Let m and n be the number of points at which the function f(x) = x, x^3, x^5,, x^21 is not differentiable and not continuous, respectively. Then m + n is equal to

Let m and n be the number of points at which the function f(x) =… — Solution | JEE Mathematics

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Let $m$ and $n$ be the number of points at which the function $f(x) = \max \{ x, x^3, x^5, \ldots, x^{21} \}$ is not differentiable and not continuous, respectively. Then $m + n$ is equal to

A
$3$
B
$4$
C
$5$
D
$6$

Answer

Correct answer:C

Step-by-step solution

the solution unavailable

Given: $f(x) = \max\{x, x^3, x^5, \ldots, x^{21}\}$ .

Find: The value of $m+n$ , where $m$ is the number of points where $f$ is not differentiable and $n$ is the number of points where $f$ is not continuous.

Working could not be extracted from the solution. Using the given answer information, the correct option is C. Therefore, $m+n = 5$ .

Common mistakes

Assuming the maximum of continuous functions can be discontinuous. This is wrong because the maximum of finitely many continuous functions is always continuous. Therefore, take $n=0$ .
Checking only the endpoints $x=-1$ and $x=1$ and ignoring switching points between the curves. Non-differentiability occurs where the dominating power changes, so compare the odd powers across intervals.
Treating all intersections of $x, x^3, x^5, \ldots, x^{21}$ as distinct for every pair. This is wrong because many powers intersect at the same common points, so count unique points only.

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Answer

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