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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If I = ∫ 1 + x ^ 4 (1 - x ^ 4 )(3)/(2) dx = 1 f(x) + C (where, C is...

If I=∫1+x4(1−x4)32dxI = ∫ \dfrac{1 + {x}^{4}}{\left(1 - {x}^{4}\right)\dfrac{3}{2}}dx=1f(x)+C= \dfrac{1}{\sqrt{f\left(x\right)}} + C (where, CC is the constant of integration)andf(2)=−154and f\left(2\right) = \dfrac{- 15}{4}, then the value of 2f(12)2f\left(\dfrac{1}{\sqrt{2}}\right) is

Correct answer

3

Step-by-step solution

Given integral is
I=∫1x3+x(1x2−x2)32dxI = ∫ \dfrac{\dfrac{1}{{x}^{3}} + x}{{\left(\dfrac{1}{{x}^{2}} - {x}^{2}\right)}^{\dfrac{3}{2}}}dx
Let 1x2−x2=t\dfrac{1}{{x}^{2}} - {x}^{2} = t
⇒(−2x3−2x)dx=dt⇒ \left(- \dfrac{2}{{x}^{3}} - 2x\right)dx = dt
∴I=−12∫dtt32∴ I = - \dfrac{1}{2}∫ \dfrac{dt}{{t}^{\dfrac{3}{2}}}
=t−12+C= {t}^{- \dfrac{1}{2}} + C
=11x2−x2+C= \dfrac{1}{\sqrt{\dfrac{1}{{x}^{2}} - {x}^{2}}} + C
∴f(x)=1x2−x2∴ f\left(x\right) = \dfrac{1}{{x}^{2}} - {x}^{2}
⇒f(12)=2−12=32⇒ f\left(\dfrac{1}{\sqrt{2}}\right) = 2 - \dfrac{1}{2} = \dfrac{3}{2}
⇒2f(12)=3⇒ 2f\left(\dfrac{1}{\sqrt{2}}\right) = 3
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