Q78 / 180 · KCET 2024 · Mathematics

Let $(g \circ f)(x)=\sin x$ and $f \circ g(x)=(\sin \sqrt{x})^2$ Then,

  • A
    $f(x)=\sin ^2 x, g(x)=x$
  • B
    $f(x)=\sin \sqrt{x}, g(x)=\sqrt{x}$
  • Correct answer
    C
    $f(x)=\sin ^2 x, g(x)=\sqrt{x}$
  • D
    $f(x)=\sin \sqrt{x}, g(x)=x^2$

Solution

$\because g(f(x))=\sin x$ and $f(g(x))=(\sin \sqrt{x})^2$ (a) $f(x)=\sin ^2 x$ and $g(x)=x$ Now, $f(g(x))=f(x)=\sin ^2 x$ and $g(f(x))=g\left(\sin ^2 x\right)=\sin ^2 x$ (b) $f(x)=\sin \sqrt{x}$ and $g(x)=\sqrt{x}$ Now, $f(g(x))=f(\sqrt{x})=\sin \sqrt{\sqrt{x}}=\sin (x)^4$ and $a(f(x))=a(\sin \sqrt{x})=\sqrt{\sin \sqrt{x}}$ (c) $f(x)=\sin ^2 x$ and $g(x)=\sqrt{x}$ Now, $f(g(x))=f(\sqrt{x})=\sin ^2 \sqrt{x}$ and $g(f(x))=g\left(\sin ^2 x\right)=\sqrt{\sin ^2 x}=|\sin x|$ (d) $f(x)=\sin \sqrt{x}$ and $g(x)=x^2$ Now, $f(g(x))=f\left(x^2\right)=\sin |x|$ and $g(f(x))=g(\sin \sqrt{x})=(\sin \sqrt{x})^2=\sin ^2 \sqrt{x}$

Hence, the correct answer is option: 'C'.

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