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

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

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
$${1 \over 2}$$
3
WB JEE 2022
MCQ (Single Correct Answer)
+1
-0.25

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
$$a = 2,b = 4$$
4
WB JEE 2022
MCQ (Single Correct Answer)
+1
-0.25

Let $$f(x) = \int\limits_{\sin x}^{\cos x} {{e^{ - {t^2}}}dt}$$. Then $$f'\left( {{\pi \over 4}} \right)$$ equals

A
$$\sqrt {{1 \over e}}$$
B
$$- \sqrt {{2 \over e}}$$
C
$$\sqrt {{2 \over e}}$$
D
$$- \sqrt {{1 \over e}}$$
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