JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 12)
Let $$f:(a,b) \to [1,\infty )$$ be a continuous function and g : R $$\to$$ R be defined as $$g(x) = \left\{ {\matrix{
0 & , & {x < a} \cr
{\int_a^x {f(t)dt} } & , & {a \le x \le b} \cr
{\int_a^b {f(t)dt} } & , & {x > b} \cr
} } \right.$$ Then,
g(x) is continuous but not differentiable at a
g(x) is differentiable on R
g(x) is continuous but not differentiable at b
g(x) is continuous and differentiable at either a or b but not both
Explanation
Given that $$f:(a,b) \to [1,\infty )$$
and $$g(x) = \left\{ {\matrix{ 0 & , & {x < a} \cr {\int_a^x {f(t)dt} } & , & {a \le x \le b} \cr {\int_a^b {f(t)dt} } & , & {x > b} \cr } } \right.$$
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, $$g'(x) = \left\{ {\matrix{ 0 & , & {x < a} \cr {f(x)} & , & {a < x < b} \cr 0 & , & {x > b} \cr } } \right.$$
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.
and $$g(x) = \left\{ {\matrix{ 0 & , & {x < a} \cr {\int_a^x {f(t)dt} } & , & {a \le x \le b} \cr {\int_a^b {f(t)dt} } & , & {x > b} \cr } } \right.$$
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, $$g'(x) = \left\{ {\matrix{ 0 & , & {x < a} \cr {f(x)} & , & {a < x < b} \cr 0 & , & {x > b} \cr } } \right.$$
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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