Let f(x) = 2 + | x – 1 | and g(x) = min. (f(t)) where 1 \leq t \leq x \forall x \in \left[\right. 1 , \infty \left.\right) and number of points of non‑derivability of g(x) in \left[\right. 1 , \infty \left.\right) be m. If h(x) = x | x2 – 3x | e| x – 3 | + \left[\right. \frac{3 + sgn \textrm{ } \left(\right. x^{2} - 1 - 5 x \left.\right)}{4 + x^{2}} \left]\right. + \left{\right. \frac{1}{3} + \left[\right. \frac{x^{2} - 2 x}{3 \left|\right. x \left|\right. + 1} \left]\right. \left.\right} is discontinuous at n number of points then find (m + n). [Note: [z] & {z} denotes greatest integer function less than or equal & fractional part function of z.]

Engineering

Mathematics

Important Points on Derivability

Question

Let f(x) = 2 + | x – 1 | and g(x) = min. (f(t)) where $1 \leq t \leq x \forall x \in [1, \infty)$ and number of points of non‑derivability of g(x) in $[1, \infty)$ be m.

If h(x) = x | x² – 3x | e^{| x – 3 |} + $[\frac{3 + sgn (x^{2} - 1 - 5 x)}{4 + x^{2}}] + {\frac{1}{3} + [\frac{x^{2} - 2 x}{3 | x | + 1}]}$ is discontinuous at n number of points then find (m + n).

[Note: [z] & {z} denotes greatest integer function less than or equal & fractional part function of z.]

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$Clearly, f (t) = {\begin{cases} 3 - t, t < 1 \\ t + 1, t \geq 1 \end{cases}$

$∴$ g(x) = $2 \forall x \geq 1$

$∴$ Number of points of non‑derivability = 0.

and h(x) = x | x | | x – 3 | e^{| x – 3 |} + 0 + $\frac{1}{3}$ always continuous.

$∴$ m + n = 0.