1
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
If   x2 + y2 + sin y = 4, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at the point ($$-$$2,0) is :
A
$$-$$ 34
B
$$-$$ 32
C
4
D
$$-$$ 2
2
JEE Main 2017 (Online) 9th April Morning Slot
+4
-1
Let f be a polynomial function such that

f (3x) = f ' (x) . f '' (x), for all x $$\in$$ R. Then :
A
f (2) + f ' (2) = 28
B
f '' (2) $$-$$ f ' (2) = 0
C
f '' (2) $$-$$ f (2) = 4
D
f (2) $$-$$ f ' (2) + f '' (2) = 10
3
JEE Main 2017 (Online) 8th April Morning Slot
+4
-1
If y = $${\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}},$$

then (x2 $$-$$ 1) $${{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is equal to :
A
125 y
B
124 y2
C
225 y2
D
225 y
4
JEE Main 2017 (Offline)
+4
-1
If for $$x \in \left( {0,{1 \over 4}} \right)$$, the derivatives of

$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals
A
$${{{3x\sqrt x } \over {1 - 9{x^3}}}}$$
B
$${{{3x} \over {1 - 9{x^3}}}}$$
C
$${{3 \over {1 + 9{x^3}}}}$$
D
$${{9 \over {1 + 9{x^3}}}}$$
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