Let and . Then the number of points, where the function gof is discontinuous, is ________. [2026]

Question

Let $f (x) = {\begin{matrix} x^{2} + 8; & x < 0, \\ x^{2} - 4; & x \geq 0 \end{matrix}$ and $g (x) = {\begin{matrix} {(x - 8)}^{1 / 3}; & x < 0, \\ {(x + 4)}^{1 / 2}; & x \geq 0 \end{matrix}$ . Then the number of points, where the function gof is discontinuous, is ________. [2026]

Smart Achievers · Accepted Answer

(3)

$g (f (x)) = {\begin{matrix} (f {(x) - 8)}^{1 / 3}; & f (x) < 0, \\ (f {(x + 4)}^{1 / 2}; & f (x) \geq 0 \end{matrix}$

$g (f (x)) = {\begin{matrix} x & x \in (- \infty, - 2) \\ {(x^{3} + 12)}^{1 / 2} & x \in [- 2, 0) \\ {(x^{2} - 12)}^{1 / 3} & [0, 2) \\ x & x \in [2, \infty) \end{matrix}$

The function is discontinuous at x = –2, 0, 2

Hence, the number of point of discontinuity is 3.

Solution

Q.
Let $f (x) = {\begin{matrix} x^{2} + 8; & x < 0, \\ x^{2} - 4; & x \geq 0 \end{matrix}$ and $g (x) = {\begin{matrix} {(x - 8)}^{1 / 3}; & x < 0, \\ {(x + 4)}^{1 / 2}; & x \geq 0 \end{matrix}$ . Then the number of points, where the function gof is discontinuous, is ________. [2026]

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Solution

Q. Let f(x)={x2+8;x<0,x2–4;x≥0 and g(x)={(x–8)1/3;x<0,(x+4)1/2;x≥0. Then the number of points, where the function gof is discontinuous, is ________. [2026]

Ans. (3) g(f(x))={(f(x)–8)1/3;f(x)<0,(f(x+4)1/2;f(x)≥0 g(f(x))={xx∈(–∞,–2)(x3+12)1/2x∈[–2,0)(x2–12)1/3[0,2)xx∈[2,∞) The function is discontinuous at x = –2, 0, 2 Hence, the number of point of discontinuity is 3.

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Q.
Let $f (x) = {\begin{matrix} x^{2} + 8; & x < 0, \\ x^{2} - 4; & x \geq 0 \end{matrix}$ and $g (x) = {\begin{matrix} {(x - 8)}^{1 / 3}; & x < 0, \\ {(x + 4)}^{1 / 2}; & x \geq 0 \end{matrix}$ . Then the number of points, where the function gof is discontinuous, is ________. [2026]