Question

Let the function $f:[0,8]\to R$ be a differentiable function. Then which of the following is/are correct ?

• A. There exist $\alpha ,\beta \in (0,8)$ such that $f^2\left(8\right)=f^2\left(0\right)+16f^{\prime }\left(\alpha \right)f\left(\beta \right);$
• B. There exists $\gamma ,\delta \in (0,2)$ such that$\underset{0}{\overset{8}{\int }}f\left(x\right)dx=3\left[{\gamma }^{2}f\right({\gamma }^3)+{\delta }^{2}f({\delta }^3\left)\right];$
• C. If $f\left(x\right)=x^5-2x^3-2,$ then it has a real root between $0$ and $2$
• D. If $f\left(x\right)=2x^3+3x^2+6x+1,$ then has $3$ real roots.

Answer 1

Answer: (A), (B), (C)

If $f\left(x\right)=x^5-2x^3-2,$ then it has a real root between $0$ and $2$

We know that $f\left(x\right)$ is differentiable so it is continuous.
By Intermediate mean value theorem, there exists $\beta \in (0,8)$ such that
$f\left(\beta \right)=\frac{f\left(8\right)+f\left(0\right)}{2}\cdots \left(1\right)$

By Lagrange's mean value theorem, there exists $\alpha \in (0,8)$ such that
$f^{\prime }\left(\alpha \right)=\frac{f\left(8\right)-f\left(0\right)}{8-0}\cdots \left(2\right)$

Multiplying $\left(1\right)$ and $\left(2\right)$, we get
$f^{\prime }\left(\alpha \right)\times f\left(\beta \right)=\frac{f\left(8\right)-f\left(0\right)}{8-0}\times \frac{f\left(8\right)+f\left(0\right)}{2}$

Hence, for some $\alpha ,\beta \in (0,8)$
$f^2\left(8\right)=f^2\left(0\right)+16f^{\prime }\left(\alpha \right)f\left(\beta \right)$

Now,
$\underset{0}{\overset{8}{\int }}f\left(x\right)dx$
Substituting $x=z^3\Rightarrow dx=3z^{2}dz$
$\underset{0}{\overset{8}{\int }}f\left(x\right)dx=\underset{0}{\overset{2}{\int }}3z^{2}f\left(z^3\right)dz$
Assuming a function
$F\left(t\right)=\underset{0}{\overset{t}{\int }}3z^{2}f\left(z^3\right)dz,t\in [0,2]$
As $F\left(t\right)$ is integral function of a continuous function, so it is differentiable.
Now, by Lagrange's mean value theorem for $c\in (0,2)$
$F^{\prime }\left(c\right)=\frac{F\left(2\right)-F\left(0\right)}{2-0}$
We know that $F\left(0\right)=0$
$\Rightarrow F^{\prime }\left(c\right)=\frac{1}{2}\underset{0}{\overset{2}{\int }}3z^{2}f\left(z^3\right)dz\cdots \left(3\right)$

By Intermediate mean value theorem on $F^{\prime }\left(t\right)$
For $0<\gamma <c<\delta <2$
$F^{\prime }\left(c\right)=\frac{F^{\prime }\left(\gamma \right)+F^{\prime }\left(\delta \right)}{2}$
Using $\left(3\right)$, we get
$\frac{1}{2}\underset{0}{\overset{2}{\int }}3z^{2}f\left(z^3\right)dz=\frac{F^{\prime }\left(\gamma \right)+F^{\prime }\left(\delta \right)}{2}\Rightarrow \underset{0}{\overset{2}{\int }}3z^{2}f\left(z^3\right)dz=3\left[{\gamma }^{2}f\right({\gamma }^3)+{\delta }^{2}f({\delta }^3\left)\right]$
Hence, for some $\gamma ,\delta \in (0,2)$
$\underset{0}{\overset{8}{\int }}f\left(x\right)dx=3\left[{\gamma }^{2}f\right({\gamma }^3)+{\delta }^{2}f({\delta }^3\left)\right]$

$f\left(x\right)=x^5-2x^3-2f\left(0\right)=-2f\left(2\right)=32-16-2=14$
So, $f\left(0\right)\cdot f\left(2\right)<0$
Therefore there exists at least one root between $0$ and $2$

$f\left(x\right)=2x^3+3x^2+6x+1f\left(0\right)=1f(-1)=-10f\left(0\right)\cdot f(-1)<0$
So, one roots between $0$ and $-1$
$f^{\prime }\left(x\right)=6x^2+6x+6f^{\prime }\left(x\right)=6(x^2+x+1)>0$
Therefore only one root is possible.