MATHEMATICS (BASIC MATHS) ONLINE QUIZ FOR IIT-JEE

The solution set of the inequality \frac{x}{{x-2}} > \frac{1}{{x + 3}} is :

If x(x + 1) (x – 2) (x – 3) \geqslant 0 then exhaustive set of values of x is :

The set of real values of ‘x’ satisfying \frac{2}{{{x^2}-x + 1}}-\frac{1}{{x + 1}}-\frac{{2x-1}}{{{x^3} + 1}}\,\, \geqslant \,\,0

The equation :

\frac{{2{x^3}-3{x^2} + x + 1}}{{2{x^3}-3{x^2}-x-1}} = \frac{{3{x^3}-{x^2} + 5x-13}}{{3{x^3}-{x^2}-5x + 13}} has

The set of values of x satisfying \frac{{{{(x-3)}^4}{{(5x-9)}^2}(x + 1)}}{{{{(2x + 1)}^6}(x-3)}}\,\, \geqslant \,0, is

\frac{{{{\left( {x-4} \right)}^{2005}}{{(x + 8)}^{2008}}(x + 1)}}{{{x^{2006}}{{(x-2)}^3}{{(x + 3)}^5}(x-6){{(x + 9)}^{2010}}}} \leqslant 0

The number of solution of the given equation : x\, + \,\frac{1}{{x\, - 2}}\, = \,2\, + \,\frac{1}{{x\, - \,2}}

Smallest positive integral value of x satisfying \frac{{({x^2} - 1)({x^2} - 4){{(x - 3)}^2}(x - 5)}}{{(x + 3)(x - 6)}} > 0 is :

The solution set of  \frac{{2x}}{{2{x^2} + 5x + 2}} > \frac{1}{{x + 1}} is :

The values of x for which \frac{1}{{x - 1}} > \frac{1}{{x + 1}} belongs to