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

Question 1

If x + $\frac{1}{x}$ = $2 c o s θ$ , then $x^{3}$ + $\frac{1}{x^{3}}$ =

$c o s 3 θ$

$2 c o s 3 θ$

$\frac{1}{2}$ $c o s 3 θ$

$\frac{1}{3}$ $c o s 3 θ$

SOLUTION

Solution : B

We have x + $\frac{1}{x}$ = $2 c o s θ$ ,

Now $x^{3}$ + $\frac{1}{x^{3}}$ = ${(x + \frac{1}{x})}^{3}$ - $3 x \frac{1}{x} (x + \frac{1}{x})$

= $(2 c o s θ)^{3} - 3 (2 c o s θ) = 8 c o s^{3} θ - 6 c o s θ$

= $2 (4 c o s^{3} θ - 3 c o s θ) = 2 c o s 3 θ$ .

Trick: Put x = 1 $\Rightarrow$ $θ = 0^{\circ}$ .

Then $x^{3}$ + $\frac{1}{x^{3}}$ = 2 = $2 c o s 3 θ$ .

Question 2

If $c o s (θ - α)$ = a, $s i n (θ - β)$ = b,

then $c o s^{2} (α - β)$ + 2ab $s i n (α - β)$ is equal to

$4 a^{2} b^{2}$

$a^{2} - b^{2}$

$a^{2} + b^{2}$

- $a^{2} b^{2}$

SOLUTION

Solution : C

We have $s i n (α - β)$ = $s i n (θ - β - ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ θ - α)$

= $s i n (θ - β) c o s (θ - α) - c o s (θ - β) s i n (θ - α)$

= ba - $\sqrt{1 - b^{2}} \sqrt{1 - a^{2}}$

And $c o s (α - β) = c o s (θ - β - ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ θ - α)$

= $c o s (θ - β) c o s (θ - α) + s i n (θ - β) s i n (θ - α)$

= $a \sqrt{1 - b^{2}} + b \sqrt{1 - a^{2}}$

∴ Given expression is $c o s^{2} (α - β) + 2 a b s i n (α - β)$

= $(a \sqrt{1 - b^{2}} + b \sqrt{1 - a^{2}})^{2}$ + 2ab{ $a b - \sqrt{1 - a^{2}} \sqrt{1 - b^{2}}$ }

= $a^{2} + b^{2}$ .

Trick: Put $α = 30^{\circ}$ , $β = 60^{\circ}$ and $θ = 90^{\circ}$ ,

then a = $\frac{1}{2}$ , b = $\frac{1}{2}$

∴ $c o s^{2} (α - β) + 2 a b s i n (α - β)$ = $\frac{3}{4}$ + $\frac{1}{2}$ × (- $\frac{1}{2}$ ) = $\frac{1}{2}$

which is given by option (c).

Question 3

If $a {cos}^{3} α + 3 a cos α {sin}^{2} α = m$ and

$a {sin}^{3} α + 3 a {cos}^{2} α sin α = n$ , Then $(m + n)^{\frac{2}{3}} + (m - n)^{\frac{2}{3}}$

is equal to

$2 a^{2}$

$2 a^{\frac{1}{3}}$

$2 a^{\frac{2}{3}}$

$2 a^{3}$

SOLUTION

Solution : C

Adding and subtracting the given relation,

we get $(m + n) = a {cos}^{3} α + 3 a cos α {sin}^{2} α$

+ $3 a {cos}^{2} α . sin α + a {sin}^{3} α$

= $a (cos α + sin α)^{3}$

and similarly $(m - n) = a (cos α - sin α)^{3}$

Thus, $(m + n)^{\frac{2}{3}} + (m - n)^{\frac{2}{3}}$
$= a^{\frac{2}{3}}$ { $(cos α + sin α)^{2} + (cos α - sin α)^{2}$ }
$= a^{\frac{2}{3}}$ $[2 ({cos}^{2} α + {sin}^{2} α)]$ = $2 a^{\frac{2}{3}}$ .

Question 4

The minimum value of $3 s i n θ + 4 c o s θ$ is

-5

SOLUTION

Solution : D

Minimum value of $(3 s i n θ + 4 c o s θ)$ is - $\sqrt{3^{2} + 4^{2}}$ i.e., -5.

Question 5

The minimum value of $9 t a n^{2} θ + 4 c o t^{2} θ$ is

SOLUTION

Solution : D

A.M. $\geq$ G.M.

$\Rightarrow$ $\frac{9 t a n^{2} θ + 4 c o t^{2} θ}{2}$ $\geq$ $\sqrt{4 c o t^{2} θ . 9 t a n^{2} θ}$

$\Rightarrow$ $9 t a n^{2} θ + 4 c o t^{2} θ$ $\geq$ 12

Therefore, the minimum value is 12.

Question 6

If sec $θ$ = $\frac{5}{4}$ , then tan $\frac{θ}{2}$ =

$\frac{1}{3}$

$\frac{3}{4}$

$\frac{1}{4}$

$\frac{1}{9}$

SOLUTION

Solution : A

Given that sec $θ$ = $\frac{5}{4}$

sec $θ$ = $\frac{1 + t a n^{2} (\frac{θ}{2})}{1 - t a n^{2} (\frac{θ}{2})}$ $\Rightarrow$ $\frac{5}{4}$ = $\frac{1 + t a n^{2} (\frac{θ}{2})}{1 - t a n^{2} (\frac{θ}{2})}$

$\Rightarrow$ $5 - 5 t a n^{2} (\frac{θ}{2}) = 4 + 4 t a n^{2} (\frac{θ}{2})$

$\Rightarrow$ $9 t a n^{2} (\frac{θ}{2})$ = 1 $\Rightarrow$ $t a n (\frac{θ}{2})$ = $\frac{1}{3}$ .

Question 7

$\frac{c o t^{2} 15^{\circ} - 1}{c o t^{2} 15^{\circ} + 1}$ =

$\frac{1}{2}$

$\frac{\sqrt{3}}{2}$

$\frac{3 \sqrt{3}}{4}$

$\sqrt{3}$

SOLUTION

Solution : B

$\frac{c o t^{2} 15^{\circ} - 1}{c o t^{2} 15^{\circ} + 1}$ = $\frac{\frac{c o s^{2} 15^{\circ}}{s i n^{2} 15^{\circ}} - 1}{\frac{c o s^{2} 15^{\circ}}{s i n^{2} 15^{\circ}} + 1}$

= $\frac{c o s^{2} 15^{\circ} - s i n^{2} 15^{\circ}}{c o s^{2} 15^{\circ} + s i n^{2} 15^{\circ}}$ = $c o s (30^{\circ})$ = $\frac{\sqrt{3}}{2}$

Question 8

If sin A = $\frac{4}{5}$ and cos B = - $\frac{12}{13}$ , where A and B lie in first

and third quadrant respectively, then cos(A + B) =

$\frac{56}{65}$

- $\frac{56}{65}$

$\frac{16}{65}$

- $\frac{16}{65}$

SOLUTION

Solution : D

We have sin A = $\frac{4}{5}$ and cos B = - $\frac{12}{13}$

Now, cos(A + B) = cos A cos B - sin A sin B

= $\sqrt{1 - \frac{16}{25}}$ $(- \frac{12}{13})$ - $\frac{4}{5}$ $(- \sqrt{1 - \frac{144}{169}})$

= - $\frac{3}{5}$ × $\frac{12}{13}$ - $\frac{4}{5}$ $(- \frac{5}{13})$ = - $\frac{16}{65}$

(Since A lies in first quadrant and B lies in third quadrant).

Question 9

If A + B = $225^{\circ}$ , then $\frac{c o t A}{1 + c o t A}$ . $\frac{c o t B}{1 + c o t B}$ =

-1

$\frac{1}{2}$

SOLUTION

Solution : D

$\frac{c o t A}{1 + c o t A}$ . $\frac{c o t B}{1 + c o t B}$ = $\frac{1}{(1 + t a n A) (1 + t a n B)}$

= $\frac{1}{t a n A + t a n B + 1 + t a n A t a n B}$

[ ∵ tan(A + B) = $t a n 225^{\circ}$ ]

$\Rightarrow$ tanA + tan B = 1 - tan A tan B

= $\frac{1}{1 - t a n A t a n B + 1 + t a n A t a n B}$ = $\frac{1}{2}$ .

Question 10

If tan A = - $\frac{1}{2}$ and tan B = - $\frac{1}{3}$ , then A + B =

[IIT 1967; MNR 1987; MP PET 1989]

$\frac{π}{4}$

$\frac{3 π}{4}$

$\frac{5 π}{4}$

None of these

SOLUTION

Solution : B

We have tan A = - $\frac{1}{2}$ and tan B = - $\frac{1}{3}$

Now, tan(A + B) = $\frac{t a n A + t a n B}{1 - t a n A t a n B}$ = $\frac{- \frac{1}{2} - \frac{1}{3}}{1 - \frac{1}{2} . \frac{1}{3}}$ = -1

$\Rightarrow$ tan(A + B) = tan $\frac{3 π}{4}$ . Hence, A + B = $\frac{3 π}{4}$ .

Question 11

If $t a n θ$ = $\frac{s i n α - c o s α}{s i n α + c o s α}$ , then $s i n α + c o s α$ and

$s i n α - c o s α$ must be equal to

$\sqrt{2} c o s θ, \sqrt{2} s i n θ$

$\sqrt{2} s i n θ, \sqrt{2} c o s θ$

$\sqrt{2} s i n θ, \sqrt{2} s i n θ$

$\sqrt{2} c o s θ, \sqrt{2} c o s θ$

SOLUTION

Solution : A

We have $t a n θ$ = $\frac{s i n α - c o s α}{s i n α + c o s α}$

$\Rightarrow$ $t a n θ$ = $\frac{s i n (α - \frac{π}{4})}{c o s (α - \frac{π}{4})}$ $\Rightarrow$ $t a n θ$ = $t a n (α - \frac{π}{4})$

$\Rightarrow$ $θ$ = $α$ - $\frac{π}{4}$ $\Rightarrow$ $α$ = $θ$ + $\frac{π}{4}$

Hence, $s i n α + c o s α$ = $s i n (θ + \frac{π}{4}) + c o s (θ + \frac{π}{4})$

= $\sqrt{2} c o s θ$

And $s i n α - c o s α$ = $s i n (θ + \frac{π}{4})$ - $c o s (θ + \frac{π}{4})$

= $\frac{1}{\sqrt{2}}$ $s i n θ$ + $\frac{1}{\sqrt{2}}$ $c o s θ$ - $\frac{1}{\sqrt{2}}$ $c o s θ$ + $\frac{1}{\sqrt{2}}$ $s i n θ$

= $\frac{2}{\sqrt{2}}$ $s i n θ$ = $\sqrt{2} s i n θ = \sqrt{2} s i n θ$ .

Question 12

If $c o s^{6} α + s i n^{6} α + K s i n^{2} 2 α = 1$ , then K=

$\frac{4}{3}$

$\frac{3}{4}$

$\frac{1}{2}$

SOLUTION

Solution : B

Since $c o s^{6} α + s i n^{6} α + K s i n^{2} 2 α = 1$ using formula $a^{3} + b^{3} = (a + b)^{3} - 3 a b (a + b)$ and on solving, we get the required result i.e., $K = \frac{3}{4}$

Question 13

$t a n α + 2 t a n 2 α + 4 t a n 4 α + 8 c o t 8 α$ =

[IIT 1988; MP PET 1991]

$t a n α$

$t a n 2 α$

$c o t α$

$c o t 2 α$

SOLUTION

Solution : C

= $t a n α + 2 t a n 2 α + 4 t a n 4 α + 8 c o t 8 α$
$t a n α + 2 t a n 2 α + 4 [\frac{s i n 4 α}{c o s 4 α} + 2 \frac{c o s 8 α}{s i n 8 α}]$

= $t a n α + 2 t a n 2 α + 4 [\frac{c o s 4 α c o s 8 α + s i n 4 α s i n 8 α + c o s 4 α c o s 8 α}{s i n 8 α c o s 4 α}]$

= $t a n α + 2 t a n 2 α + 4 [\frac{c o s 4 α + c o s 4 α c o s 8 α}{s i n 8 α c o s 4 α}]$

= $t a n α + 2 t a n 2 α + 4 [\frac{c o s 4 α (1 + c o s 8 α)}{c o s 4 α s i n 8 α}]$

= $t a n α + 2 t a n 2 α + 4 [\frac{2 c o s^{2} 4 α}{2 s i n 4 α c o s 4 α}]$

= $t a n α + 2 t a n 2 α + 4 c o t 4 α$
= $t a n α + 2 (t a n 2 α + 2 c o t 4 α)$
= $t a n α + 2 [\frac{s i n 2 α}{c o s 2 α} + 2 \frac{c o s 4 α}{s i n 4 α}]$
= $t a n α + 2 [\frac{c o s 2 α (1 + c o s 4 α)}{s i n 4 α c o s 2 α}]$
= $t a n α + 2 c o t 2 α = \frac{s i n α}{c o s α} + \frac{2 c o s 2 α}{s i n 2 α}$
= $\frac{c o s α + c o s α c o s 2 α}{s i n 2 α c o s α}$
= $\frac{1 + c o s 2 α}{s i n 2 α} = \frac{2 c o s^{2} α}{2 s i n α c o s α} = c o t α$

Question 14

$c o s^{2} 48^{\circ} - s i n^{2} 12^{\circ} =$

$\frac{\sqrt{5} - 1}{4}$

$\frac{\sqrt{5} + 1}{8}$

$\frac{\sqrt{3} - 1}{4}$

$\frac{\sqrt{3} + 1}{2 \sqrt{2}}$

SOLUTION

Solution : B

$c o s^{2} A - s i n^{2} B$ = cos(A + B). Cos(A - B)

$∴ c o s^{2} 48^{\circ} - s i n^{2} 12^{\circ} = c o s 60^{\circ} . c o s 36^{\circ}$

= $\frac{1}{2} (\frac{\sqrt{5} + 1}{4}) = \frac{\sqrt{5} + 1}{8}$

Question 15

If $\frac{3 π}{4} < α < π$ , then $\sqrt{c o s e c^{2} α + 2 c o t α}$ is equal to

$1 + c o t α$

$1 - c o t α$

$- 1 - c o t α$

$- 1 + c o t α$

SOLUTION

Solution : C

$\sqrt{c o s e c^{2} α + 2 c o t α} = \sqrt{1 + c o t^{2} α + 2 c o t α} = | 1 + c o t α |$
But $\frac{3 π}{4} < α < π \Rightarrow c o t α < - 1 \Rightarrow 1 + c o t α < 0$
Hence, $| 1 + c o t α | = - (1 + c o t α)$

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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