Question

Let S = {\( x∈(−\frac{ π}{ 2} , \frac{π}{2} )\) : 9^1−tan2 x + 9^tan2 x=10 } and \(β=∑_{ x∈s} t a n ^2 (\frac{ x}{ 3} )\) , then \(\frac{1}{6}\) ( β − 14 )²  is equal to

• A. 8
• B. 16
• C. 32
• D. 34

Answer 1

The Correct Option is C

To solve the given problem, we begin by analyzing the expression \(9^{1-\tan^2 x} + 9^{\tan^2 x} = 10\). Our task is to find the set \(S = \{ x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \}\) where this equation holds true.

Define \(a = \tan^2 x\). Thus, the equation becomes \(9^{1-a} + 9^a = 10\). Rewriting it: \[ 9 \cdot 9^{-a} + 9^a = 10 \quad \Rightarrow \quad 9 \cdot \frac{1}{9^a} + 9^a = 10. \] Substituting \(y = 9^a\), the equation converts to: \[ 9 \cdot \frac{1}{y} + y = 10 \quad \Rightarrow \quad 9 + y^2 = 10y. \]

Reorganize to find: \[ y^2 - 10y + 9 = 0. \] Solving this quadratic equation using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -10, \) and \(c = 9\): \[ y = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}. \] Thus, \(y = 9\) or \(y = 1\).

\(y = 9\) implies \(9^a = 9\) which gives \(a = 1\) and \(\tan^2 x = 1\) leading to \(\tan x = \pm 1\).

Considering \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), solutions for \(\tan x = \pm 1\) are \(x = \frac{\pi}{4}, -\frac{\pi}{4}\).

Next, for \(y = 1\), \(9^a = 1\) gives \(a = 0\) implying \(\tan^2 x = 0\), hence \(\tan x = 0\) and \(x = 0\).

Thus, the set \(S\) is \(\{-\frac{\pi}{4}, 0, \frac{\pi}{4}\}\).

Now calculate \(\beta = \sum_{x \in S} \tan^2 \left(\frac{x}{3}\right)\): - For \(x = 0\), \(\frac{0}{3} = 0 \implies \tan^2(0) = 0\). - For \(x = \frac{\pi}{4}\), \(\frac{\pi}{12} \implies \tan \left(\frac{\pi}{12}\right) = \tan(15^\circ) = 2 - \sqrt{3}\). \[ \tan^2 \left(\frac{\pi}{12}\right) = (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}. \] - For \(x = -\frac{\pi}{4}\), \(\frac{-\pi}{12} \implies \tan \left(-\frac{\pi}{12}\right) = -(2 - \sqrt{3}) = \sqrt{3} - 2\). \[ \tan^2 \left(-\frac{\pi}{12}\right) = (\sqrt{3} - 2)^2 = 7 - 4\sqrt{3}. \]

Therefore, \(\beta = 0 + (7 - 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 - 8\sqrt{3}\).

Finally, calculate \(\frac{1}{6}(\beta - 14)^2\): \[ \beta - 14 = 14 - 8\sqrt{3} - 14 = -8\sqrt{3}. \] \[ (\beta - 14)^2 = (-8\sqrt{3})^2 = 192. \] \[ \frac{1}{6} \times 192 = 32. \]

Therefore, the final answer is \(32\).