Let $\mathrm{f}:[0,1] \rightarrow \mathrm{R}$ be a twice differentiable function in $(0,1)$ such that $\mathrm{f}(0)=3$ and $\mathrm{f}(1)=5$. If the line $\mathrm{y}=2 \mathrm{x}+3$ intersects the graph of f at only two distinct points in $(0,1)$, then the least number of points $\mathrm{x} \in(0,1)$, at which $\mathrm{f}^{\prime \prime}(\mathrm{x})=0$, is ____.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Application of Derivatives

Let

f : [0, 1] \to R

be a twice differentiable function in

(0, 1)

such that

f (0) = 3

and

f (1) = 5

. If the line

y = 2 x + 3

intersects the graph of f at only two distinct points in

(0, 1)

, then the least number of points

x \in (0, 1)

, at which

f^{''} (x) = 0

, is ____.

Answer: 2

The line

y = 2 x + 3

passes through

(0, 3)

and

(1, 5)

, which are the endpoints of the graph of

f

.

Since the line intersects the graph at exactly two distinct points in

(0, 1)

, there must be at least three points

a, b, c \in (0, 1)

where

f^{'} (x) = 2

(the slope of the line).

By Rolle's theorem applied to

f^{'} (x) - 2

, there must be at least two points where

f^{''} (x) = 0

: one in

(a, b)

and one in

(b, c)

.

Therefore, the least number is 2.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer