i n β$ = $c o s {π - (\frac{π}{4} + γ)} = - c o s (\frac{π}{4} + γ)$
$\Rightarrow \sqrt{p} \sqrt{1 - p} - \sqrt{1 - p} \sqrt{p} = - (\frac{1}{\sqrt{2}} \sqrt{1 - q} - \frac{1}{\sqrt{2}} . \sqrt{q})$
$\Rightarrow 0 = \sqrt{1 - q} - \sqrt{q} \Rightarrow 1 - q = q \Rightarrow q = \frac{1}{2}$ .

Question 14

$c o t^{- 1} [(c o s α)^{\frac{1}{2}}] - t a n^{- 1} [(c o s α)^{\frac{1}{2}}] = x$ , then $s i n x =$

t a n^{2} (\frac{α}{2})

c o t^{2} (\frac{α}{2})

t a n α

c o t (\frac{α}{2})

SOLUTION

Solution : A

$t a n^{- 1} [\frac{1}{\sqrt{c o s α}}] - t a n^{- 1} [\sqrt{c o s α}] = x$
$\Rightarrow t a n^{- 1} ⎡ ⎢ ⎣ \frac{\frac{1}{\sqrt{c o s α}} - \sqrt{c o s α}}{1 + \frac{\sqrt{c o s α}}{\sqrt{c o s α}}} ⎤ ⎥ ⎦ = x \Rightarrow t a n x = \frac{1 - c o s α}{2 \sqrt{c o s α}}$
$∴ s i n x = \frac{1 - c o s α}{1 + c o s α} = \frac{2 s i n^{2} \frac{α}{2}}{2 c o s^{2} \frac{α}{2}} = t a n^{2} (\frac{α}{2})$ .

Question 15

If $t a n^{- 1} (x) + t a n^{- 1} (y) + t a n^{- 1} (z) = π$ , then $\frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x} =$

0

1

\frac{1}{x y z}

x y z

SOLUTION

Solution : B

$t a n^{- 1} (x) + t a n^{- 1} (y) + t a n^{- 1} (z) = π$
$\Rightarrow t a n^{- 1} x + t a n^{- 1} y = π - t a n^{- 1} z$
$\Rightarrow \frac{x + y}{1 - x y} = - z \Rightarrow x + y = - z + x y z$
$\Rightarrow x + y + z = x y z$
Dividing by xyz, we get
$\frac{1}{y z} + \frac{1}{x z} + \frac{1}{x y} = 1.$
Note: Students should remember this question as a formula.

Free Objective Test 02 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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