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 …
What is the formula to find number of one one function ? Read More »
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 = …
Find the domain of the function f(x) = \(1\over x + 2\). Read More »
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 = …
Find the domain and range of function f(x) = \(x-2\over 3-x\). Read More »
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 …
Find the range of the function \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\) Read More »
