QJEEAdv14Ma6cd2: JEE Advanced 2014 Mathematics - Class 12 JEEAdv14Ma6cd2 Solution: Let be a continuous function and g : R R be defined as g ( x )…

Question

Verified Solution • 2025–26 Session moderate

Let

f:(a,b) \to [1,\infty )

be a continuous function and g : R

\to

R be defined as

g (x) = {\begin{matrix} 0 & , & x < a \\ \int_{a}^{x} f (t) d t & , & a \leq x \leq b \\ \int_{a}^{b} f (t) d t & , & x > b \end{matrix}

Then,

Accepted Answer

Given that

f:(a,b) \to [1,\infty )

and

g (x) = {\begin{matrix} 0 & , & x < a \\ \int_{a}^{x} f (t) d t & , & a \leq x \leq b \\ \int_{a}^{b} f (t) d t & , & x > b \end{matrix}

Now,

g({a^ - }) = 0 = g({a^ + }) = g(a)

[as

g({a^ + }) = \mathop {\lim }\limits_{x \to {a^ + }} \int_a^x {f(t)dt = 0}

and

g(a) = \int_a^a {f(t)dt = 0}

]

g({b^ - }) = g({b^ + }) = g(b) = \int_a^b {f(t)dt}

\Rightarrow

g is continuous,

\forall

x

\in

R.

Now,

g^{'} (x) = {\begin{matrix} 0 & , & x < a \\ f (x) & , & a < x < b \\ 0 & , & x > b \end{matrix}

g'(a^$-$) = 0 but g'(a⁺) = f(a)

\ge

1

[

\because

Range of f(x) is [1,

\infty

),

\forall

x

\in

[a, b]]

\Rightarrow

g is non-differentiable at x = a

and g'(b⁺) = 0

but g'(b^$-$) = f(b)

\ge

1

\Rightarrow

g is not differentiable at x = b.

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