Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :

Let f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right), 0 < x < 1. Then :

If y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right), then {{dy} \over {dx}} at x = {{5\pi } \over 6} is :

The population P = P(t) at time 't' of a certain species follows the differential equation {{dP} \over {dt}} = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is :

If a curve y = f(x) passes through the point (1, 2) and satisfies x {{dy} \over {dx}} + y = b{x^4}, then for what value of b, \int\limits_1^2 {f(x)dx = {{62} \over 5}}?

If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is {{{x^2} - 4x + y + 8} \over {x - 2}}, then this curve also passes through the point :

The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after {k \over {{{\log }_e}\left( {{6 \over 5}} \right)}} hours, then {\left( {{k \over {{{\log }_e}2}}} \right)^2} is equal to :

If y = y(x) is the solution of the differential equation, {{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0, then the maximum value of the function y(x) over R is equal to:

If y = y(x) is the solution of the differential equation {{dy} \over {dx}} + (tan x) y = sin x, 0 \le x \le {\pi \over 3}, with y(0) = 0, then y\left( {{\pi \over 4}} \right) equal to :

Let C1 be the curve obtained by the solution of differential equation 2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0. Let the curve C2 be the solution of {{2xy} \over {{x^2} - {y^2}}} = {{dy} \over {dx}}. If both the curves pass through (1, 1), then the area enclosed by the curves C1 and C2 is equal to :

Which of the following is true for y(x) that satisfies the differential equation {{dy} \over {dx}} = xy $$-$$ 1 + x $$-$$ y; y(0) = 0 :

If the curve y = y(x) is the solution of the differential equation $$2(x^2+x^{5/4})dy-y(x+x^{1/4})dx=2x^{9/4}dx$$, x > 0 which passes through the point \left( {1,1 - {4 \over 3}{{\log }_e}2} \right), then the value of y(16) is equal to :

Let y = y(x) be the solution of the differential equation \cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0. Then, y\left( {{\pi \over 3}} \right) is equal to :

The differential equation satisfied by the system of parabolas y2 = 4a(x + a) is :

Let y = y(x) be the solution of the differential equation {{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right), 0 < x < 2.1, with y(2) = 0. Then the value of {{dy} \over {dx}} at x = 1 is equal to :

Let y = y(x) be the solution of the differential equation x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx, $$-1\le x\le 1$$, y\left( {{1 \over 2}} \right) = {\pi \over 6}. Then the area of the region bounded by the curves x = 0, x = {1 \over {\sqrt 2 }} and y = y(x) in the upper half plane is :

Let y = y(x) be the solution of the differential equation {e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0, y(1) = $$-$$1. Then the value of (y(3))2 is equal to :

Let y = y(x) satisfies the equation {{dy} \over {dx}} - |A| = 0, for all x > 0, where A = \left[ {\matrix{ y & {\sin x} & 1 \cr 0 & { - 1} & 1 \cr 2 & 0 & {{1 \over x}} \cr } } \right]. If $$y(\pi )=\pi +2$$, then the value of y\left( {{\pi \over 2}} \right) is :

Let y = y(x) be the solution of the differential equation $$\cos e{c}^{2}xdy+2dx=(1+y\cos 2x)\cos e{c}^{2}xdx$$, with y\left( {{\pi \over 4}} \right) = 0. Then, the value of $$(y(0)+1{)}^2$$ is equal to :

Let y = y(x) be the solution of the differential equation {{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0 then, the minimum value of $$y(x),x\in \left(-\sqrt{2},\sqrt{2}\right)$$ is equal to :