Question

If $f\left(x\right)=\sum _{r=0}^n{a}_r{x}^r;$ for $a_rϵR;rϵN;n\ge 3$
If $f\left(x\right)\ne 0$ for $xϵ(\alpha ,\beta )$
Then, if $(\alpha <t<\beta )$,

• A. $f\left(x\right)$ is continuous and Differentiable over $(\alpha ,\beta )$ atleast
• B. $(x-\alpha )(x-\beta )f\left(x\right)$ is continuous and Differentiable over $(\alpha ,\beta )$ atleast
• C. $(x-\alpha )(x-\beta )f\left(x\right)$ is continuous, but Not differentiable over $(\alpha ,\beta )$
• D. $\frac{f^{\prime }\left(t\right)}{f\left(t\right)}=\frac{1}{\alpha -t}+\frac{1}{\beta -t}$, for atleast one ${}^{\prime }{t}^{\prime }$ in $(\alpha ,\beta )$

Answer 1

Answer: (A), (B), (D)

$\frac{f^{\prime }\left(t\right)}{f\left(t\right)}=\frac{1}{\alpha -t}+\frac{1}{\beta -t}$, for atleast one ${}^{\prime }{t}^{\prime }$ in $(\alpha ,\beta )$
$(x-\alpha )(x-\beta )f\left(x\right)$ is continuous and Differentiable over $(\alpha ,\beta )$ atleast
$f\left(x\right)$ is continuous and Differentiable over $(\alpha ,\beta )$ atleast

$f\left(x\right)$ is a polynomial function

So, it is continuous and differentiable atleast over $(\alpha ,\beta )$
$\Rightarrow$ $\therefore \left(OptionA\right)$

Consider $g\left(x\right)=(x-\alpha )(x-\beta )f\left(x\right)$

Now, $g\left(x\right)$ is also a Polynomial function

So, $g\left(x\right)$ is also continuous and Differentiable atleast on $(\alpha ,\beta )\Rightarrow \left(OptionB\right)$
Further $g\left(\alpha \right)=0$ and $g\left(\beta \right)=0$
Hence, by Rolle's theorem, there exists a point ${}^{\prime }{t}^{\prime }ϵ(\alpha ,\beta )$
Such that $g^{\prime }\left(t\right)=0$.
But $g^{\prime }\left(x\right)=\frac{d}{dx}\left(\right(x-\alpha \left)\right(x-\beta \left)\right(f\left(x\right)\left)\right)$

$g^{\prime }\left(x\right)=(x-\alpha )(x-\beta )f^{\prime }\left(x\right)+(x-\alpha ).1.f(x{)}_1.(x-\beta \left)f\right(x)$

Put $x=t$,

$g^{\prime }\left(t\right)=(t-\alpha )(t-\beta )f^{\prime }\left(t\right)+(t-\alpha )f\left(t\right)+(t-\beta )f\left(t\right)$

$\Rightarrow 0=(t-\alpha )(t-\beta )f^{\prime }\left(t\right)+(t-\alpha )f\left(t\right)+(t-\beta )f\left(t\right)$

$\Rightarrow \frac{f^{\prime }\left(t\right)}{f\left(t\right)}=\frac{1}{t-\alpha }+\frac{1}{t-\beta }$
$\left(OptionD\right)$, (For atleast one ${}^{\prime }{t}^{\prime }$ in $(\alpha ,\beta )$).