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IIT-JEE 2011 Paper 2 Offline

MCQ (More than One Correct Answer)

If $$f(x) = \left\{ {\matrix{ { - x - {\pi \over 2},} & {x \le - {\pi \over 2}} \cr { - \cos x} & { - {\pi \over 2} < x \le 0} \cr {x - 1} & {0 < x \le 1} \cr {\ln x} & {x > 1} \cr } } \right.$$, then

A
f(x) is continuous at x = $$-$$ $$\pi$$/2.
B
f(x) is not differentiable at x = 0.
C
f(x) is differentiable at x = 1.
D
f(x) is differentiable at x = $$-$$3/2.

Explanation

$$\mathop {\lim }\limits_{x \to {{{\pi ^ - }} \over 2}} f(x) = 0 = f( - \pi /2)$$

$$\mathop {\lim }\limits_{x \to {{{\pi ^ + }} \over 2}} f(x) = \cos \left( { - {\pi \over 2}} \right) = 0$$

$$f'(x) = \left\{ {\matrix{ { - 1,} & {x \le - \pi /2} \cr {\sin x,} & { - \pi /2 < x \le 0} \cr {1,} & {0 < x \le 1} \cr {1/x,} & {x > 1} \cr } } \right.$$

Clearly, f(x) is not differentiable at x = 0 as f'(0$$-$$) = 0 and f'(0+) = 1.

f(x) is differentiable at x = 1 as $$f'({1^{ - 1}}) = f'({1^ + }) = 1$$.

2

IIT-JEE 2011 Paper 1 Offline

MCQ (More than One Correct Answer)

Let f : R $$\to$$ R be a function such that $$f(x + y) = f(x) + f(y),\,\forall x,y \in R$$. If f(x) is differentiable at x = 0, then

A
f(x) is differentiable only in a finite interval containing zero.
B
f(x) is continuous $$\forall x \in R$$.
C
f'(x) is constant $$\forall x \in R$$.
D
f(x) is differentiable except at finitely many points.

Explanation

Set x = 0 in the functional equation to obtain

$$f(0) = f(0) + f(0)$$ $$\therefore$$ $$f(0) = 0$$

$$f'(x) = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x + 0)} \over h} = \mathop {\lim }\limits_{h \to 0} {{f(x) + f(h) - f(x) - f(0)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over h} = f'(0)$$

Thus, $$f'(x) = \lambda $$ (say). Also $$f(x) = \lambda x + \mu $$

As $$f(x) = 0$$ we have $$\mu = 0$$ $$\therefore$$ $$f(x) = \lambda x$$.

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