Question

Let $g(x)=\cos \left( {{x}^{2}} \right)\ and\ f(x)=\sqrt{x}$ , and $\alpha ,\beta (\alpha <\beta )$ be the roots of the quadratic equation $18{{x}^{2}}-9\pi x+{{\pi }^{2}}=0$ . Then the area (in sq. units) bounded by the curve $y=(gof)(x)$ and the lines $x=\alpha$, $x=\beta$ and y=0, is
(A). $\dfrac{1}{2}\left( \sqrt{3}-\sqrt{2} \right)$
(B). $\dfrac{1}{2}\left( \sqrt{2}-1 \right)$
(C). $\dfrac{1}{2}\left( \sqrt{3}-1 \right)$
(D). $\dfrac{1}{2}\left( \sqrt{3}+1 \right)$

Answer 1

HINT: - The most important formulae that would be used in solving this question are given as follows
The quadratic formula to get the roots of a quadratic equation
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
(Where the quadratic equation is $a{{x}^{2}}+bx+c=0$ )
Also, the formula to find the area under the curve which means area enclosed by the function from above and by the x-axis from below is
$Area=\int\limits_{a}^{b}{f(x)dx}$
(Where a and b are the lower and upper limits respectively)

Complete step-by-step solution -
In this question, first we will find the function that has been formed and is mentioned as
$y=(gof)(x)$ which is known as composite functions.
Then, we will find the roots of the given quadratic equation that is
$18{{x}^{2}}-9\pi x+{{\pi }^{2}}=0$
Now, the roots of this given quadratic equation will act as the limits of the integral which will be evaluated to get the area under the function $y=(gof)(x)$ .
On, evaluating the integral, we will get the required solution.
As mentioned in the question, we have to find the area that is under the curve of the composite function within the limits which are the roots of the given quadratic equation.
Now, as mentioned in the hint, we will first find the composite function that is $y=(gof)(x)$ where the functions are as follows
$g(x)=\cos \left( {{x}^{2}} \right)\ and\ f(x)=\sqrt{x}$
Now,

& y=(gof)(x) \\\ \Rightarrow & y=\cos \left( {{\left\\{ \sqrt{x} \right\\}}^{2}} \right) \\\ \Rightarrow & y=\cos x \\\ \end{aligned}$$ Hence, the above mentioned in the composite function that is required. Now, for solving the quadratic equation, we will use the quadratic formula that is given in the hint as follows $$\begin{aligned} & x=\dfrac{-(-9\pi )\pm \sqrt{{{(-9\pi )}^{2}}-4\times 18\times {{\pi }^{2}}}}{2\times 18} \\\ \Rightarrow & x=\dfrac{9\pi \pm \sqrt{81{{\pi }^{2}}-72{{\pi }^{2}}}}{36} \\\ \Rightarrow & x=\dfrac{9\pi \pm \sqrt{9{{\pi }^{2}}}}{36} \\\ \Rightarrow & x=\dfrac{9\pi \pm 3\pi }{36} \\\ \Rightarrow & x=\dfrac{12\pi }{36},\dfrac{6\pi }{36} \\\ \Rightarrow & x=\dfrac{\pi }{3},\dfrac{\pi }{6} \\\ \end{aligned}$$ Hence, we get the roots of the quadratic equation as mentioned above. Now, we know that these are going to be the limits of the integral as mentioned in the question and hence, we can form the definite integral as follows using the formula that is given in the hint $$\begin{aligned} & Area=\int\limits_{a}^{b}{f\left( x \right)dx} \\\ \Rightarrow & Area=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\cos \left( x \right)dx} \\\ \Rightarrow & Area=\left[ \sin x \right]_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} \\\ \Rightarrow & Area=\left[ \sin \dfrac{\pi }{3}-\sin \dfrac{\pi }{6} \right]=\dfrac{\sqrt{3}}{2}-\dfrac{1}{2} \\\ \Rightarrow & Area=\dfrac{1}{2}\left( \sqrt{3}-1 \right)\ sq.\ units \\\ \end{aligned}$$ Hence, this is the required area which is mentioned above. NOTE:- The students can make an error if they don’t know the method or the formula to get the roots of a quadratic equation. Also, if the students don’t know the method of finding the area under the curve which means area enclosed by the function from above and by the x-axis from below. Another important thing to keep in mind while solving this question is that it might be possible that the students may make a lot of mistakes in calculations and for that the students must be extra careful while solving the questions.