Let f be a twice differentiable function such that f(x)= _x^0 (t-x)dt- _x^0f(t) tdt,x (-( …

Question

Let $f$ be a twice differentiable function such that

$f(x)={\int }_{x}^{0}	an (t-x)dt-{\int }_{x}^{0}f(t)	an tdt,x\in (-\frac{\pi }{2},\frac{\pi }{2})$

Then ${f}^{''}(\frac{\pi }{6})+12{f}^{'}(-\frac{\pi }{6})+f(\frac{\pi }{6})$ is equal to \_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

Let

$f(x)={\int }_{x}^{0}	an (t-x)\, dt-{\int }_{x}^{0}f(t)	an t\, dt,\, x\in (-\frac{\pi }{2},\frac{\pi }{2}).$

We have to find

${f}^{''}(\frac{\pi }{6})+12{f}^{'}(-\frac{\pi }{6})+f(\frac{\pi }{6}).$

Simplify the first integral: Consider

${\int }_{x}^{0}	an (t-x)\, dt.$

Put

$u=t-x\, \implies \, du=dt.$

When $t=0$, $u=-x$, and when $t=x$, $u=0$.

So,

${\int }_{x}^{0}	an (t-x)\, dt={\int }_{0}^{-x}	an u\, du.$

Now,

$\int 	an u\, du=-\ln (\cos u).$

Hence,

${\int }_{0}^{-x}	an u\, du={[-\ln (\cos u)]}_{0}^{-x}=-\ln (\cos 0)+\ln (\cos (-x)).$

Since $\cos 0=1$ and $\cos (-x)=\cos x$,

${\int }_{x}^{0}	an (t-x)\, dt=\ln (\cos x).$

Therefore the given equation becomes

$f(x)=\ln (\cos x)-{\int }_{x}^{0}f(t)	an t\, dt.$

So,

$f(x)+{\int }_{x}^{0}f(t)	an t\, dt=\ln (\cos x).$

Differentiate the equation: Differentiate both sides with respect to $x$.

Using the Fundamental Theorem of Calculus,

$\frac{d}{dx}({\int }_{x}^{0}f(t)	an t\, dt)=f(x)	an x.$

Also,

$\frac{d}{dx}(\ln (\cos x))=-	an x.$

Thus,

${f}^{'}(x)+f(x)	an x=-	an x.$

So,

${f}^{'}(x)=-(f(x)+1)	an x.$

Find $f(x)$ explicitly: Now observe that $f(x)=-1$ satisfies the differential equation:

${f}^{'}(x)=0,\, -(f(x)+1)	an x=0.$

Also from the original equation, put $x=0$:

$f(0)={\int }_{0}^{0}	an (t-0)\, dt-{\int }_{0}^{0}f(t)	an t\, dt=0.$

So,

$f(0)=0.$

Let us solve the differential equation properly.

From

${f}^{'}(x)+(f(x)+1)	an x=0,$

put

$y(x)=f(x)+1.$

Then

${y}^{'}(x)+y(x)	an x=0.$

This gives

$\frac{{y}^{'}}{y}=-	an x.$

Integrating,

$\ln |y|=\int -	an x\, dx=\ln (\cos x)+C.$

Hence,

$y=C\cos x.$

Therefore,

$f(x)+1=C\cos x.$

Using $f(0)=0$,

$0+1=C\cos 0=C.$

So $C=1$, and hence

$f(x)+1=\cos x$

which gives

$f(x)=\cos x-1.$

Find derivatives: Since

$f(x)=\cos x-1,$

we get

${f}^{'}(x)=-\sin x,$

and

${f}^{''}(x)=-\cos x.$

Now compute required values:

1. ${f}^{''}(\frac{\pi }{6})$
${f}^{''}(\frac{\pi }{6})=-\cos (\frac{\pi }{6})=-\frac{\sqrt{3}}{2}.$

2. ${f}^{'}(-\frac{\pi }{6})$
${f}^{'}(-\frac{\pi }{6})=-\sin (-\frac{\pi }{6})=\frac{1}{2}.$

So,

$12{f}^{'}(-\frac{\pi }{6})=12\cdot \frac{1}{2}=6.$

3. $f(\frac{\pi }{6})$
$f(\frac{\pi }{6})=\cos (\frac{\pi }{6})-1=\frac{\sqrt{3}}{2}-1.$

Add them: ${f}^{''}(\frac{\pi }{6})+12{f}^{'}(-\frac{\pi }{6})+f(\frac{\pi }{6})$

$=-\frac{\sqrt{3}}{2}+6+(\frac{\sqrt{3}}{2}-1)$

$=6-1=5.$

Hence, the required value is

$5.$

Solutions · 1