Let f :[ a , b ] [1, ) be a continuous function and let g : R R b… - Vidyadip

MCQ

Let $f :[ a , b ] \rightarrow[1, \infty)$ be a continuous function and let $g : R \rightarrow R$ be defined as

$g(x)=\left\{\begin{array}{ccc}0 & \text { if } & x < a, \\ \int_a^x f(t) d t & \text { if } & a \leq x \leq b, \\ \int_a^b f(t) d t & \text { if } & x > b .\end{array}\right.$, Then

$(A)$ $g(x)$ is continuous but not differentiable at a

$(B)$ $g(x)$ is differentiable on $R$

$(C)$ $g(x)$ is continuous but not differentiable at $b$

$(D)$ $g(x)$ is continuous and differentiable at either a or $b$ but not both

A
$(B,D)$
B
$(B,C)$
✓
$(A,C)$
D
$(A,D)$

✓

Answer

Correct option: C.

$(A,C)$

c
It may be discontinuous at $x = a$ or $x = b$

$\lim _{x \rightarrow a^{-}} g(x)=0 $

$\lim _{x \rightarrow a^{+}} g(x)=\lim _{x \rightarrow a^{+}} \int_a^x f(t) d$ $t=\int_a^a f(t) d t=0 $

$g(a)=\int_a^a f(t) d t=0$

Similarly at $x=b$ we will get continuous

So $g(x)$ is continuous $\forall x \in R$

$g^{\prime}(x)=\left[\begin{array}{cc}0 & x < a \\ f(x) & a \leq x \leq b \\ 0 & x > b\end{array}\right.$

$g^{\prime}\left(a^{-}\right)=0 \quad\quad g^{\prime}\left(b^{-}\right)=f(b) $

$g^{\prime}\left(a^{+}\right)=f(a) \quad\quad g^{\prime}\left(b^{+}\right)=0$

Since $f(x)$ co-domain is $[1, \infty) f(a) and f(b)$ can never be zero.

Hence it is non derivable at $x = a \& x = b$.

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