NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

WB JEE 2009

MCQ (Single Correct Answer)

$$f(x) = x + |x|$$ 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

MCQ (Single Correct Answer)

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

MCQ (Single Correct Answer)

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

MCQ (Single Correct Answer)

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.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12