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 $c\left(\in\frac{1}{3}, 3\right)8\cdot t$
$\frac{f \left(3\right)-f \left(\frac{1}{3}\right)}{3-\frac{1}{3}}=f '\left(c\right)$
$\Rightarrow f '\left(c\right)=\frac{0-\frac{32}{27}}{\frac{8}{3}}=\frac{-4}{9}$
$\Rightarrow 6c^{2}-10c-4=\frac{-4}{9}$
$\Rightarrow 6c^{2}-10c-\frac{32}{9}=0$
$\Rightarrow c=\frac{45 \pm61.26}{54}$
$\Rightarrow c=\frac{45\pm61.26}{54}\approx1.9\, \left(\because c \in\left(\frac{1}{3}, 3\right)\right)$
(ii) $f(x) = (x - 2)(x - 3)$, $x \in [2,3]$
$\therefore f(x) = x^2 - 5x + 6$ and $f' (x) = 2x - 5\quad \ldots(i)$
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
$f '\left(c\right)=0$
$\Rightarrow 2c-5=0$
$\Rightarrow c=\frac{5}{2}$
(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 $f '(c) = 0$.
Now, $f(x) = x^2 - 5x + 9$
$\Rightarrow f'(x) = 2x - 5$
$\therefore f '\left(c\right)=0$
$\Rightarrow 2c-5=0$
$\Rightarrow c=\frac{5}{2} \in\left(1, 4\right)$
(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
$\frac{f \left(6\right)-f \left(0\right)}{6-0}=f '\left(c\right)$
$\Rightarrow \frac{0-0}{6-0}=f '\left(c\right)$
$\Rightarrow f '\left(c\right)=0$
Now $f \left(6x\right)=x^{2}-x^{3}$
$\Rightarrow f '\left(x\right)=12x-3x^{2}$
$\Rightarrow f '\left(c\right)=12c-3c^{2}$
$\Rightarrow 0=12c-3c^{2}$
$\Rightarrow 3c^{2}=12c$
$\Rightarrow c=4$ which lies in the interval $(0,6)$.