Question:medium
Let \(f:[-2a,2a]\to\mathbb{R}\) be a thrice differentiable function and define
\[
g(x)=f(a+x)+f(a-x).
\]
If \(m\) is the minimum number of roots of \(g'(x)=0\) in the interval \((-a,a)\) and
\(n\) is the minimum number of roots of \(g''(x)=0\) in the interval \((-a,a)\),
then \(m+n\) is equal to:
Let \(f:[-2a,2a]\to\mathbb{R}\) be a thrice differentiable function and define
\[
g(x)=f(a+x)+f(a-x).
\]
If \(m\) is the minimum number of roots of \(g'(x)=0\) in the interval \((-a,a)\) and
\(n\) is the minimum number of roots of \(g''(x)=0\) in the interval \((-a,a)\),
then \(m+n\) is equal to:
Show Hint
If a function is even, its derivative is odd and must vanish at the origin.
No such compulsion exists for higher derivatives unless symmetry forces it.
Updated On: Mar 5, 2026
- \(1\)
- \(2\)
- \(4\)
- \(5\)
Show Solution
The Correct Option is A
Solution and Explanation
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