Let f: [0, 1] → R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5 . If the line y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1) , then the least number of points x

Correct answer: 2 Let f: [0, 1] → R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5 . If the line y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1) , then the least number of points x ∈ (0, 1), at which f ^ ' ' (x) = 0 , is Now plotting the diagram of given data we have, Given f(x) cuts y = 2x + 3 at two distinct point between x ∈ (0, 1) , So, f ^ ' (a) = f ^ ' (b) = f ^ ' (c) = 2 as slope of given line y = 2x + 3 is 2 So we can conclude that f ^ ' ' (x) is zero for atleast x _ 1 ∈ (a, b) and x _ 2 ∈(b, c)

JEE Main 202228 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 28 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2022 · 28 July · Morning Shift · Mathematics · Application of Derivatives

Let f: [0, 1] → R be a twice differentiable function in (0, 1) such...

Let

f: \left[0, 1\right] \to R

be a twice differentiable function in \left(0, 1\right)

such that

f\left(0\right) = 3

and

f\left(1\right) = 5

. If the line

y = 2x + 3

intersects the graph of

at only two distinct points in \left(0, 1\right)

, then the least number of points

x \in \left(0, 1\right), at which

{f}^{' '}\left(x\right)

= 0

, is

Let f: [0, 1] → R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5 . If the line y = 2x + 3

Correct answer

Step-by-step solution

Let $f: \left[0, 1\right] \to R$ be a twice differentiable function in \left(0, 1\right) $such that$ f\left(0\right) = 3 $and$ f\left(1\right) = 5 $. If the line$ y = 2x + 3 $intersects the graph of$ f $at only two distinct points in \left(0, 1\right)$ , then the least number of points $x \in \left(0, 1\right), at which$ {f}^{' '}\left(x\right) $= 0$ , is $Now plotting the diagram of given data we have,<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/a16d3b2d-9e9a-4c7a-bba4-1fc660cf7d3a-4e693ce9-c855-4624-a288-4ae27cdf7822-image.png" style="width: 236px; height: 247px;" />Given$ f\left(x\right) cuts $y = 2x + 3$ at two distinct point between $x \in \left(0, 1\right)$ ,
So, ${f}^{'}\left(a\right) = {f}^{'}\left(b\right) = {f}^{'}\left(c\right) = 2$ {as slope of given line $y = 2x + 3$ is $2$ }
So we can conclude that ${f}^{' '}\left(x\right) is zero$ for atleast{x}_{1} \in \left(a, b\right)\text{ and $} {x}_{2} \in\left(b, c\right)$

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