1
JEE Main 2016 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If   f(x) is a differentiable function in the interval (0, $$\infty $$) such that f (1) = 1 and

$$\mathop {\lim }\limits_{t \to x} $$   $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$ equal to :
A
$${{13} \over 6}$$
B
$${{23} \over 18}$$
C
$${{25} \over 9}$$
D
$${{31} \over 18}$$
2
JEE Main 2016 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If   $$2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,$$

then $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$ is equalto :
A
log4
B
$${\pi \over 2}$$ + log2
C
log2
D
$${\pi \over 2}$$ $$-$$ log4
3
JEE Main 2016 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The minimum distance of a point on the curve y = x2−4 from the origin is :
A
$${{\sqrt {19} } \over 2}$$
B
$$\sqrt {{{15} \over 2}} $$
C
$${{\sqrt {15} } \over 2}$$
D
$$\sqrt {{{19} \over 2}} $$
4
JEE Main 2016 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If   $$\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k,$$

where k is a constant of integration, then A + B +C equals :
A
$${{21} \over 5}$$
B
$${{16} \over 5}$$
C
$${{7} \over 10}$$
D
$${{27} \over 10}$$
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