JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 18)
Let $$f:[0,1] \rightarrow \mathbf{R}$$ 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 $$f^{\prime \prime}(x)=0$$, is ____________.
Answer
2
Explanation
If a graph cuts $$y = 2x + 5$$ in (0, 1) twice then its concavity changes twice.
$$\therefore$$ $$f'(x) = 0$$ at at least two points.
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