If c = frac{1}{2} and f(x) = 2x - x^2,then th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the function $f(x) = 2x - x^2$,the derivative is $f^{\prime}(x) = 2 - 2x$.According to $LMVT$,there exists a point $c \in (a, b)$ such that $f

Vedclass · Accepted Answer

$(0, 1)$

Vedclass · Answer

(C) For the function $f(x) = 2x - x^2$,the derivative is $f^{\prime}(x) = 2 - 2x$.According to $LMVT$,there exists a point $c \in (a, b)$ such that $f^{\prime}(c) = \frac{f(b) - f(a)}{b - a}$.Given $c = \frac{1}{2}$,we have $f^{\prime}(\frac{1}{2}) = 2 - 2(\frac{1}{2}) = 2 - 1 = 1$.Now,calculate the slope of the secant line: $\frac{f(b) - f(a)}{b - a} = \frac{(2b - b^2) - (2a - a^2)}{b - a} = \frac{2(b - a) - (b^2 - a^2)}{b - a} = \frac{2(b - a) - (b - a)(b + a)}{b - a} = 2 - (a + b)$.Equating the two: $1 = 2 - (a + b)$,which implies $a + b = 1$.For $LMVT$ to be applicable on $(a, b)$,the function must be continuous on $[a, b]$ and differentiable on $(a, b)$. Since $f(x)$ is a polynomial,it is continuous and differentiable everywhere.Checking the options,for $(0, 1)$,$a=0$ and $b=1$,so $a+b = 1$. This satisfies the condition $c = \frac{1}{2} \in (0, 1)$.

If $c = \frac{1}{2}$ and $f(x) = 2x - x^2$,then the interval $(a, b)$ of $x$ in which the Lagrange's Mean Value Theorem $(LMVT)$ is applicable for the function $f(x)$ is:

Similar Questions

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be such that $f(0)=0$ and $|f^{\prime}(x)| \leq 5$ for all $x$. Then $f(1)$ is in

$A$ value of $c$ according to the Lagrange's mean value theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ is

$f:[1,3] \rightarrow R$ is a function defined as $f(x)=x^3+a x^2+b x$. If $f(1)-f(3)=0$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$,then $a-b$ is equal to

Let $f(x)=(x-1)(x-2)(x-3)$,where $x \in [0,4]$. Find the values of $c$ if Lagrange's Mean Value Theorem $(LMVT)$ can be applied.

If a function $f$ is differentiable on $R$ such that $f^{\prime}(x) \leq 4$ for all $x \in R$; and if $f(2)=-6$ and $f(6)=8$,then the value of $f(4)$ belongs to the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module