Free Geometry Set I - 02 Practice Test - CAT

Question 1

The value of k, for which
(cos x + sin x) $^{2}$ + k sin x cos x - 1 = 0 is an identity, is

A. -1

B. -2

C. 0

D. 1

SOLUTION

Solution : B

(b) Given, (cos x + sin x) $^{2}$ + k sin x cos x - 1 = 0, $\forall$ x
   $\Rightarrow$ cos $^{2}$ x +sin $^{2}$ x + 2cos x sin x + k sin x cos x - 1 = 0, $\forall$ x
      $\Rightarrow$ (k + 2)cos x sin x = 0, $\forall$ x
   $\Rightarrow$ k = -2

Question 2

If $t a n (A - B) = 1, s e c (A + B) = \frac{2}{\sqrt{3}},$ then the smallest positive value of B is,

\frac{25}{24} π

\frac{19}{24} π

\frac{13}{24} π

\frac{11}{24} π

SOLUTION

Solution : B

tan(A-B)=1

$A - B = (2 n + 1) \frac{π}{4}$

n=0 since question is the smallest value possible

$A - B = \frac{π}{4}$

$A + B = 2 π - \frac{π}{6} = 11 \frac{π}{6}$

Subtract both the equations

$2 B = 11 \frac{π}{6} - \frac{π}{4}$

$2 B = (22 - 3) \frac{π}{12}$

$B = 19 \frac{π}{24}$

Question 3

If $\frac{s i n^{4} A}{a} + \frac{c o s^{4} A}{b} = \frac{1}{a + b},$ then the value of $\frac{s i n^{8} A}{a^{3}} + \frac{c o s^{8} A}{b^{3}}$ is equal to

\frac{1}{(a + b)^{3}}

\frac{a^{3} b^{3}}{(a + b)^{3}}

\frac{a^{2} b^{2}}{(a + b)^{2}}

D. None of these

SOLUTION

Solution : A

(a) It is given that $\frac{s i n^{4} A}{a} + \frac{c o s^{4} A}{b} = \frac{1}{a + b}$

$\Rightarrow \frac{(1 - c o s 2 A)^{2}}{4 a} + \frac{(1 + c o s 2 A)^{2}}{4 b} = \frac{1}{a + b}$

$\Rightarrow b (a + b) (1 - 2 c o s 2 A + c o s^{2} 2 A) + a (a + b) (1 + 2 c o s 2 A + c o s^{2} 2 A) = 4 a b$

$\Rightarrow {b (a + b) + a (a + b)} c o s^{2} 2 A + 2 (a + b) (a - b) c o s 2 A$

Question 4

A. 15

B. 6

C. 1

D. 0

SOLUTION

Solution : B

Question 5

D. both b) and c)

SOLUTION

Solution : D

Question 6

For a positive integer n, let
$f_{n} (θ)$ = $(tan \frac{θ}{2})$ (1 + sec $θ$ )(1 + sec 2 $θ$ )(1 + sec 4 $θ$ ) .........
(1 + sec 2 $^{n}$ $θ$ ). Then

f_{2} (\frac{π}{16})

= 1

f_{3} (\frac{π}{32})

= 1

f_{4} (\frac{π}{64})

= 1

D. All of above

SOLUTION

Solution : D

(a,b,c) $f_{n} (θ)$ = $\frac{sin (θ / 2)}{cos (θ / 2)}$ $[\frac{2 {cos}^{2} θ / 2}{cos θ} . \frac{2 {cos}^{2} θ}{cos 2 θ} . \frac{2 {cos}^{2} 2 θ}{cos 4 θ}]$
   Combine first two factors, $f_{n} (θ)$ = $\frac{s i n θ}{c o s θ}$ $[\frac{2 {cos}^{2} θ}{cos 2 θ} . \frac{2 {cos}^{2} 2 θ}{cos 4 θ}]$
   Again combine first two factors,
   $f_{n} (θ)$ = tan 2 $θ$ $[\frac{2 {cos}^{2} 2 θ}{c o s 4 θ} . . . . . . .]$ = tan (2 $^{n}$ $θ$ )
$∴$ $f_{2} (\frac{π}{16})$ = tan $\frac{4 π}{16}$ = tan $(\frac{π}{4})$ = 1
   $f_{3} (\frac{π}{32})$ = tan $\frac{8 π}{32}$ = tan $(\frac{π}{4})$ = 1
   $f_{4} (\frac{π}{64})$ = tan $\frac{16 π}{64}$ = tan $(\frac{π}{4})$ = 1
   $f_{5} (\frac{π}{128})$ = tan $\frac{32 π}{128}$ = tan $(\frac{π}{4})$ = 1

Question 7

Let n be a positive integer such that sin $\frac{π}{2^{n}}$ + cos $\frac{π}{2^{n}}$ = $\frac{\sqrt{n}}{2}$ . Then

A. 6

\leq

\leq

B. 4 < n

\leq

C. 4

\leq

n < 8

D. 4 < n < 8

SOLUTION

Solution : B

(b) sin $\frac{π}{2^{n}}$ + cos $\frac{π}{2^{n}}$ = $\frac{\sqrt{n}}{2}$
   $\Rightarrow$ $\sqrt{2}$ $(sin \frac{π}{2^{n}} . cos \frac{π}{4} + cos \frac{π}{2^{n}} . sin \frac{π}{4})$ = $\frac{\sqrt{n}}{2}$
   $\Rightarrow$ $\sqrt{2}$ sin $(\frac{π}{4} + \frac{π}{2^{n}})$ = $\frac{\sqrt{n}}{2}$
   Since sin $(\frac{π}{4} + \frac{π}{2^{n}}) \leq 1$
   $∴$ $\frac{\sqrt{n}}{2}$ $\leq$ $\sqrt{2}$ $\Rightarrow$ $\sqrt{2}$ $\leq$ 2 $\sqrt{2}$ $\Rightarrow$ n $\leq$ 8 .
  Again   $\frac{\sqrt{n}}{2}$ = $\sqrt{2}$ sin $(\frac{π}{4} + \frac{π}{2^{n}})$ > $\sqrt{2}$ . $\frac{1}{\sqrt{2}}$ = 1
$(∵ sin (\frac{π}{4} + \frac{π}{2^{n}}) sin \frac{π}{4})$
   $∴$ n > 4, Hence, 4 < n $\leq$ 8.

Question 8

If k = sin $\frac{π}{18}$ .sin $\frac{5 π}{18}$ .sin $\frac{7 π}{18}$ , then the numerical value of k is

\frac{1}{4}

\frac{1}{8}

\frac{1}{16}

D. None of these

SOLUTION

Solution : B

(b) We have k = sin $\frac{π}{18}$ .sin $\frac{5 π}{18}$ .sin $\frac{7 π}{18}$
= cos $(\frac{π}{2} - \frac{π}{18})$ cos $(\frac{π}{2} - \frac{5 π}{18})$ cos $(\frac{π}{2} - \frac{7 π}{18})$
= cos $\frac{π}{9}$ cos $\frac{2 π}{9}$ cos $\frac{4 π}{9}$ = $\frac{s i n 2^{3} \frac{π}{9}}{2^{3} s i n \frac{π}{9}}$ = $\frac{s i n \frac{8 π}{9}}{8 s i n \frac{π}{9}}$
= $\frac{s i n (π - \frac{π}{9})}{8 s i n \frac{π}{9}}$ = $\frac{1}{8}$

Question 9

A. 0

B. 1

C. -1

D. 2

SOLUTION

Solution : A

Question 10

If $α$ , $β$ , $γ$ , $δ$ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive
quantity k, then the value of 4 sin $\frac{α}{2}$ + 3 sin $\frac{β}{2}$ + 2 sin $\frac{γ}{2}$ + sin $\frac{δ}{2}$ is equal to

A. 2

\sqrt{1 - k}

\frac{1}{2}

\sqrt{1 + k}

C. 2

\sqrt{1 + k}

D. None of these

SOLUTION

Solution : C

(c) Given $α$ < $β$ < $γ$ < $δ$ and sin $α$ = sin $β$ = sin $γ$ = sin $δ$ = k. Also $α$ , $β$ , $γ$ , $δ$ are smallest positive angles satisfying above two conditions.
$∴$ We can take $β$ = $π$ - $α$ , $γ$ = 2 $π$ + $α$ , $δ$ = 3 $π$ - $α$
Given expression
= 4 sin $\frac{α}{2}$ + 3 sin $(\frac{π}{2} - \frac{α}{2})$ + 2 sin $(π + \frac{α}{2})$ + 3 sin $(\frac{3 π}{2} - \frac{α}{2})$
= 4 sin $\frac{α}{2}$ + 3 cos $\frac{α}{2}$ - 2 sin $\frac{α}{2}$ - cos $\frac{α}{2}$ = 2 $(sin \frac{α}{2} + cos \frac{α}{2})$
= 2 $\sqrt{{(sin \frac{1}{2} α + cos \frac{1}{2} α)}^{2}}$ = 2 $\sqrt{1 + sin α}$ = 2 $\sqrt{1 + k}$ .

Question 11

If (sec A + tan A)(sec B + tan B)(sec C + tan C) = (sec A - tan A)(sec B - tan B)(sec C - tan C), then each side is equal to

A. 1

B. -1

C. 0

D. None of these

SOLUTION

Solution : A

(a,b) If L=M, then L $^{2}$ = LM or ML = M $^{2}$
Both LM = ML = 1 as sec $^{2}$ A - tan $^{2}$ A = 1
$∴$ L $^{2}$ = M $^{2}$ = 1.

Question 12

If tan $θ$ = $\sqrt{\frac{3}{2}}$ , the sum of the infinite series
1 + 2(1 - cos $θ$ ) + 3(1 - cos $θ$ ) $^{2}$ + 4(1 - cos $θ$ ) $^{3}$ +........ $\infty$ is

\frac{2}{3}

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

\frac{5}{2 \sqrt{2}}

\frac{5}{2}

SOLUTION

Solution : D

(d)  1 + 2(1 - cos $θ$ ) + 3(1 - cos $θ$ ) $^{2}$ + 4(1 - cos $θ$ ) $^{3}$ +........ to $\infty$
   = (1 - x) $^{- 1}$ , [Let (1 - cos $θ$ ) = x]
   $∴$ Series = (1 - 1 + cos $θ$ ) $^{- 2}$ = sec $^{2}$ $θ$
   = (1 + tan $^{2}$ $θ$ ) = 1 + $\frac{3}{2}$ = $\frac{5}{2}$

Question 13

If $x_{1}, x_{2}, x_{3}, . . . . . ., x_{n}$ are in A.P. whose common difference is a, then the value of sin $α$ (sec $x_{1}$ sec $x_{2}$ + sec $x_{2}$ sec $x_{3}$ +....... + sec $x_{n - 1}$ , sec $x_{n}$ )=

\frac{sin (n - 1) α}{cos x_{1} cos x_{n}}

\frac{sin n α}{cos x_{1} cos x_{n}}

C. sin (n - 1)

α

cos x

_{1}

cos x

_{n}

D. sin n

α

cosx

_{1}

cosx

_{n}

SOLUTION

Solution : A

(a) We have sin $α$ sec $x_{1}$ sec $x_{2}$ + sin $α$ sec $x_{2}$ sec $x_{3}$ +....... + sin $α$ sec $x_{n - 1}$ , sec $x_{n}$ )
= $\frac{sin (x_{2} - x_{1})}{cos x_{1} cos x_{2}}$ + $\frac{sin (x_{3} - x_{2})}{cos x_{2} cos x_{3}}$ + ........ + $\frac{sin (x_{n} - x_{n - 1})}{cos x_{n - 1} cos x_{n}}$
   = tan x $_{2}$ - tan x $_{1}$ + tan x $_{3}$ - tan x $_{2}$ + .......... + tan x $_{n}$ - tan x $_{n -}$
   = tan x $_{n}$ - tan x $_{1}$ = $\frac{sin (x_{n} - x_{n - 1})}{cos x_{n - 1} cos x_{1}}$ = $\frac{sin (n - 1) α}{cos x_{n} cos x_{1}}$ {( $∵$ x $_{n}$ = x $_{1}$ + (n - 1) $α$ )}

Question 14

Let $α$ , $β$ be such that $π$ < ( $α$ - $β$ ) < 3 $π$ . If sin $α$ + sin $β$ = - $\frac{21}{65}$ and cos $α$ + cos $β$ = - $\frac{27}{65}$ , then the value of \
cos $\frac{α - β}{2}$

\frac{- 6}{65}

\frac{3}{\sqrt{130}}

\frac{6}{65}

D. -

\frac{3}{\sqrt{130}}

SOLUTION

Solution : D

(d) sin $α$ + sin $β$ = - $\frac{21}{65}$ , cos $α$ + cos $β$ = - $\frac{27}{65}$
Now (sin $α$ + sin $β$ ) $^{2}$ + (cos $α$ + cos $β$ ) $^{2}$ = ${(- \frac{21}{65})}^{2}$ + ${(- \frac{27}{65})}^{2}$
$\Rightarrow$ 2 + 2 sin $α$ sin $β$ + 2 cos $α$ cos $β$ = $\frac{441}{65^{2}}$ + $\frac{729}{65^{2}}$
$\Rightarrow$ 2 + 2[cos ( $α$ - $β$ )] = $\frac{1170}{(65)^{2}}$ $\Rightarrow$ 2.2 cos $^{2}$ $(\frac{α + β}{2}) = \frac{1170}{(65)^{2}}$
$\Rightarrow$ cos $(\frac{α - β}{2})$ = $\frac{3 \sqrt{130}}{130}$ = $\frac{3}{\sqrt{130}}$
Therefore cos $(\frac{α - β}{2})$ = $\frac{- 3}{\sqrt{130}}$ , ${$ $∵$ $\frac{π}{2}$ < $\frac{α - β}{2}$ < $\frac{3 π}{2}$ $}$

Question 15

A. 1

B. 2

C. 0

SOLUTION

Solution : C

Option C is the correct answer.

Question 16

Let A $_{0}$ A $_{1}$ A $_{2}$ A $_{3}$ A $_{4}$ A $_{5}$ be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A $_{0}$ A $_{1}$ , A $_{0}$ A $_{2}$ and A $_{0}$ A $_{4}$ is

\frac{3}{4}

B. 3

\sqrt{3}

C. 3

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

SOLUTION

Solution : C

(c) Each triangle is an equilateral triangle

Hence A $_{0}$ A $_{1}$ = 1
     A $_{0}$ A $_{0}^{2}$ = A $_{0}$ A $_{1}^{2}$ + A $_{1}$ A $_{0}^{2}$ - 2A $_{0}$ A $_{1}$ A $_{1}$ A $_{2}$ cos 120 $^{\circ}$
   = 1 + 1 - 2.1.1(- $\frac{1}{2}$ ) = 3
   $\Rightarrow$ A $_{0}$ A $_{2}$ = $\sqrt{3}$ = A $_{0}$ A $_{4}$
   $∴$ A $_{0}$ A $_{1}$ $\times$ A $_{0}$ A $_{2}$ $\times$ A $_{0}$ A $_{4}$ = 1. $\sqrt{3}$ . $\sqrt{3}$

Question 17

3 $[s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)]$ - $[s i n^{6} (\frac{π}{2} + α) + s i n^{6} (5 π - α)]$ =

sin 4 $α$ + sin 6 $α$

SOLUTION

Solution : B

(b) 3 $[s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)]$ - $[s i n^{6} (\frac{π}{2} + α) + s i n^{6} (5 π - α)]$
$= 3 (- c o s α)^{4} + (- s i n α)^{4} - 2 c o s^{6} α + s i n^{6} α$
$= 3 (c o s^{2} α + s i n^{2} α)^{2} - 2 s i n^{2} α c o s^{2} α - 2 (c o s^{2} α + s i n^{2} α)^{3} - 3 s i n^{2} α c o s^{2} α (c o s^{2} α + s i n^{2} α)$
$= 3 - 6 s i n^{2} α c o s^{2} α - 2 + 6 s i n^{2} α c o s^{2} α = 3 - 2 = 1$
Trick: Put $α = 0$ , the value of expressions remains 1 i.e., it is independent of $α$

Question 18

If $α \in (0, \frac{π}{2})$ then $\sqrt{x^{2} + x} + \frac{t a n^{2} α}{\sqrt{x^{2} + x}}$ is always greater than or equal to

2 t a n α

B. 1

C. 2

s e c^{2} α

SOLUTION

Solution : A

(a) $\sqrt{x^{2} + x} + \frac{t a n^{2} α}{\sqrt{x^{2} + x}} \geq 2 t a n α$ $(A . M \leq G . M)$ .

Question 19

The maximum value of cos $α_{1}$ .cos $α_{2}$ ........cos $α_{n}$ , under the restrictions 0 $\leq$ $α_{1}$ , $α_{2}$ ,......... $α_{n}$ $\leq$ $\frac{π}{2}$ and
cot $α_{1}$ .cot $α_{2}$ ........cot $α_{n}$ = 1 is

\frac{1}{2^{n / 2}}

\frac{1}{2^{n}}

\frac{1}{2 n}

D. 1

SOLUTION

Solution : A

(a) Here (cot $α_{1}$ ).(cot $α_{2}$ )....(cot $α_{n}$ )= 1
$∴$ cos $α_{1}$ .cos $α_{2}$ ........cos $α_{n}$ = sin $α_{1}$ .sin $α_{2}$ ........sin $α_{n}$
Now, (cos $α_{1}$ .cos $α_{2}$ ........cos $α_{n}$ ) $^{2}$
= (cos $α_{1}$ .cos $α_{2}$ ........cos $α_{n}$ ) (cos $α_{1}$ .cos $α_{2}$ ........cos $α_{n}$ )
= (cos $α_{1}$ .cos $α_{2}$ ........cos $α_{n}$ ) (sin $α_{1}$ .sin $α_{2}$ ........sin $α_{n}$ )
= $\frac{1}{2^{n}}$ sin 2 $α_{1}$ .sin 2 $α_{2}$ ........sin 2 $α_{n}$
But each of sin 2 $α_{i}$ $\leq$ 1
(cos $α_{1}$ .cos $α_{2}$ ........cos $α_{n}$ ) $^{2}$ $\leq$ $\frac{1}{2^{n}}$
But each of cos $α_{i}$ , is positive.
$∴$ cos $α_{1}$ .cos $α_{2}$ ........cos $α_{n}$ $\leq$ $\sqrt{\frac{1}{2^{n}}}$ = $\frac{1}{2^{n / 2}}$ .

Question 20

If $A = s i n^{8} θ + c o s^{14} θ$ , then for all real value of $θ$

A \geq 1

0 < A \leq 1

1 < 2 A \leq 3

D. None of these

SOLUTION

Solution : B

Option B is the correct answer.

Question 21

If $θ$ is an acute angle and $s i n θ = \frac{p - 6}{8 - p},$ then p must satisfy

6 \leq p < 8

6 \leq p < 7

3 \leq p \leq 4

4 \leq p < 7

SOLUTION

Solution : B

(b) $θ$ is an acute angle so $0^{\circ} \leq θ < 90^{\circ}$
$∴ 0 \leq \frac{p - 6}{8 - p} < 1$ => $0 \leq (p - 6) < (8 - p) => 6 \leq p < 7$ .

Question 22

$I f A, B, C$ be the angle of a triangle, the $\sum \frac{c o t A + c o t B}{t a n A + t a n B}$

A. 1

B. 2

C. -1

D. -2

SOLUTION

Solution : A

Option A is the correct answer.

Question 23

If $α + β = \frac{π}{2}$ and $β + α = α$ , then $t a n α e q u a l s$

2 (t a n β + t a n γ)

t a n β + t a n γ

t a n β + 2 t a n γ

2 t a n β + t a n γ

SOLUTION

Solution : C

(c) $α + β = \frac{π}{2} => t a n β = c o t α$
      $t a n (β + γ) = t a n α => t a n α = \frac{t a n β + t a n γ}{1 - t a n β t a n γ}$
      $=> t a n α = \frac{c o t α + t a n γ}{1 - c o t α t a n γ}$
      $=> t a n α - t a n γ = c o t α + t a n γ$
      $=> t a n α = t a n β + 2 t a n γ$

Question 24

If $∣ ∣ ∣ a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c) ∣ ∣ ∣ \leq \frac{1}{2} k,$ then $k^{2}$ is equal to

b^{2} + (a - c)^{2}

a^{2} + (b - c)^{2}

c^{2} + (a - b)^{2}

D. None of these

SOLUTION

Solution : A

(a) $a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c)$
       = $\frac{1}{2} [- a c o s 2 θ + b s i n 2 θ + c c o s 2 θ]$
       = $\frac{1}{2} [b s i n 2 θ - (a - c) c o s 2 θ]$
       $∵ | b s i n 2 θ - (a - c) c o s 2 θ | \leq \sqrt{b^{2} + (a - c)^{2}}$
       $∴ ∣ ∣ ∣ \frac{1}{2} b s i n 2 θ - (a - c) c o s 2 θ ∣ ∣ ∣ \leq \frac{1}{2} \sqrt{b^{2} + (a - c)^{2}}$
       $∴ K = \sqrt{b^{2} + (a - c)^{2}}$

Question 25

If for all real values of $x, \frac{4 x^{2} + 1}{64 x^{2} - 96 x s i n α + 5} < \frac{1}{32}$ , then $α$ lies in the interval.

(0, \frac{π}{3})

(\frac{π}{3}, \frac{2 π}{3})

(\frac{2 π}{3}, π)

(\frac{4 π}{3}, \frac{5 π}{3})

SOLUTION

Solution : D

Option D is the correct answer.

Free Geometry Set I - 02 Practice Test - CAT

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

Question 16

SOLUTION

Question 17

SOLUTION

Question 18

SOLUTION

Question 19

SOLUTION

Question 20

SOLUTION

Question 21

SOLUTION

Question 22

SOLUTION

Question 23

SOLUTION

Question 24

SOLUTION

Question 25

SOLUTION

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