1
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
Let f : [a, b] $$ \to $$ R be such that f is differentiable in (a, b), f is continuous at x = a and x = b and moreover f(a) = 0 = f(b). Then
A
there exists at least one point c in (a, b) such that f'(c) = f(c)
B
f'(x) = f(x) does not hold at any point in (a, b)
C
at every point of (a, b), f'(x) > f(x)
D
at every point of (a, b), f'(x) < f(x)
2
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
Let f : R $$ \to $$ R be a twice continuously differentiable function such that f(0) = f(1) = f'(0) = 0. Then
A
f''(0) = 0
B
f''(c) = 0 for some c$$ \in $$R
C
if c $$ \ne $$ 0, then f''(c) $$ \ne $$ 0
D
f'(x) > 0 for all x $$ \ne $$ 0
3
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
If $$\int {{e^{\sin x}}} .\left[ {{{x{{\cos }^3}x - \sin x} \over {{{\cos }^2}x}}} \right]dx = {e^{\sin x}}f(x) + c$$, where c is constant of integration, then f(x) is equal to
A
sec x $$-$$ x
B
x $$-$$ sec x
C
tan x $$-$$ x
D
x $$-$$ tan x
4
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
If $$\int {f(x)} \sin x\cos xdx = {1 \over {2({b^2} - {a^2})}}\log (f(x)) + c$$, where c is the constant of integration, then f(x) is equal to
A
$${2 \over {({b^2} - {a^2})\sin 2x}}$$
B
$${2 \over {ab\sin 2x}}$$
C
$${2 \over {({b^2} - {a^2})\cos 2x}}$$
D
$${2 \over {ab\cos 2x}}$$
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