Geometry Set I(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

GEOMETRY SET I MCQs

Total Questions : 110 | Page 8 of 11 pages

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

$\frac{3}{4}$
3 $\sqrt{3}$
3
$\frac{3 \sqrt{3}}{2}$

Discuss Question

Answer: Option C. -> 3
:
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 72.

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

0
1
3
sin 4 $α$ + sin 6 $α$

Discuss Question

Answer: Option B. -> 1
:
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 73.

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

$2 t a n α$
1
2
$s e c^{2} α$

Discuss Question

Answer: Option A. ->

2 t a n α

:
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 74.

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

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

Discuss Question

Answer: Option A. ->

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

:
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 75.

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

$A \geq 1$
$0 < A \leq 1$
$1 < 2 A \leq 3$
None of these

Discuss Question

Answer: Option B. ->

0 < A \leq 1

:
B
Option B is the correct answer.

Question 76.

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

$6 \leq p < 8$
$6 \leq p < 7$
$3 \leq p \leq 4$
$4 \leq p < 7$

Discuss Question

Answer: Option B. ->

6 \leq p < 7

:
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 77.

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

1
2
-1
-2

Discuss Question

Answer: Option A. -> 1
:
A
Option A is the correct answer.

Question 78.

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

$2 (t a n β + t a n γ)$
$t a n β + t a n γ$
$t a n β + 2 t a n γ$
$2 t a n β + t a n γ$

Discuss Question

Answer: Option C. ->

t a n β + 2 t a n γ

:
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 79.

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

$b^{2} + (a - c)^{2}$
$a^{2} + (b - c)^{2}$
$c^{2} + (a - b)^{2}$
None of these

Discuss Question

Answer: Option A. ->

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

:
A

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