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

Question 1

$1 + c o s 56^{\circ} + c o s 58^{\circ} - c o s 66^{\circ} =$

$2 c o s 28^{\circ} c o s 29^{\circ} c o s 33^{\circ}$

$4 c o s 28^{\circ} c o s 29^{\circ} c o s 33^{\circ}$

$4 c o s 28^{\circ} c o s 29^{\circ} s i n 33^{\circ}$

$2 c o s 28^{\circ} c o s 29^{\circ} s i n 33^{\circ}$

SOLUTION

Solution : C

$1 + c o s 56^{\circ} + c o s 58^{\circ} - c o s 66^{\circ} =$
= $2 c o s^{2} 28^{\circ} + 2 s i n 62^{\circ} . s i n 4^{\circ}$
= $2 c o s^{2} 28^{\circ} + 2 c o s 28^{\circ} . s i n 4^{\circ}$
= $2 c o s 28^{\circ} (c o s 28^{\circ} + c o s 86^{\circ})$
= $2 c o s 28^{\circ} .2 c o s 57^{\circ} . c o s 29^{\circ}$
= $4 c o s 28^{\circ} c o s 29^{\circ} s i n 33^{\circ}$
Aliter: Apply the condition identify

$c o s A + c o s B - c o s C = - 1 + 4 c o s \frac{A}{2} c o s \frac{B}{2} s i n \frac{c}{2}$
$[∴ 56^{\circ} + 58^{\circ} + 66^{\circ} = 180^{\circ}]$
We get the value of required expression equal to $4 c o s 28^{\circ} c o s 29^{\circ} s i n 33^{\circ}$ .

Question 2

$\sqrt{3} c o s e c 20^{\circ} - s e c 20^{\circ}$ =

$\frac{2 s i n 20^{\circ}}{s i n 40^{\circ}}$

$\frac{4 s i n 20^{\circ}}{s i n 40^{\circ}}$

SOLUTION

Solution : C

$\sqrt{3} c o s e c 20^{\circ} - s e c 20^{\circ} = \frac{\sqrt{3}}{s i n 20^{\circ}} - \frac{1}{c o s 20^{\circ}}$
$= \frac{\sqrt{3} c o s 20^{\circ} - s i n 20^{\circ}}{s i n 20^{\circ} c o s 20^{\circ}} = \frac{2 [\frac{\sqrt{3}}{2} c o s 20^{\circ} - \frac{1}{2} s i n 20^{\circ}]}{\frac{2}{2} s i n 20^{\circ} c o s 20^{\circ}}$
$= \frac{4 c o s (20^{\circ} + 30^{\circ})}{s i n 40^{\circ}} = \frac{4 c o s 50^{\circ}}{s i n 40^{\circ}} = \frac{4 s i n 40^{\circ}}{s i n 40^{\circ}} = 4$ .

Question 3

$s i n 20^{\circ} s i n 40^{\circ} s i n 60^{\circ} s i n 80^{\circ}$ =

- $\frac{3}{16}$

$\frac{5}{16}$

$\frac{3}{16}$

- $\frac{5}{16}$

SOLUTION

Solution : C

$s i n 20^{o} s i n 40^{\circ} s i n 60^{\circ} s i n 80^{\circ}$
= $\frac{1}{2} s i n 20^{\circ} s i n 60^{\circ} (2 s i n 40^{\circ} s i n 80^{\circ})$
= $\frac{1}{2} s i n 20^{\circ} s i n 60^{\circ} (c o s 40^{\circ} - c o s 120^{\circ})$
= $\frac{1}{2} . \frac{\sqrt{3}}{2} s i n 20^{\circ} (1 - 2 s i n^{2} 20^{\circ} + \frac{1}{2})$
= $\frac{\sqrt{3}}{4} s i n 20^{\circ} (\frac{3}{2} - 2 s i n^{2} 20^{\circ})$
= $\frac{\sqrt{3}}{8} (3 s i n 20^{\circ} - 4 s i n^{3} 20^{\circ})$
= $\frac{\sqrt{3}}{8} s i n 60^{\circ} = \frac{\sqrt{3}}{8} . \frac{\sqrt{3}}{2} = \frac{3}{16}$

Question 4

If $s i n x + c o s e c x = 2$ , then $s i n^{n} x + c o s e c^{n} x$ is equal to

$2^{n - 1}$

$2^{n - 2}$

SOLUTION

Solution : A

$s i n x + c o s e c x = 2 \Rightarrow (s i n x - 1)^{2} = 0 \Rightarrow s i n x = 1$
$∴ s i n^{n} x + c o s e c^{n} x = 1 + 1 = 2$

Question 5

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

sin $(\frac{A - B}{2})$

tan $(\frac{A - B}{2})$

cot $(\frac{A - B}{2})$

cos $(\frac{A - B}{2})$

SOLUTION

Solution : B

We have sin A = n sin B $\Rightarrow$ $\frac{n}{1}$ = $\frac{s i n A}{s i n B}$

$\Rightarrow$ $\frac{n - 1}{n + 1}$ $\frac{s i n A - s i n B}{s i n A + s i n B}$ = $\frac{2 c o s \frac{A + B}{2} s i n \frac{A - B}{2}}{2 s i n \frac{A + B}{2} c o s \frac{A - B}{2}}$

= tan $\frac{A - B}{2}$ cot $\frac{A + B}{2}$

$\Rightarrow$ $\frac{n - 1}{n + 1}$ tan $(\frac{A + B}{2})$ = tan $\frac{A - B}{2}$ .

Question 6

$\frac{s i n 2 A}{1 + c o s 2 A}$ . $\frac{c o s A}{1 + c o s A}$ =

tan $\frac{A}{2}$

cot $\frac{A}{2}$

sec $\frac{A}{2}$

cosec $\frac{A}{2}$

SOLUTION

Solution : A

$(\frac{s i n 2 A}{1 + c o s 2 A})$ $(\frac{c o s A}{1 + c o s A})$

= $\frac{2 s i n A c o s A}{2 c o s^{2} A}$ $\frac{c o s A}{1 + c o s A}$ = $\frac{s i n A}{1 + c o s A}$ = tan $\frac{A}{2}$

Question 7

If a tan $θ$ = b, then a cos 2 $θ$ + b sin 2 $θ$ =

-a

-b

SOLUTION

Solution : A

Given that tan $θ$ = $\frac{b}{a}$ .

Now, a cos 2 $θ$ + b sin 2 $θ$ = a $(\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ})$ + b $(\frac{2 t a n θ}{1 + t a n^{2} θ})$

Putting $t a n θ$ = $\frac{b}{a}$ , we get

= a $⎛ ⎝ \frac{1 - \frac{b^{2}}{a^{2}}}{1 + \frac{b^{2}}{a^{2}}} ⎞ ⎠$ + b $⎛ ⎝ \frac{2 \frac{b}{a}}{1 + \frac{b^{2}}{a^{2}}} ⎞ ⎠$ = a $(\frac{a^{2} - b^{2}}{a^{2} + b^{2}})$ + b $(\frac{2 b a}{a^{2} + b^{2}})$

= $\frac{1}{(a^{2} + b^{2})}$ $a^{3} - a b^{2} + 2 a b^{2}$ = $\frac{a (a^{2} + b^{2})}{a^{2} + b^{2}}$ = a.

Question 8

If cos 2 B = $\frac{c o s (A + C)}{c o s (A - C)}$ , then tan A, tan B, tan C are in

A.P.

G.P.

H.P.

None of these

SOLUTION

Solution : B

cos 2B = $\frac{c o s (A + C)}{c o s (A - C)}$ = $\frac{c o s A c o s C - s i n A s i n C}{c o s A c o s C + s i n A s i n C}$

$\Rightarrow$ $\frac{1 - t a n^{2} B}{1 + t a n^{2} B}$ = $\frac{1 - t a n A t a n C}{1 + t a n A t a n C}$

$\Rightarrow$ $1 + t a n^{2} B - t a n A t a n C - t a n A t a n C t a n^{2} B$

= $1 - t a n^{2} B + t a n A t a n C - t a n A t a n C t a n^{2} B$

$\Rightarrow$ $2 t a n^{2} B = 2 t a n A t a n C$ $\Rightarrow$ $t a n^{2} B$ = tan A tan C

Hence, tan A, tan B and tan C will be in G.P.

Question 9

If cos A = $\frac{\sqrt{3}}{2}$ , then tan 3A =

$\frac{1}{2}$

$\infty$

SOLUTION

Solution : D

We have cos A = $\frac{\sqrt{3}}{2}$ $\Rightarrow$ A = $30^{\circ}$

$\Rightarrow$ tan 3A = $t a n 90^{\circ}$ = $\infty$ .

Question 10

If tan $\frac{A}{2}$ = $\frac{3}{2}$ , then $\frac{1 + c o s A}{1 - c o s A}$ =

-5

$\frac{9}{4}$

$\frac{4}{9}$

SOLUTION

Solution : D

Given that tan $\frac{A}{2}$ = $\frac{3}{2}$ .

$\frac{1 + c o s A}{1 - c o s A}$ = $\frac{2 c o s^{2} \frac{A}{2}}{2 s i n^{2} \frac{A}{2}}$ = $c o t^{2} \frac{A}{2}$ = ${(\frac{2}{3})}^{2}$ = $\frac{4}{9}$

Question 11

If cos $θ$ = $\frac{3}{5}$ and cos $ϕ$ = $\frac{4}{5}$ , where $θ$ and $ϕ$ are

positive acute angles, then cos $\frac{θ - ϕ}{2}$ =

$\frac{7}{\sqrt{2}}$

$\frac{7}{5 \sqrt{2}}$

$\frac{7}{\sqrt{5}}$

$\frac{7}{2 \sqrt{5}}$

SOLUTION

Solution : B

We have cos $θ$ = $\frac{3}{5}$ and cos $ϕ$ = $\frac{4}{5}$ .

Therefore $c o s (θ - ϕ) = c o s θ c o s ϕ + s i n θ s i n ϕ$

= $\frac{3}{5}$ . $\frac{4}{5}$ + $\frac{4}{5}$ . $\frac{3}{5}$ = $\frac{24}{25}$

But $2 c o s^{2} (\frac{θ - ϕ}{2}) = 1 + c o s (θ - ϕ)$ = 1 + $\frac{24}{25}$ = $\frac{49}{50}$

∴ $c o s^{2} (\frac{θ - ϕ}{2})$ = $\frac{49}{50}$ . Hence, $c o s (\frac{θ - ϕ}{2})$ = $\frac{7}{5 \sqrt{2}}$ .

Question 12

The maximum value of $c o s^{2} (\frac{π}{3} - x) - c o s^{2} (\frac{π}{3} + x)$ is

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

$\frac{1}{2}$

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

$\frac{3}{2}$

SOLUTION

Solution : C

$c o s^{2} (\frac{π}{3} - x)$ - $c o s^{2} (\frac{π}{3} + x)$

= ${c o s (\frac{π}{3} - x) + c o s (\frac{π}{3} + x)}$ ${c o s (\frac{π}{3} - x) - c o s (\frac{π}{3} + x)}$

= ${2 c o s \frac{π}{3} c o s x}$ ${2 s i n \frac{π}{3} s i n x}$

= sin $\frac{2 π}{3}$ sin 2x = $\frac{\sqrt{3}}{2}$ sin 2x

Its maximum value is $\frac{\sqrt{3}}{2}$ , {- 1 $\leq$ sin 2x $\leq$ 1}.

Question 13

$\frac{1}{4}$ $[\sqrt{3} c o s 23^{\circ} - s i n 23^{\circ}]$ =

$c o s 43^{\circ}$

$c o s 7^{\circ}$

$c o s 53^{\circ}$

None of these

SOLUTION

Solution : D

$\frac{1}{4}$ $\sqrt{3} c o s 23^{\circ} - s i n 23^{\circ}$

= $\frac{1}{2}$ $c o s 30^{\circ} c o s 23^{\circ} - s i n 30^{\circ} s i n 23^{\circ}$

= $\frac{1}{2}$ $c o s (30^{\circ} + 23^{\circ})$ = $\frac{1}{2}$ $c o s 53^{\circ}$ .

Question 14

The circular wire of diameter 10cm is cut and placed along the circumference of a circle of diameter 1 metre. The angle subtended by the wire at the centre of the circle is equal to

$\frac{π}{4}$ radian

$\frac{π}{3}$ radian

$\frac{π}{5}$ radian

$\frac{π}{10}$ radian

SOLUTION

Solution : C

Given, diameter of circular wire = 10cm, therefore length of wire = $10 π$ .

Hence required angle = $\frac{a r c}{r a d i u s} = \frac{10 π}{50} = \frac{π}{5}$ radian.

Question 15

The value of ${sin}^{2} 5^{\circ} + {sin}^{2} 10^{\circ} + {sin}^{2} 15^{\circ} +$ ..............

$+ {sin}^{2} 85^{\circ} + {sin}^{2} 90^{\circ}$ is equal to

$7$

$8$

$9$

$\frac{19}{2}$

SOLUTION

Solution : D

Given expression is

${sin}^{2} 5^{\circ} + {sin}^{2} 10^{\circ} + {sin}^{2} 15^{\circ} + . . . . . . . . . . . . . . + {sin}^{2} 85^{\circ} + {sin}^{2} 90^{\circ}$ .

We know that $sin 90^{\circ} = 1 or {sin}^{2} 90^{\circ} = 1$

Similarly, $sin 45^{\circ} =$ $\frac{1}{\sqrt{2}}$ or ${sin}^{2} 45^{\circ} = \frac{1}{2}$ and the angles are in A.P. of 18 terms. We also know that

${sin}^{2} 85^{\circ} = [sin (90^{\circ} - 5^{\circ})]^{2} = {cos}^{2} 5^{\circ}$ .

Therefore from the complementary rule, we find ${sin}^{2} 5^{\circ} + {sin}^{2} 85^{\circ} = {sin}^{2} 5^{\circ} + {cos}^{2} 5^{\circ} = 1$

Therefore,

${sin}^{2} 5^{\circ} + {sin}^{2} 10^{\circ} + {sin}^{2} 15^{\circ} + . . . . . . . . . . . . . + {sin}^{2} 85^{\circ} + {sin}^{2} 90^{\circ}$

$= (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) + 1 + \frac{1}{2} = 9 \frac{1}{2}$ .

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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