Question

If \(g(x)=\cos x^2\), \(f(x)=\sqrt{x}\) and \(\alpha,\beta\;(\alpha<\beta)\) are the roots of \(18x^2-9\pi x+\pi^2=0\), then the area bounded by the curve \(y=(g\circ f)(x)\) and the lines \(x=\alpha,\;x=\beta\) and \(y=0\) is:

• A. \(\dfrac{\sqrt{3}}{2}\)
• B. \(\dfrac{\sqrt{3}+1}{2}\)
• C. \(\dfrac{\sqrt{3}-1}{2}\)
• D. \(\dfrac{1}{2}\)

Hint:

Always simplify composite functions first. Here, \(g(\sqrt{x})=\cos((\sqrt{x})^2)=\cos x\), which converts the problem into a standard definite integral.

Answer 1

The Correct Option is C

Step 1: Find the composite function \((g\circ f)(x)\). \[ (g\circ f)(x)=g(f(x))=\cos\left((\sqrt{x})^2\right)=\cos x \] So the curve is: \[ y=\cos x \]
Step 2: Find the limits of integration \(\alpha,\beta\). Given quadratic equation: \[ 18x^2-9\pi x+\pi^2=0 \] Discriminant: \[ \Delta=(9\pi)^2-4(18)(\pi^2)=81\pi^2-72\pi^2=9\pi^2 \] Roots: \[ x=\frac{9\pi\pm3\pi}{36} \] \[ \alpha=\frac{6\pi}{36}=\frac{\pi}{6}, \qquad \beta=\frac{12\pi}{36}=\frac{\pi}{3} \] 
Step 3: Set up the area integral. Required area: \[ A=\int_{\alpha}^{\beta}\cos x\,dx =\int_{\pi/6}^{\pi/3}\cos x\,dx \] 
Step 4: Evaluate the integral. \[ A=\sin x\Big|_{\pi/6}^{\pi/3} =\sin\frac{\pi}{3}-\sin\frac{\pi}{6} \] \[ =\frac{\sqrt{3}}{2}-\frac{1}{2} =\frac{\sqrt{3}-1}{2} \] 
Hence, the required area is \[ \boxed{\dfrac{\sqrt{3}-1}{2}} \]