Question

Fill in the blanks.
(i) The value of $c$ in $LMVT$ for the function $f\left(x\right)=2x^3-5x^2-4x+3$, $x\in \left[\frac{1}{3},3\right]$ is P.
(ii) The value of $c$ in Rolle's theorem for the function $f(x)=(x-2)(x-3)$ in $[2,3]$ is Q.
(iii) The value of $c$ in Rolle's theorem for the function $f(x)=x^2-5x+9$, $x\in [1,4]$ is R.
(iv) The value of $c$ in $LMVT$ for the function $f(x)=6x^2-x^3$, $x\in [0,6]$ is S.

	P	Q	R	S
(a)	$1.5$	$3/2$	$5/2$	$4$
(b)	$1.9$	$5/2$	$3/2$	$2$
(c)	$1.5$	$2$	$1$	$2$
(d)	$1.9$	$5/2$	$5/2$	$4$

• A. (a)
• B. (b)
• C. (c)
• D. (d)

Answer 1

Answer: (D)

(i) $f(x)=2x^3-5x^2-4x+3$ being a polynomial function is continous on $\left[\frac{1}{3},3\right]$ and differentiable on $\left(\frac{1}{3},3\right)$
$\therefore$ By $LMVT$, we have $c\left(\in \frac{1}{3},3\right)8\cdot t$
$\frac{f\left(3\right)-f\left(\frac{1}{3}\right)}{3-\frac{1}{3}}=f^{\prime }\left(c\right)$
$\Rightarrow f^{\prime }\left(c\right)=\frac{0-\frac{32}{27}}{\frac{8}{3}}=\frac{-4}{9}$
$\Rightarrow 6c^2-10c-4=\frac{-4}{9}$
$\Rightarrow 6c^2-10c-\frac{32}{9}=0$
$\Rightarrow c=\frac{45\pm 61.26}{54}$
$\Rightarrow c=\frac{45\pm 61.26}{54}\approx 1.9\left(\because c\in \left(\frac{1}{3},3\right)\right)$
(ii) $f(x)=(x-2)(x-3)$, $x\in [2,3]$
$\therefore f(x)=x^2-5x+6$ and $f^{\prime }(x)=2x-5\dots (i)$
Now, $f(x)$ is continuous on $[2,3]$ and differentiable on $(2,3)$.
Also, $f(2)=f(3)=0$
$\therefore$ By Rolle’s theorem we have $c\in (2,3)$ such that
$f^{\prime }\left(c\right)=0$
$\Rightarrow 2c-5=0$
$\Rightarrow c=\frac{5}{2}$
(iii) The function $f(x)=x^2-5x+9$ being a polynomial function is continuous on $[1,4]$ and differentiable on $(1,4)$.
Now, $f(1)=1^2-5(1)+9=1-5+9=5$
and $f(4)=4^2-5(4)+9=16-20+9=5$
$\therefore f(1)=f(4)$
Thus, the function satisfies all the conditions of the Rolle's theorem.
$\therefore$ We have $c\in (1,4)$ such that $f^{\prime }(c)=0$.
Now, $f(x)=x^2-5x+9$
$\Rightarrow f^{\prime }(x)=2x-5$
$\therefore f^{\prime }\left(c\right)=0$
$\Rightarrow 2c-5=0$
$\Rightarrow c=\frac{5}{2}\in \left(1,4\right)$
(iv) $f(x)$ being polynomial function is continuous on $[0,6]$ and differentiable on $(0,6)$.
$\therefore$ By $LMVT$, we have $c\in (0,6)$ such that
$\frac{f\left(6\right)-f\left(0\right)}{6-0}=f^{\prime }\left(c\right)$
$\Rightarrow \frac{0-0}{6-0}=f^{\prime }\left(c\right)$
$\Rightarrow f^{\prime }\left(c\right)=0$
Now $f\left(6x\right)=x^2-x^3$
$\Rightarrow f^{\prime }\left(x\right)=12x-3x^2$
$\Rightarrow f^{\prime }\left(c\right)=12c-3c^2$
$\Rightarrow 0=12c-3c^2$
$\Rightarrow 3c^2=12c$
$\Rightarrow c=4$ which lies in the interval $(0,6)$.