Question

Let $\int \frac{x^{1/2}}{\sqrt{1-x^3}}dx = \frac{2}{3}gof(x) + C$ (where $gof(x)$ is the composite function defined by $(gof)x = g(f(x))$, $C$ is constant of integration and $gof(0) = 0$, then which of the following is/are correct?

• A. If $f(x) = \sin^{-1}x$, then $g(x) = x^{3/2}$
• B. If $f(x) = x^{3/2}$, then $g(x) = \sin^{-1}x$
• C. If $f(x) = x^{2/3}$, then $g(x) = \sin^{-1}(x^{9/4})$
• D. If $f(x) = \sqrt{x}$, then $g(x) = \sin^{-1}(x^3)$

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

The integral $\int \frac{x^{1/2}}{\sqrt{1-x^3}}dx$ is solved by substituting $u = x^{3/2}$. This substitution leads to $du = \frac{3}{2}x^{1/2}dx$, so $x^{1/2}dx = \frac{2}{3}du$. The integral becomes $\int \frac{1}{\sqrt{1-u^2}} \frac{2}{3}du = \frac{2}{3}\int \frac{1}{\sqrt{1-u^2}}du = \frac{2}{3}\sin^{-1}(u) + C$. Substituting back $u=x^{3/2}$, we get $\frac{2}{3}\sin^{-1}(x^{3/2}) + C$. Comparing this with $\frac{2}{3}gof(x) + C$, we require $gof(x) = \sin^{-1}(x^{3/2})$. The condition $gof(0)=0$ must also be satisfied. Let's check the options: A. $f(x) = \sin^{-1}x$, $g(x) = x^{3/2}$. $gof(x) = g(f(x)) = (\sin^{-1}x)^{3/2}$. This is not $\sin^{-1}(x^{3/2})$. B. $f(x) = x^{3/2}$, $g(x) = \sin^{-1}x$. $gof(x) = g(f(x)) = \sin^{-1}(x^{3/2})$. $gof(0) = \sin^{-1}(0^{3/2}) = 0$. Correct. C. $f(x) = x^{2/3}$, $g(x) = \sin^{-1}(x^{9/4})$. $gof(x) = g(f(x)) = \sin^{-1}((x^{2/3})^{9/4}) = \sin^{-1}(x^{3/2})$. $gof(0) = \sin^{-1}(0^{3/2}) = 0$. Correct. D. $f(x) = \sqrt{x} = x^{1/2}$, $g(x) = \sin^{-1}(x^3)$. $gof(x) = g(f(x)) = \sin^{-1}((x^{1/2})^3) = \sin^{-1}(x^{3/2})$. $gof(0) = \sin^{-1}(0^{3/2}) = 0$. Correct.