Question

Let $f'(x) = \dfrac{{192{x^3}}}{{(2 + {{\sin }^4}\pi x)}}$for all $x \in R$ with $f(\dfrac{1}{2}) = 0$. If $m \leqslant \int\limits_{\dfrac{1}{2}}^1 {f(x)dx} \leqslant M$, then the possible values of m and M are,

A.$m = 13,M = 24$

B.$m = \dfrac{1}{4},M = \dfrac{1}{2}$

C.$m = - 11,M = 0$

D.$m = 1,M = 12$

Answer 1

Firstly the condition for as $x \in [\dfrac{1}{2},1]$ so putting the limiting value of x we can calculate the ranging value of $f'(x)$.

Hence, after getting the idea of the value of $f'(x)$ integrate both sides and get the range of $f(x)$. Hence, after getting the range of value of $f(x)$ so we will get the values of m and M. Hence, our required answer will be obtained.

Complete step-by-step answer:

As the given equation is $f'(x) = \dfrac{{192{x^3}}}{{(2 + {{\sin }^4}\pi x)}}$for all $x \in R$ with $f(\dfrac{1}{2}) = 0$,

So, calculate the value of $f'(x)$ for $x \in [\dfrac{1}{2},1]$.

Hence at $x = \dfrac{1}{2}$,

$f'(\dfrac{1}{2}) = \dfrac{{192{{(\dfrac{1}{2})}^3}}}{{(2 + {{\sin }^4}(\dfrac{\pi }{2}))}}$

Now using $\sin \dfrac{\pi }{2} = 1$, and simplification we get,

$\Rightarrow f'(\dfrac{1}{2}) = \dfrac{{(\dfrac{{192}}{8})}}{{(2 + 1)}}$

On solving we get,

$\Rightarrow f'(\dfrac{1}{2}) = \dfrac{{24}}{3} = 8$

Now, calculate with above same procedure at $x = 1$,

$f'(1) = \dfrac{{192{{(1)}^3}}}{{(2 + {{\sin }^4}(\pi ))}}$

Using, $\sin \pi = 0$, we get,

$\Rightarrow f'(1) = \dfrac{{192}}{{(2)}} = 96$

Hence the range of values of $f'(x)$ can be given as,

$8 \leqslant f'(x) \leqslant 96$

Now, we can apply the integration in the above inequality for the value of $x \in [\dfrac{1}{2},x]$

$\Rightarrow \int\limits_{\dfrac{1}{2}}^x 8 .dx \leqslant \int\limits_{\dfrac{1}{2}}^x {f'(x).dx} \leqslant \int\limits_{\dfrac{1}{2}}^x {96.} dx$

On integrating we get,

$\Rightarrow \left. {8x} \right|_{\dfrac{1}{2}}^x \leqslant \left. {f(x)} \right|_{\dfrac{1}{2}}^x \leqslant \left. {96x} \right|_{\dfrac{1}{2}}^x$

Now, on applying limits we get ,

$\Rightarrow 8x - 4 \leqslant f(x) - f(\dfrac{1}{2}) \leqslant 96x - 48$

As the value of $f(\dfrac{1}{2}) = 0$ so the above inequality will be,

$\Rightarrow 8x - 4 \leqslant f(x) \leqslant 96x - 48$

Now integrate the $f(x)$ with the condition given in the question as,

$\Rightarrow \int\limits_{\dfrac{1}{2}}^1 {8x - 4.dx} \leqslant \int\limits_{\dfrac{1}{2}}^1 {f(x).dx} \leqslant \int\limits_{\dfrac{1}{2}}^1 {96x - 48.dx}$

Hence, on integrating both sides of the inequality the maxima and minima value of the function can be obtained.

$\Rightarrow 4{x^2}|_{\dfrac{1}{2}}^1 - 4x|_{\dfrac{1}{2}}^1 \leqslant \int\limits_{\dfrac{1}{2}}^1 {f(x).dx} \leqslant 48{x^2}|_{\dfrac{1}{2}}^1 - 48x|_{\dfrac{1}{2}}^1$

On applying limits we get,

$\Rightarrow 4(1 - \dfrac{1}{4}) - 4(1 - \dfrac{1}{2}) \leqslant \int\limits_{\dfrac{1}{2}}^1 {f(x).dx} \leqslant 48(1 - \dfrac{1}{4}) - 48(1 - \dfrac{1}{2})$

On further simplifying, we get,

$\Rightarrow 4(\dfrac{3}{4}) - 4(\dfrac{1}{2}) \leqslant \int\limits_{\dfrac{1}{2}}^1 {f(x).dx} \leqslant 48(\dfrac{3}{4}) - 48(\dfrac{1}{2})$

$\Rightarrow 3 - 2 \leqslant \int\limits_{\dfrac{1}{2}}^1 {f(x).dx} \leqslant 36 - 24$

$\Rightarrow 1 \leqslant \int\limits_{\dfrac{1}{2}}^1 {f(x).dx} \leqslant 12$

From here we can obtain the idea about maximum and minimum value of the function.

So, option (D) is the correct answer.

Note: Estimate the range of x and from there calculate the range of $f'(x)$and $f(x)$. Hence integrate properly and the required answer will be obtained.
And use the concept of estimated value of function properly. If you did not know the value of $f(x)$ then calculate the value of the known function which is just bigger or smaller than $f(x)$.