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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 = ∫ dx x ^ 2 ⋅ 1 + x ^ 2 = (f(x))/(x) + C, ( x > 0) (where, C...

If I=∫dxx2⋅1+x2=\text{I} = ∫ \dfrac{\text{dx}}{{\text{x}}^{2} ⋅ \sqrt{1 + {\text{x}}^{2}}} =f(x)x+C,\dfrac{f\left(x\right)}{x} + C,\left(\text{\forall x} > 0\right)(where,(where,Cistheconstantofintegration)is the constant of integration) and f\left(1\right) = - \sqrt{2},thenthevalueof∣f(3)∣then the value of \left|f\left(\sqrt{3}\right)\right| is

Correct answer

2

Step-by-step solution

Given integral is I=∫dxx31+1x2I = ∫ \dfrac{dx}{{x}^{3}\sqrt{1 + \dfrac{1}{{x}^{2}}}}
Let 1+1x2=t21 + \dfrac{1}{{x}^{2}} = {t}^{2}
⇒−2x3dx=2tdt⇒ \dfrac{- 2}{{x}^{3}}dx = 2tdt
⇒I=−∫tdtt⇒ I = - ∫ \dfrac{tdt}{t}
=−x2−1x+C= - \dfrac{\sqrt{{x}^{2} - 1}}{x} + C
⇒f(x)=−x2+1⇒ f\left(x\right) = - \sqrt{{x}^{2} + 1}
∴∣f(3)∣=∣−3+1∣=2∴ \left|f\left(\sqrt{3}\right)\right| = \left|- \sqrt{3 + 1}\right| = 2
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