Function Questions

Solution : If A and B are two sets having m and n elements respectively such that m $\le$ n, then the total number of one-one functions from A to B is $^nC_m \times m!$ where m! is m factorial. For example, Let set A have 3 elements and set B have 4 elements, then […]

Solution : We have, f(x) = $1\over x + 2$ Clearly f(x) assumes real values for all real values for all x except for the values of x satisfying x + 2 = 0  i.e. x = -2. Hence, Domain(f) = R – {-2} Similar Questions If y = 2[x] + 3 & y =

Solution : y = 3[x – 2] + 5 = 3[x] – 1 so 3[x] – 1 = 2[x] + 3 [x] = 4 $\implies$ 4 $\le$ x < 5 then y = 11 so x + y will lie in the interval [15, 16) so [x + y] = 15 Similar Questions Find the

Solution : we have,  f(x) = $x-2\over 3-x$ Domain of f : Clearly f(x) is defined for all x satisfying 3 – x $\ne$ 0 i.e. x $\ne$ 3 Hence, Domain of f is R – {3} Range of f : Let y = f(x), i.e.  y = $x-2\over 3-x$ $\implies$ 3y – xy =

Solution : f(x) = $e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}$ Period of x – [x] = 1 Period of $|cos\pi x|$ = 1 Period of $|cos2\pi x|$ = $1\over 2$ ………………………………. Period of $|cosn\pi x|$ = $1\over n$ So period of f(x) will be L.C.M of all period = 1. Similar Questions If y

Solution : Given f(x) = $log_a(x + \sqrt{(x^2+1)})$ f'(x) = $log_ae\over {\sqrt{1+x^2}}$ > 0 which is strictly increasing functions. Thus, f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible. Interchanging x & y $\implies$  $log_a(y + \sqrt{(y^2+1)})$ = x $\implies$  $y + \sqrt{(y^2+1)}$ = $a^x$ ……..(1) and  $\sqrt{(y^2+1)}$ –

Solution : Now 1 $\le$ $16sin^2x$ + 1) $\le$ 17 0 $\le$ $log_2(16sin^2x+1)$ $\le$ $log_217$ 2 – $log_217$ $\le$ 2 – $log_2(16sin^2x+1)$ $\le$ 2 Now consider 0 < 2 – $log_2(16sin^2x+1)$ $\le$ 2 -$\infty$ < $log_{\sqrt{2}}(2-log_2(16sin^2x+1))$ $\le$ $log_{\sqrt{2}}2$ = 2 the range is (-$\infty$, 2] Similar Questions If y = 2[x] + 3 & y