Let the functions f- - - and g- - - be defined as-fx - x - 2- x - 0 x2- x - 0- and gx - x3- x - 1 3x - 2- x - 1-Then- the number of points in - where fog x is non differentiable is equal to -

Fog(x)={ x^3 + 2, amp;x lt; 0 nbsp; x^6, amp;0 ≤ x lt; 1 nbsp; (3x 2)^2, amp;x ≥ 1 nbsp; . Clearly, fog(x) is discontinuous at x = 0 ...

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Byju's Answer

Standard XII

Mathematics

Points of Discontinuity

Let the funct...

Question

Let the functions $f : R \to R$ and $g : R \to R$ be defined as:
$f (x) = {\begin{matrix} x + 2, & x < 0 x^{2}, & x \geq 0 \end{matrix} and g (x) = {\begin{matrix} x^{3}, & x < 1 3 x - 2, & x \geq 1 \end{matrix}$
Then, the number of points in $R$ where $(f o g) (x)$ is non differentiable is equal to :

1

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Solution

The correct option is A $1$
$f o g (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{3} + 2, & x < 0 x^{6}, & 0 \leq x < 1 (3 x - 2)^{2}, & x \geq 1 \end{matrix}$
Clearly, $f o g (x)$ is discontinuous at $x = 0$ then non - differentiable at $x = 0$
Now, at $x = 1$
R.H.D.
$= lim h \to 0^{+} \frac{f (1 + h) - f (1)}{h} = lim h \to 0^{+} \frac{(3 (1 + h) - 2)^{2} - 1}{h} = 6$
L.H.D.
$= lim h \to 0^{-} \frac{f (1 - h) - f (1)}{- h} = lim h \to 0^{-} \frac{(1 - h)^{6} - 1}{- h} = 6$
Number of point of non - differentiability is $1.$

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Let the functions f:R→R and g:R→R be defined as: f(x)={x+2,x<0x2,x≥0and g(x)={x3,x<13x−2,x≥1 Then, the number of points in R where (fog)(x) is non differentiable is equal to :