Let f:R R be a function such that f(x)+3f(( )/(2)-x)= x,x R.

Question

Let $f:R	o R$ be a function such that $f(x)+3f(\frac{\pi }{2}-x)=\sin x,x\in R$. Let the maximum value of $f$ on $R$ be $\alpha $. If the area of the region bounded by the curves $g(x)={x}^{2}$ and $h(x)=\beta {x}^{3},\beta >0$, is ${\alpha }^{2}$, then $30{\beta }^{3}$ is equal to $____$ .

Answer: 16

Accepted Answer

$f(x)+3f(\frac{\pi }{2}-x)=\sin x\, \ldots (1)$

Now replace $x$ by $\frac{\pi }{2}-x$ in (1). This is a standard NCERT step to get a second equation in the same two terms.

$\Rightarrow f(\frac{\pi }{2}-x)+3f(x)=\cos x\, \ldots (2)$

Now we solve the two linear equations (1) and (2) to find $f(x)$.

Multiply (1) by $3$:

$3f(x)+9f(\frac{\pi }{2}-x)=3\sin x\, \ldots (3)$

Subtract (2) from (3):

$(3f(x)-3f(x))+(9-1)f(\frac{\pi }{2}-x)=3\sin x-\cos x$

$8f(\frac{\pi }{2}-x)=3\sin x-\cos x$

Now put $x	o \frac{\pi }{2}-x$ in this result to get $f(x)$:

$8f(x)=3\cos x-\sin x$

$f(x)=\frac{1}{8}(3\cos x-\sin x)$

To find the maximum value, write $3\cos x-\sin x$ in the form $R\cos (x+\phi )$, where $R=\sqrt{{3}^{2}+(-1{)}^{2}}=\sqrt{10}$. So the maximum value of $3\cos x-\sin x$ is $\sqrt{10}$.

${f}_{max}=\frac{\sqrt{10}}{8}$

$y=g(x)\&y=h(x)$ intersect as shown in the figure

The curves are $g(x)={x}^{2}$ and $h(x)=\beta {x}^{3}$ with $\beta >0$. Their points of intersection come from ${x}^{2}=\beta {x}^{3}$:

${x}^{2}=\beta {x}^{3}\Rightarrow {x}^{2}(1-\beta x)=0\Rightarrow x=0x=\frac{1}{\beta }$

For $0\beta x^3$(since$\beta x<1$),sotheareabetweenthecurvesistheintegralof$(x^2-\beta x^3)$from$0$to$\frac{1}{\beta}$.

$\begin{gathered} 	herefore =\Delta=|{\int }_{\frac{1}{\beta }}^{0}(\beta {x}^{3}-{x}^{2})dx| \ =|{[\frac{\beta {x}^{4}}{4}-\frac{{x}^{3}}{3}]}_{1/\beta }^{0}| \ =|(\frac{\beta }{4}\cdot \frac{1}{{\beta }^{4}}-\frac{1}{3}\cdot \frac{1}{{\beta }^{3}})| \ =|(\frac{1}{4{\beta }^{3}}-\frac{1}{3{\beta }^{3}})|=\frac{1}{12{\beta }^{3}}={\alpha }^{2} \ \Rightarrow \frac{1}{12{\beta }^{3}}={(\frac{\sqrt{10}}{8})}^{2}=\frac{10}{64}=\frac{5}{32} \ \Rightarrow {\beta }^{3}=\frac{1}{12}\cdot \frac{32}{5}=\frac{8}{15} \ \Rightarrow 30{\beta }^{3}=30\cdot \frac{8}{15}=16 \end{gathered}$

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