1
JEE Main 2016 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If the function

f(x) = $$\left\{ {\matrix{ { - x} & {x < 1} \cr {a + {{\cos }^{ - 1}}\left( {x + b} \right),} & {1 \le x \le 2} \cr } } \right.$$

is differentiable at x = 1, then $${a \over b}$$ is equal to :
A
$${{\pi - 2} \over 2}$$
B
$${{ - \pi - 2} \over 2}$$
C
$${{\pi + 2} \over 2}$$
D
$$ - 1 - {\cos ^{ - 1}}\left( 2 \right)$$
2
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
$${{27} \over 10}$$
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   $$2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,$$

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
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