1
WB JEE 2011
MCQ (Single Correct Answer)
+1
-0.25

If the function $$f(x) = \left\{ {\matrix{ {{{{x^2} - (A + 2)x + A} \over {x - 2}},} & {for\,x \ne 2} \cr {2,} & {for\,x = 2} \cr } } \right.$$ is continuous at x = 2, then

A
A = 0
B
A = 1
C
A = $$-$$1
D
A = 2
2
WB JEE 2011
MCQ (Single Correct Answer)
+1
-0.25

$$f(x) = \left\{ {\matrix{ {[x] + [ - x],} & {when\,x \ne 2} \cr {\lambda ,} & {when\,x = 0} \cr } } \right.$$

If f(x) is continuous at x = 2, the value of $$\lambda$$ will be

A
$$-$$ 1
B
1
C
0
D
2
3
WB JEE 2011
MCQ (Single Correct Answer)
+1
-0.25

For the function $$f(x) = {e^{\cos x}}$$, Rolle's theorem is

A
applicable when $${\pi \over 2} \le x \le {{3\pi } \over 2}$$
B
applicable when $$0 \le x \le {\pi \over 2}$$
C
applicable when $$0 \le x \le \pi $$
D
applicable when $${\pi \over 4} \le x \le {\pi \over 2}$$
4
WB JEE 2011
MCQ (Single Correct Answer)
+1
-0.25

$$f(x) = \left\{ {\matrix{ {0,} & {x = 0} \cr {x - 3,} & {x > 0} \cr } } \right.$$

The function f(x) is

A
increasing when x $$\ge$$ 0
B
strictly increasing when x > 0
C
strictly increasing at x = 0
D
not continuous at x = 0 and so it is not increasing when x > 0
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