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

Hint: 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 \[f(x)\].