1
WB JEE 2022
MCQ (Single Correct Answer)
+1
-0.25
Change Language

Let f be derivable in [0, 1], then

A
there exists $$c \in (0,1)$$ such that $$\int\limits_0^c {f(x)dx = (1 - c)f(c)} $$
B
there does not exist any point $$d \in (0,1)$$ for which $$\int\limits_0^d {f(x)dx = (1 - d)f(d)} $$
C
$$\int\limits_0^c {f(x)dx} $$ does not exist, for any $$c \in (0,1)$$
D
$$\int\limits_0^c {f(x)dx} $$ is independent of $$c,c \in (0,1)$$
2
WB JEE 2022
MCQ (Single Correct Answer)
+1
-0.25
Change Language

Let $$\int {{{{x^{{1 \over 2}}}} \over {\sqrt {1 - {x^3}} }}dx = {2 \over 3}g(f(x)) + c} $$ ; then

(c denotes constant of integration)

A
$$f(x) = \sqrt x ,g(x) = {x^{{3 \over 2}}}$$
B
$$f(x) = {x^{{3 \over 2}}},g(x) = {\sin ^{ - 1}}x$$
C
$$f(x) = \sqrt x ,g(x) = {\sin ^{ - 1}}x$$
D
$$f(x) = {\sin ^{ - 1}}x,g(x) = {x^{{3 \over 2}}}$$
3
WB JEE 2022
MCQ (Single Correct Answer)
+1
-0.25
Change Language

The value of $$\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{\sin x}}} \over {{{(\cos x)}^{\sin x}} + {{(\sin x)}^{\cos x}}}}dx} $$ is

A
$${\pi \over 4}$$
B
0
C
$${\pi \over 2}$$
D
$${1 \over 2}$$
4
WB JEE 2022
MCQ (Single Correct Answer)
+1
-0.25
Change Language

Let $$\mathop {\lim }\limits_{ \in \to 0 + } \int\limits_ \in ^x {{{bt\cos 4t - a\sin 4t} \over {{t^2}}}dt = {{a\sin 4x} \over x} - 1,\left( {0 < x < {\pi \over 4}} \right)} $$. Then a and b are given by

A
$$a = 2,b = 2$$
B
$$a = {1 \over 4},b = 1$$
C
$$a = - 1,b = 4$$
D
$$a = 2,b = 4$$
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