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1

### WB JEE 2009

$$f(x) = x + |x|$$ is continuous for

A
x $$\in$$ ($$-$$ $$\infty$$, $$\infty$$)
B
x $$\in$$ ($$-$$ $$\infty$$, $$\infty$$) $$-$$ {0}
C
only x > 0
D
no value of x

## Explanation

For x > 0, f(x) = 2x is a straight line which is continuous.

Also for x < 0, f(x) = 0 is x-axis which is continuous.

Now there is x = 0 at which continuity should be checked.

$$\therefore$$ x > 0, f(0+) = $$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} 2x = 0$$

x < 0, f(0$$-$$) = $$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} 0 = 0$$

Also f(0) = 0 $$\Rightarrow$$ f(0) = f(0) = f(0+)

$$\therefore$$ f(x) is continuous for x $$\in$$ ($$-$$$$\infty$$, $$\infty$$)

2

### WB JEE 2009

Let $$f(x) = {{\sqrt {x + 3} } \over {x + 1}}$$, then the value of $$\mathop {\lim }\limits_{x \to - 3 - 0} f(x)$$ is

A
0
B
does not exist
C
1/2
D
$$-$$1/2

## Explanation

$$\mathop {\lim }\limits_{x \to - 3 - 0} f(x) = \mathop {\lim }\limits_{x \to - 3 - 0} {{\sqrt {x + 3} } \over {x + 1}}$$

Let x = $$-$$3 $$-$$ h, h $$\to$$ 0 and h > 0

$$\therefore$$ $$\mathop {\lim }\limits_{x \to - 3 - 0} f(x) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {( - 3 - h + 3)} } \over { - 3 - h + 1}} = \mathop {\lim }\limits_{h \to 0} {{\sqrt { - h} } \over { - 2 - h}}$$

$$= {{\mathop {\lim }\limits_{h \to 0} \sqrt { - h} } \over {\mathop {\lim }\limits_{h \to 0} - 2 - h}} = {{imaginary\,number} \over { - 2 - 0}}$$ ($$\because$$ h > 0 $$\therefore$$ $$\sqrt { - h}$$ is imaginary)

= is not real number.

3

### WB JEE 2008

A function f(x) is defined as follows for real x

$$f(x) = \left\{ {\matrix{ {1 - {x^2}} & , & {for\,x < 1} \cr 0 & , & {for\,x = 1} \cr {1 + {x^2}} & , & {for\,x > 1} \cr } } \right.$$

Then

A
f(x) is not continuous at x = 1
B
f(x) is continuous but not differentiable at x = 1
C
f(x) is both continuous and differentiable at x = 1
D
f(x) is continuous everywhere but differentiable nowhere.

## Explanation

To check continuity at x = 1

L.H.L. = $$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(1 - h) = \mathop {\lim }\limits_{h \to 0} 1 - {(1 - h)^2}$$

$$= 1 - 1 = 0$$.

R.H.L. $$= \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} 1 + {(1 + h)^2}$$

$$= \mathop {\lim }\limits_{h \to 0} 1 + {(1 + h)^2} = 2$$.

$$\therefore$$ L.H.L. $$\ne$$ R.H.L. $$\ne$$ f(1)

$$\therefore$$ f(x) is not continuous at x = 1.

4

### WB JEE 2008

The $$\mathop {\lim }\limits_{x \to 2} {5 \over {\sqrt 2 - \sqrt x }}$$ is

A
10$$\sqrt2$$
B
+ $$\infty$$
C
$$-$$ $$\infty$$
D
does not exist

## Explanation

$$\mathop {\lim }\limits_{x \to 2} {5 \over {\sqrt 2 - \sqrt x }}$$

when x < 2, let x = 2 $$-$$ h and h $$\to$$ 0

$$\therefore$$ L.H.L. = $$\mathop {\lim }\limits_{x \to {2^ - }} {5 \over {\sqrt 2 - \sqrt x }} = \mathop {\lim }\limits_{h \to 0} {5 \over {\sqrt 2 - \sqrt {2 - h} }}$$

$$= \mathop {\lim }\limits_{h \to 0} {{5(\sqrt 2 + \sqrt {2 - h} )} \over {2 - 2 + h}} = \mathop {\lim }\limits_{h \to 0} {{5(\sqrt 2 + \sqrt {2 - h} )} \over h} = + \infty$$

When x > 2, let x = 2 + h and h $$\to$$ 0

$$\therefore$$ R.H.L. $$= \mathop {\lim }\limits_{x \to {2^ + }} {5 \over {\sqrt 2 - \sqrt x }} = \mathop {\lim }\limits_{h \to 0} {5 \over {\sqrt 2 - \sqrt {2 + h} }}$$

$$= \mathop {\lim }\limits_{h \to 0} {{5(\sqrt 2 + \sqrt {2 + h} )} \over {2 - 2 + h}} = \mathop {\lim }\limits_{h \to 0} {{5(\sqrt 2 + \sqrt {2 + h} )} \over { - h}} = - \infty$$

$$\because$$ R.H.L. $$\ne$$ L.H.L. $$\therefore$$ limit does not exist.

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