The function \(f(x) = x 1 + xtan x \,\) , \(0≤ x≤π/2\) is maximum when

\[ f'(x) = 1(1 + x tan x) - x(x sec 2 x + tan x)/(1 + x tan x) 2 \] \[ = 1 - x 2 sec 2 x/(1 + x tan x) 2 \] Here we see that when \(x 2 sec 2 x 0\), and when \(x 2 sec 2 x > 1\), \(f'(x) < 0\). Hence, the function is maximum at \(x sec x = 1\). \[ cos x = x \]

The function \(f(x) = x 1 + xtan x \,\) ,... - Solved

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Question 45

Mathematics Maxima and Minima Hard

The function \(f(x) = \cfrac{x}{{1 + x\tan x}}\,\) , \(0\leq x\leq\frac{\pi}{2}\) is maximum when

(A) \(x=secx\)

(B) \(x= an x\)

(C) \(x=\cos x\)

(D) None of these

✓ Correct Answer: Option C Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

\[ f'(x) = \frac{1(1 + x \tan x) - x(x \sec^2 x + \tan x)}{(1 + x \tan x)^2} \] \[ = \frac{1 - x^2 \sec^2 x}{(1 + x \tan x)^2} \] Here we see that when \(x^2 \sec^2 x < 1\), \(f'(x) > 0\), and when \(x^2 \sec^2 x > 1\), \(f'(x) < 0\). Hence, the function is maximum at \(x \sec x = 1\). \[ \Rightarrow \cos x = x \]

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