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UPSC NDA I 2018 Mathematics Solved Paper

Question : 119 of 120

Marks: +1, -0

f(x) = e^{\tan x} + \ln(\sec x) - e^{\ln x}

x = \frac{\pi}{4}

Solution:

Concept:

Differentiate the function after simplifying

e^{\ln x}

x

, then apply standard derivative rules for exponential, logarithmic, and trigonometric functions.

Explanation:

First, simplify:

f(x) = e^{\tan x} + \ln(\sec x) - e^{\ln x} = e^{\tan x} + \ln(\sec x) - x

Differentiate term by term:

\frac{d}{dx}\left(e^{\tan x}\right) = e^{\tan x} \cdot \sec^2 x

(using chain rule).

\frac{d}{dx}\left(\ln(\sec x)\right) = \frac{1}{\sec x} \cdot \sec x \tan x = \tan x

\frac{d}{dx}(-x) = -1

Thus

f'(x) = e^{\tan x} \sec^2 x + \tan x - 1

Now evaluate at

x = \frac{\pi}{4}

\tan\left(\frac{\pi}{4}\right) = 1

\sec\left(\frac{\pi}{4}\right) = \sqrt{2}

, so

\sec^2\left(\frac{\pi}{4}\right) = 2

, and

e^{\tan(\pi/4)} = e^1 = e

Substitute:

f'\left(\frac{\pi}{4}\right) = e \cdot 2 + 1 - 1 = 2e

Answer:

2e

, which corresponds to option C.

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