1
JEE Main 2013 (Offline)
+4
-1
If $$y = \sec \left( {{{\tan }^{ - 1}}x} \right),$$ then $${{{dy} \over {dx}}}$$ at $$x=1$$ is equal to :
A
$${1 \over {\sqrt 2 }}$$
B
$${1 \over 2}$$
C
$$1$$
D
$$\sqrt 2$$
2
AIEEE 2011
+4
-1
$${{{d^2}x} \over {d{y^2}}}$$ equals:
A
$$- {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
B
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$
C
$$- \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
D
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$
3
AIEEE 2010
+4
-1
Let $$f:\left( { - 1,1} \right) \to R$$ be a differentiable function with $$f\left( 0 \right) = - 1$$ and $$f'\left( 0 \right) = 1$$. Let $$g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}$$. Then $$g'\left( 0 \right) =$$
A
$$-4$$
B
$$0$$
C
$$-2$$
D
$$4$$
4
AIEEE 2009
+4
-1
Let $$y$$ be an implicit function of $$x$$ defined by $${x^{2x}} - 2{x^x}\cot \,y - 1 = 0$$. Then $$y'(1)$$ equals
A
$$1$$
B
$$\log \,2$$
C
$$-\log \,2$$
D
$$-1$$
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