Match the following Column – I with Column – II Column – I Column – II A)If I=∫sinx−cosx|sinx−cosx| dx, where π4 <x<3π8 then I is equal to(P)sinx B)∫x2(x3+1)(x3+2) dx=13f(x3+1x3+2)+C, then f(x) is equal to (Q)x + C C)If ∫sin−1x. cos−1x dx=f−1(x)[π2x−x f−1(x)−21−x2] + 2x + C, then f(x) is equal to (Assume f(x) to be invertible) (R)ln |x|D)∫dxx f(x)=f(f(x))+C, then f(x) is equal to (S)sin−1x

A) If π4 cosx∴∫sinx−cosx|sinx−cosx| dx=∫1 dx=x+C.B) ∫x2dx(x3+1)(x3+2)=13∫3x2(1x3+1−1x3+2)dx=13ln|x3+1x3+2|+CC) ∫sin−1x cos−1x dx=∫[π2sin−1x−(sin−1x)2] dx⇒π2(x sin−1x+1−x2)−{x(sin−1x)2+sin−1x1−x2−x}+C (by parts) ⇒sin−1x[π2x−xsin−1x−21−x2]+π21−x+2x+C∴ f−1(x)=sin−1x, f(x)=sinxD) ∫dxx ln|x|=ln |ln|x||+C∴f(x)=ln|x|

Match the following Column – I with Column – II Column – I Column – II A)If I=∫sinx−cosx|sinx−cosx| dx, where π4 <x<3π8 then I is equal to(P)sinx B)∫x2(x3+1)(x3+2) dx=13f(x3+1x3+2)+C, then f(x) is equal to (Q)x + C C)If ∫sin−1x. cos−1x dx=f−1(x)[π2x−x f−1(x)−21−x2] + 2x + C, then f(x) is equal to (Assume f(x) to be invertible) (R)ln |x|D)∫dxx f(x)=f(f(x))+C, then f(x) is equal to (S)sin−1x

A) If π4 cosx∴∫sinx−cosx|sinx−cosx| dx=∫1 dx=x+C.B) ∫x2dx(x3+1)(x3+2)=13∫3x2(1x3+1−1x3+2)dx=13ln|x3+1x3+2|+CC) ∫sin−1x cos−1x dx=∫[π2sin−1x−(sin−1x)2] dx⇒π2(x sin−1x+1−x2)−{x(sin−1x)2+sin−1x1−x2−x}+C (by parts) ⇒sin−1x[π2x−xsin−1x−21−x2]+π21−x+2x+C∴ f−1(x)=sin−1x, f(x)=sinxD) ∫dxx ln|x|=ln |ln|x||+C∴f(x)=ln|x|

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Match the following Column – I with Column – II
Column – I Column – II
A) If $I = \int \frac{\sin x - \cos x}{| \sin x - \cos x |} d x, w h e r e \frac{π}{4} < x < \frac{3 π}{8}$ then I is equal to (P) sinx
B) $\int \frac{x^{2}}{(x^{3} + 1) (x^{3} + 2)} d x = \frac{1}{3} f (\frac{x^{3} + 1}{x^{3} + 2}) + C,$ then f(x) is equal to (Q) x + C
C) If $\int \sin^{- 1} x . \cos^{- 1} x d x = f^{- 1} (x) [\frac{π}{2} x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}]$ + 2x + C, then f(x) is equal to (Assume f(x) to be invertible) (R) ln |x|
D) $\int \frac{d x}{x f (x)} = f (f (x)) + C$ , then f(x) is equal to (S) $\sin^{- 1} x$

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A – P; B – Q; C – R; D – S

A – S; B – R; C – S ; D – Q

A – Q; B – R; C – P; D – R

A – P; B – Q; C – R ; D – S

answer is B.

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Detailed Solution

A) If $\frac{π}{4} < x < \frac{3 π}{8}, t h e n \sin x > \cos x$

$∴ \int \frac{\sin x - \cos x}{| \sin x - \cos x |} d x = \int 1 d x = x + C$ .

B) $\int \frac{x^{2} d x}{(x^{3} + 1) (x^{3} + 2)} = \frac{1}{3} \int 3 x^{2} (\frac{1}{x^{3} + 1} - \frac{1}{x^{3} + 2}) d x$ $= \frac{1}{3} \ln | \frac{x^{3} + 1}{x^{3} + 2} | + C$

C) $\int \sin^{- 1} x \cos^{- 1} x d x = \int [\frac{π}{2} \sin^{- 1} x - {(\sin^{- 1} x)}^{2}] d x$

$\Rightarrow \frac{π}{2} (x \sin^{- 1} x + \sqrt{1 - x^{2}}) - {x {(\sin^{- 1} x)}^{2} + \sin^{- 1} x \sqrt{1 - x^{2}} - x} + C$ (by parts)

$\Rightarrow \sin^{- 1} x [\frac{π}{2} x - x \sin^{- 1} x - 2 \sqrt{1 - x^{2}}] + \frac{π}{2} \sqrt{1 - x} + 2 x + C$

$∴ f^{- 1} (x) = \sin^{- 1} x, f (x) = \sin x$

D) $\int \frac{d x}{x \ln | x |} = \ln | \ln | x | | + C$

$∴ f (x) = \ln | x |$

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