If f ( x ) is the antiderivative of (1 + 2 x ( x + x ))(1)/(2) and f((π)/(6)) = 2 , then the value of f ( 0 ) is

Correct answer: 0 f(x) = ∫ (1 + 2tan x(tan x + sec x)) ^ (1)/(2) dx = ∫ (1 + 2 ^ 2 x + 2tan xsec x) ^ (1)/(2) dx = ∫ (1 + ^ 2 x + ^ 2 x + 2tan xsec x) ^ (1)/(2) dx = ∫ ( ^ 2 x + ^ 2 x + 2tan xsec x) ^ (1)/(2) dx = ∫ ( ( x + x) ^ 2 ) ^ (1)/(2) dx = ( x + x) + log sec x + c = x( x + x) + c f((π)/(6)) = 2 ⇒ log 2 3 ( 2 3 + 1 3 ) + c = 2 ⇒ c = 0 f(x) = (sec x(sec x + tan x)) ⇒ f(0) = ((1)(1 + 0)) = 0

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If f ( x ) is the antiderivative of (1 + 2 x ( x + x ))(1)/(2) and...

f ( x ) is the antiderivative of \left(1 + 2\tan x ( \tan x + \sec x )\right)\dfrac{1}{2} and f\left(\dfrac{\pi}{6}\right) = \log 2

, then the value of

f ( 0 )

Correct answer

Step-by-step solution

f\left(x\right) = ∫ {\left(1 + 2tan x\left(tan x + sec x\right)\right)}^{\dfrac{1}{2}}dx

= ∫ {\left(1 + 2\tan^{2} x + 2tan xsec x\right)}^{\dfrac{1}{2}}dx

= ∫ {\left(1 + \tan^{2} x + \tan^{2} x + 2tan xsec x\right)}^{\dfrac{1}{2}}dx

= ∫ {\left(\sec^{2} x + \tan^{2} x + 2tan xsec x\right)}^{\dfrac{1}{2}}dx

= ∫ {\left({\left(\sec x + \tan x\right)}^{2}\right)}^{\dfrac{1}{2}}dx

= \log \left(\sec x + \tan x\right) + log sec x + c

= \log \sec x\left(\sec x + \tan x\right) + c

f\left(\dfrac{\pi}{6}\right) = \text\log2 ⇒ log\dfrac{2}{\sqrt{3}}\left(\dfrac{2}{\sqrt{3}} + \dfrac{1}{\sqrt{3}}\right) + c = \text\log2

⇒ c = 0

f\left(x\right) = \log \left(sec x\left(sec x + tan x\right)\right)

⇒ f\left(0\right) = \log \left(\left(1\right)\left(1 + 0\right)\right) = 0

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