1
JEE Main 2016 (Offline)
+4
-1
Out of Syllabus
Consider :
f $$\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).$$

A normal to $$y =$$ f$$\left( x \right)$$ at $$x = {\pi \over 6}$$ also passes through the point:

A
$$\left( {{\pi \over 6},0} \right)$$
B
$$\left( {{\pi \over 4},0} \right)$$
C
$$(0,0)$$
D
$$\left( {0,{{2\pi } \over 3}} \right)$$
2
JEE Main 2015 (Offline)
+4
-1
Out of Syllabus
The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0$$, at $$(1,1)$$
A
meets the curve again in the third quadrant.
B
meets the curve again in the fourth quadrant.
C
does not meet the curve again.
D
meets the curve again in the second quadrant.
3
JEE Main 2015 (Offline)
+4
-1
Let $$f(x)$$ be a polynomial of degree four having extreme values
at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :
A
$$0$$
B
$$4$$
C
$$-8$$
D
$$-4$$
4
JEE Main 2014 (Offline)
+4
-1
If $$x=-1$$ and $$x=2$$ are extreme points of $$f\left( x \right) = \alpha \,\log \left| x \right|+\beta {x^2} + x$$ then
A
$$\alpha = 2,\beta = - {1 \over 2}$$
B
$$\alpha = 2,\beta = {1 \over 2}$$
C
$$\alpha = - 6,\beta = {1 \over 2}$$
D
$$\alpha = - 6,\beta = -{1 \over 2}$$
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