For how many values of x does 1/p become zero?

Correct Answer is : Only two values

Correct Answer is : Only two values

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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 75-76

Consider the following for the next items that follow:

Let

p=\frac{1}{3}-\frac{tan3x}{tanx}

and q = 1 - 3 tan2 x, 0 < x < π,

x≠\frac{\pi }{2}

Question : 76 of 120

Marks: +1, -0

\frac{1}{p}

become zero?

Solution:

Concept:

We use the identity for

\tan 3x

to simplify the given expression and set

\frac{1}{p}=0

Explanation:

Start with

p = \frac{1}{3} - \frac{\tan 3x}{\tan x}

Write

\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}

Substitute into

p

p = \frac{1}{3} - \frac{\frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}}{\tan x}

= \frac{1}{3} - \frac{3\tan x - \tan^3 x}{(1 - 3\tan^2 x)\tan x}

Combine into a single fraction:

p = \frac{(1-3\tan^2 x)\tan x - 9\tan x + 3\tan^3 x}{3(1-3\tan^2 x)\tan x}

Simplify numerator:

\tan x - 3\tan^3 x - 9\tan x + 3\tan^3 x = -8\tan x

Thus

p = \frac{-8\tan x}{3(1-3\tan^2 x)\tan x} = \frac{-8}{3(1-3\tan^2 x)}

(provided

\tan x \neq 0

Now

\frac{1}{p} = -\frac{3(1-3\tan^2 x)}{8}

Set

\frac{1}{p} = 0

-\frac{3}{8}(1-3\tan^2 x) = 0 \Rightarrow 1-3\tan^2 x = 0

\tan^2 x = \frac{1}{3}

, giving

\tan x = \pm \frac{1}{\sqrt{3}}

In the principal interval

[0, \pi)

, the solutions are

x = \frac{\pi}{6}

and

x = \frac{5\pi}{6}

Hence there are exactly two values of

x

for which

\frac{1}{p} = 0

Answer:

Only two values (Option C).

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