1
IIT-JEE 2005 Mains
MCQ (Single Correct Answer)
+3
-1

Evatuate:

$$\int_\limits{0}^{\pi} e^{|\cos x|}\left[2 \sin \left(\frac{1}{2} \cos x\right)+3 \cos \left(\frac{1}{2} \cos x\right)\right] \sin x ~d x$$

A
$${e \over 5}\left[ {\cos \left( {{1 \over 2}} \right) + \left( {{1 \over 2}} \right)\sin \left( {{1 \over 2}} \right) - 1} \right]$$
B
$$24{e \over 5}\left[ {\cos \left( {{1 \over 2}} \right) + \left( {{1 \over 2}} \right)\sin \left( {{1 \over 2}} \right) - 1} \right]$$
C
$$12{e \over 5}\left[ {\cos \left( {{1 \over 2}} \right) + \left( {{1 \over 2}} \right)\sin \left( {{1 \over 2}} \right) - 1} \right]$$
D
$$5{e \over 5}\left[ {\cos \left( {{1 \over 2}} \right) + \left( {{1 \over 2}} \right)\sin \left( {{1 \over 2}} \right) - 1} \right]$$
2
IIT-JEE 2004 Screening
MCQ (Single Correct Answer)
+3
-0.75
The value of the integral $$\int\limits_0^1 {\sqrt {{{1 - x} \over {1 + x}}} dx} $$ is
A
$${\pi \over 2} + 1$$
B
$${\pi \over 2} - 1$$
C
$$-1$$
D
$$1$$
3
IIT-JEE 2004 Screening
MCQ (Single Correct Answer)
+3
-0.75
If $$f(x)$$ is differentiable and $$\int\limits_0^{{t^2}} {xf\left( x \right)dx = {2 \over 5}{t^5},} $$ then $$f\left( {{4 \over {25}}} \right)$$ equals
A
$$2/5$$
B
$$-5/2$$
C
$$1$$
D
$$5/2$$
4
IIT-JEE 2003 Screening
MCQ (Single Correct Answer)
+3
-0.75
If $$f\left( x \right) = \int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}} dt,$$ then $$f(x)$$ increases in
A
$$(-2, 2)$$
B
no value of $$x$$
C
$$\left( {0,\infty } \right)$$
D
$$\left( { - \infty ,0} \right)$$

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