If number of points of discontinuity of f x -sgncos 5 x is equal to the number of points of non- differentiability of g x - n - m sin x where x -0- - n - m - I- then value of m iswhere- x denotes fractional part of x and sgn x denotes signum function of x

Sgn(cos 5x) is discontinuous at nbsp; 5x = π/2, 3π/2,......., 9π/2 g(x) = {n+m sin x} nbsp;={m sin x} nbsp; is non differentiable at nbsp; sin x = 1 ...

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Standard XII

Mathematics

Continuous Function

If number of ...

Question

If number of points of discontinuity of $f (x) = s g n (cos 5 x)$ is equal to the number of points of non- differentiability of g(x) = {n + m sin x} where $x \in (0, π), n, m \in I$ , then value of $m$ is
(where, {x} denotes fractional part of x and sgn(x) denotes signum function of x)

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Solution

$sgn (cos 5 x)$ is discontinuous at
$5 x = \frac{π}{2}, \frac{3 π}{2}, . . . . . . ., \frac{9 π}{2}$
$g (x) = {n + m sin x} = {m sin x}$ is non - differentiable at
$sin x = \frac{1}{m}, \frac{2}{m}, . . . \frac{m - 1}{m}, \frac{m}{m}$
So total points of non-differentiability of $g (x)$ are $2 (m - 1) + 1 = (2 m - 1)$
$∴ 2 m - 1 = 5 \Rightarrow m = 3$

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If number of points of discontinuity of f(x)=sgn(cos5x) is equal to the number of points of non- differentiability of g(x) = {n + m sin x} where x ∈ (0,π), n, m ∈ I, then value of m is (where, {x} denotes fractional part of x and sgn(x) denotes signum function of x)

sgn(cos5x) is discontinuous at 5x=π2,3π2,.......,9π2 g(x)={n+m sinx}={m sinx} is non - differentiable at sinx=1m,2m,...m−1m,mm So total points of non-differentiability of g(x) are 2(m−1)+1=(2m−1) ∴2m−1=5⇒m=3

If number of points of discontinuity of $f (x) = s g n (cos 5 x)$ is equal to the number of points of non- differentiability of g(x) = {n + m sin x} where $x \in (0, π), n, m \in I$ , then value of $m$ is
(where, {x} denotes fractional part of x and sgn(x) denotes signum function of x)