Let S = x ( - 2, 2): 9^1 - ^2 x + 9^ ^2 x = 10 b = x S ^2 ( x3), then ( - 14)^2 is equal to:

Let S = x ( - 2, 2): 9^1 - ^2 x + 9^ ^2 x = 10 b = x S ^2 ( x3),… — Solution | JEE Mathematics

MCQMediumJEE 2023Trigonometric Equations

Let $S = \left\{ x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) : 9^{1 - \tan^2 x} + 9^{\tan^2 x} = 10 \right\}$ $b = \sum_{x \in S} \tan^2 \left( \frac{x}{3} \right)$ , then $\left( \beta - 14 \right)^2$ is equal to:

A
$16$
B
$32$
C
$8$
D
$64$

Answer

Correct answer:B

Step-by-step solution

the solution unrelated to this question

Given: $S = \left\{ x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) : 9^{1 - \tan^2 x} + 9^{\tan^2 x} = 10 \right\}$ and $b = \sum_{x \in S} \tan^2\left(\frac{x}{3}\right)$ .

Find: $\left(b - 14\right)^2$ .

The provided solution discusses differentiating expressions of the form $\left(\frac{x}{e}\right)^{\beta x}$ and is unrelated to the given trigonometric set question. Therefore, the working for this question could not be extracted from the solution.

Using the provided correct answer field, the marked answer maps to option B.

Therefore, the correct option is B.

Common mistakes

Treating the symbol in the final expression as the Greek letter $\beta$ instead of the defined quantity $b$ is incorrect, because the question first defines $b$ and then asks for the same quantity in the final expression. Track the defined variable carefully and use the quantity introduced in the statement.
Solving $9^{1-\tan^2 x}+9^{\tan^2 x}=10$ by taking logarithms termwise is incorrect, because logarithms do not distribute over addition. First reduce the equation by substituting a single variable for $\tan^2 x$ and then solve the resulting algebraic relation.
Ignoring that $x \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ can lead to extra or repeated tangent values. Respect the interval restriction and list only those solutions that actually lie in the given domain before evaluating the sum.

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Answer

Step-by-step solution

Common mistakes

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