Function f(x) = {(x - 3)^2} satisfies all conditions of mean value theorem in [3, 4]. A point on y =

English

Gujarati

Hindi

5. Continuity and Differentiation

medium

The function $f(x) = {(x - 3)^2}$ satisfies all the conditions of mean value theorem in $[3, 4].$ A point on $y = {(x - 3)^2}$, where the tangent is parallel to the chord joining $ (3, 0)$ and $(4, 1)$ is

A

$\left( {{7 \over 2},{1 \over 2}} \right)$

B

$\left( {{7 \over 2},{1 \over 4}} \right)$

C

$(1, 4)$

D

$(4, 1)$

Solution

(b) Let the point be $({x_1},\,{y_1}).$ Therefore ${y_1} = {({x_1} – 3)^2}$ …..$(i)$

Now slope of the tangent at $({x_1},\,{y_1})$ is $2({x_1} – 3),$ but it is equal to $ 1.$

Therefore, $2({x_1} – 3) = 1 \Rightarrow {x_1} = \frac{7}{2}$

$\therefore {y_1} = {\left( {\frac{7}{2} – 3} \right)^2} = \frac{1}{4}$.

Hence the point is $\left( {\frac{7}{2},\frac{1}{4}} \right)$.

Standard 12

Mathematics

Share

Similar Questions

Which of the following function can satisfy Rolle's theorem ?

normal

If Rolle's theorem holds for the function $f(x) = 2{x^3} + b{x^2} + cx,\,x\, \in \,\left[ { – 1,1} \right]$ at the point $x = \frac{1}{2}$ , then $(2b+c)$ is equal to

normal

Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq$ 0 , and $f(x)=g(x) \sin x$.

$STATEMENT$ $-1: \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.and

$STATEMENT$ $-2: f^{\prime}(0)=g(0)$.

normal

(IIT-2008)

Let $f$ be any function continuous on $[\mathrm{a}, \mathrm{b}]$ and twice differentiable on $(a, b) .$ If for all $x \in(a, b)$ $f^{\prime}(\mathrm{x})>0$ and $f^{\prime \prime}(\mathrm{x})<0,$ then for any $\mathrm{c} \in(\mathrm{a}, \mathrm{b})$ $\frac{f(\mathrm{c})-f(\mathrm{a})}{f(\mathrm{b})-f(\mathrm{c})}$ is greater than

hard

(JEE MAIN-2020)

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $ [-1, 1] $ because

easy