Question

Suppose $f(x)$ be a twice differentiable polynomial function such that $f(1)=2,f(2)=8$ and $f(3)=18$, then

• A. $f^{\prime \prime }(x)=4\forall x\in R$
• B. there exist at least one $x\in (1,3)$ such that $f^{\prime \prime }(x)=4$.
• C. there exist at least one $x\in (2,3)$ such that $f^{\prime }(x)-10=0$.
• D. there exist at least one $x\in (1,3)$ such that $f^{\prime }(x)-8=0$.

Answer 1

Answer: (B), (C), (D)

Let $g(x)=f(x)-2x^2$
$\Rightarrow g(x)=0$ has 3 real roots $x=1,2$ and 3
$\Rightarrow g^{\prime }(x)=0$ has atleast 2 real roots in $(1,3)$
$\Rightarrow g^{\prime \prime }(x)-0$ has atleast 1 roal root in $(1,3)$
$\Rightarrow f^{\prime \prime }(x)-4=0$ has atleast 1 real root in (1,
3) Ans. (B)
Now applying L.M.V.T. on $f(x)$ in $(2,3)$
$\exists$ atleast one $x\in (2,3)$ such that
$f^{\prime }(x)=\frac{f(3)-f(2)}{3-2}=\frac{18-8}{1}=10$
$\Rightarrow f^{\prime }(x)=10$ has atleast one root in $(2,3)$
Ans. (C)
Again applying L.M.V.T. on $f(x)$ in $(1,3)$
$\exists$ atleast one $x\in (1,3)$ such that
$f^{\prime }(x)=\frac{f(3)-f(1)}{3-1}=\frac{18-2}{2}=8$
$\Rightarrow f^{\prime }(x)=8\text{ has atleast one root in }(1,3)$