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f(π/2) when f'(x)=1/(1+cosx), f(0)=3 | JEE

JEE Maths question with a full step-by-step solution.

Question
Let ff be continuous with f′(x)=11+cos⁡xf'(x)=\dfrac{1}{1+\cos x} and f(0)=3f(0)=3. Then f ⁣(π2)f\!\left(\dfrac{\pi}{2}\right) equals
A134\dfrac{13}{4}
B22
C44correct
DNone of these
Solution
Step 1: Integrate f′(x)f'(x). Multiply numerator and denominator by (1−cos⁡x)(1-\cos x):
11+cos⁡x=1−cos⁡x(1+cos⁡x)(1−cos⁡x)=1−cos⁡x1−cos⁡2x=1−cos⁡xsin⁡2x\dfrac{1}{1+\cos x}=\dfrac{1-\cos x}{(1+\cos x)(1-\cos x)}=\dfrac{1-\cos x}{1-\cos^{2}x}=\dfrac{1-\cos x}{\sin^{2}x}
 Step 2: Split:
1−cos⁡xsin⁡2x=1sin⁡2x−cos⁡xsin⁡2x=csc⁡2x−csc⁡xcot⁡x\dfrac{1-\cos x}{\sin^{2}x}=\dfrac{1}{\sin^{2}x}-\dfrac{\cos x}{\sin^{2}x}=\csc^{2}x-\csc x\cot x
 Step 3: Integrate:
f(x)=∫(csc⁡2x−csc⁡xcot⁡x) dx=−cot⁡x+csc⁡x+C=1−cos⁡xsin⁡x+C=tan⁡x2+Cf(x)=\int(\csc^{2}x-\csc x\cot x)\,dx=-\cot x+\csc x+C=\dfrac{1-\cos x}{\sin x}+C=\tan\dfrac{x}{2}+C
 Step 4: Apply f(0)=3f(0)=3: tan⁡0+C=C=3\tan 0+C=C=3. Step 5: Evaluate at x=π2x=\dfrac{\pi}{2}:
f ⁣(π2)=tan⁡π4+3=1+3=4f\!\left(\dfrac{\pi}{2}\right)=\tan\dfrac{\pi}{4}+3=1+3=4
 Correct answer: (3)
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