Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is {1 \over 2}. If P(1, $$\beta$$), $$\beta$$ > 0 is a point on this ellipse, then the equation of the normal to it at P is :

If the co-ordinates of two points A and B are $$\left(\sqrt{7},0\right)$$ and $$\left(-\sqrt{7},0\right)$$ respectively and P is any point on the conic, 9x2 + 16y2 = 144, then PA + PB is equal to :

Which of the following points lies on the locus of the foot of perpedicular drawn upon any tangent to the ellipse, {{{x^2}} \over 4} + {{{y^2}} \over 2} = 1 from any of its foci?

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies :

If A = {x $$\in$$ R : |x| < 2} and B = {x $$\in$$ R : |x – 2| $$\ge$$ 3}; then :

A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be:

Let $$\underset{i=1}{\overset{50}{\cup }}{X}_i=\underset{i=1}{\overset{n}{\cup }}{Y}_i=T$$ where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi’s and exactly 6 of sets Yi’s, then n is equal to :

A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be :

Let R1 and R2 be two relation defined as follows : R1 = {(a, b) $$\in$$ R2 : a2 + b2 $$\in$$ Q} and R2 = {(a, b) $$\in$$ R2 : a2 + b2 $$\notin$$ Q}, where Q is the set of all rational numbers. Then :

Consider the two sets : A = {m $$\in$$ R : both the roots of x2 – (m + 1)x + m + 4 = 0 are real} and B = [–3, 5). Which of the following is not true?

If R = {(x, y) : x, y $$\in$$ Z, x2 + 3y2 $$\le$$ 8} is a relation on the set of integers Z, then the domain of R–1 is :

\mathop {\lim }\limits_{x \to a} {{{{\left( {a + 2x} \right)}^{{1 \over 3}}} - {{\left( {3x} \right)}^{{1 \over 3}}}} \over {{{\left( {3a + x} \right)}^{{1 \over 3}}} - {{\left( {4x} \right)}^{{1 \over 3}}}}} ($$a$$ $$\ne$$ 0) is equal to :

Let $$f:\left(0,\infty \right)\to \left(0,\infty \right)$$ be a differentiable function such that f(1) = e and \mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0. If f(x) = 1, then x is equal to :

The function f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr } } \right. is :

If the function f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right. is twice differentiable, then the ordered pair (k1, k2) is equal to :

If $$\alpha$$ is positive root of the equation, p(x) = x2 - x - 2 = 0, then \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}} is equal to :

Let f : R $$\to$$ R be a function defined by f(x) = max {x, x2}. Let S denote the set of all points in R, where f is not differentiable. Then :

For all twice differentiable functions f : R $$\to$$ R, with f(0) = f(1) = f'(0) = 0

\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}} is equal to :