Solution : \(81^{log_3 5}\) + \(3^{3log_9 36}\) + \(3^{4log_9 7}\) \(\implies\) \(3^{4log_3 5}\) + \(3^{log_3 {(36)}^{3/2}}\) + \(3^{log_3 {7}^2}\) = 625 + 216 + 49 = 890. Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Find the value of \(2log{2\over 5}\) + \(3log{25\over 8}\) – \(log{625\over 128}\). If \(log_a x\) […]
Solution : \(\displaystyle{\lim_{x \to \infty}}\)\(x^2 + x + 1\over {3x^2 + 2x – 5}\) It is (\(\infty\over \infty\) form), Put x = \(1\over y\) = \(\displaystyle{\lim_{y \to 0}}\) \(1 + y + y^2\over {3 + 2y – 5y^2}\) = \(1\over 3\) Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Evaluate : \(\displaystyle{\lim_{x \to
Solution : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cosx\over {sinx(1-cosx)}\) = \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cosx(1 + cosx)\over {sinxsin^2x}\) = \(\displaystyle{\lim_{x \to 0}}\) \({x^3\over sin^3x}.cosx(1 + cosx)\) = 2 Similar Questions Evaluate the limit : \(\displaystyle{\lim_{x \to \infty}}\) \(x^2 + x + 1\over {3x^2 + 2x – 5}\) Evaluate : \(\displaystyle{\lim_{x \to \infty}}\) \(({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}\) Evaluate
Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Read More »
Solution : We have, \(tan^{-1}(1)\) + \(cos^{-1}({-1\over 2})\) + \(sin^{-1}({-1\over 2})\) = \(\pi\over 4\) + \(2\pi\over 3\) – \(\pi\over 6\) = \(3\pi\over 4\) Similar Questions Solve the equation : 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\) Prove that : \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over 3}\) + \(tan^{-1}{63\over 16}\) = \(\pi\) Evaluate \(sin^{-1}(sin10)\) Prove that : \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\)
Solution : We have, \(x^2\over 16\) – \(y^2\over 9\) = 1 Compare given equation with \(x^2\over a^2\) – \(y^2\over b^2\) = 1 a = 4 and b = 3 Since the normal to the given hyperbola whose slope is ‘m’, is y = mx \(\mp\) \({m(a^2+b^2)}\over \sqrt{a^2 – m^2b^2}\) Hence, required equation of normal is
Find the normal to the hyperbola \(x^2\over 16\) – \(y^2\over 9\) = 1 whose slope is 1. 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 : Let the equation of the ellipse be \(x^2\over a^2\) + \(y^2\over b^2\) = 1. Then, coordinates of the foci are \((\pm ae, 0)\). Therefore, ae = 2 \(\implies\) a = 4 We have \(b^2\) = \(a^2(1 – e^2)\) \(\implies\) \(b^2\) =12 Thus, the equation of the ellipse is \(x^2\over 16\) + \(y^2\over 12\)
Solution : Let the equation of the required ellipse be \(x^2\over a^2\) + \(y^2\over b^2\) = 1 ……….(i) Since the vertices of the ellipse are on y-axis. So, the coordinates of the vertices are \((0, \pm b)\). \(\therefore\) b = 10 Now, \(a^2\) = \(b^2(1 – e^2)\)
