A continuous and differentiable function y=f(x) is such that its graph cuts line y=m x+c at n distinct points. Then the minimum number of points at which f''(x)=0 is/are

Question

A continuous and differentiable function y=f(x) is such that its graph cuts line y=m x+c at n distinct points. Then the minimum number of points at which f''(x)=0 is/are (A) n-1 (B) n-3 (C) n-2 (D) cannot say

Tardigrade Admin · Accepted Answer

From LMVT, there exists at least (n-1) points where f'(x)=m Therefore, there exist at least (n-2) points where f''(x)=0 (using Rolle's theorem).

Q. A continuous and differentiable function $y = f (x)$ is such that its graph cuts line $y = m x + c$ at $n$ distinct points. Then the minimum number of points at which $f^{''} (x) = 0$ is/are

$n - 1$

$n - 3$

$n - 2$

cannot say

From LMVT, there exists at least $(n - 1)$ points where $f^{'} (x) = m$
Therefore, there exist at least $(n - 2)$ points where $f^{''} (x) = 0$ (using Rolle's theorem).

Q. A continuous and differentiable function y=f(x) is such that its graph cuts line y=mx+c at n distinct points. Then the minimum number of points at which f′′(x)=0 is/are

n−1

n−3

n−2