COMPASSLearning

Talk to us Register Login

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Trigonometric Equations

Let

S = {x \in (- \frac{π}{2}, \frac{π}{2}) : 9^{1 - t a n^{2} x} + 9^{t a n^{2} x} = 10}

and

β = \sum_{x \in s} tan^{2} (\frac{x}{3})

, then

\frac{1}{6} (β - 14)^{2}

is equal to

(A)32✓ Correct
(B)8
(C)64
(D)16

Answer: (A)

Solution

Writing

P = 9^{t a n^{2} x}

turns the equation into

\frac{9}{P} + P = 10

, that is

P^{2} - 10 P + 9 = 0

, so

P = 1

or

P = 9

. These give

tan^{2} x = 0

or

tan^{2} x = 1

, hence

x = 0, \pm \frac{π}{4}

inside

(- \frac{π}{2}, \frac{π}{2})

. Therefore

β = tan^{2} 0 + 2 tan^{2} \frac{π}{12} = 2 (2 - 3)^{2} = 2 (7 - 43) = 14 - 83

. Then

β - 14 = - 83

, and

\frac{1}{6} (β - 14)^{2} = \frac{( 8 3 ) ^{2}}{6} = \frac{192}{6} = 32

. The required value is

32

.

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Trigonometric Equations →

More JEE Main questions on Trigonometric Equations

Number of solutions of $3 cos 2 θ + 8 cos θ + 33 = 0, θ \in [- 3 π, 2 π]$ is:
JEE Main 2026View solution →
The number of elements in the set ${x \in [0, 18 0^{\circ}] : tan (x + 10 0^{\circ}) = tan (x + 5 0^{\circ}) tan x tan (x - 5 0^{\circ})}$ is
JEE Main 2026View solution →
Let $S = {x \in [- π, π] : sin x (sin x + cos x) = a$ , $a \in Z}$ . Then $n (S)$ is equal to :
JEE Main 2026View solution →
If $S = {θ \in [- π, π] : cos θ cos \frac{5 θ}{2} = cos 7 θ cos \frac{7 θ}{2}}$ , then $n (S)$ is equal to
JEE Main 2026View solution →
Let $S = {θ \in (- 2 π, 2 π) : cos θ + 1 = 3 sin θ}$ .

Then $\sum_{θ \in S} θ$ is equal to:
JEE Main 2026View solution →
The sum of all values of $θ \in [0, 2 π]$ satisfying $2 sin^{2} θ = cos 2 θ$ and $2 cos^{2} θ = 3 sin θ$ is
JEE Main 2025View solution →