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 »
Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4 \(\therefore\) mx – 2 = -1 \(\implies\) m = \(1\over 2\) Since \(3x^2+4y^2\) = 12 or \(x^2\over 4\) + \(y^2\over 3\) = 1 Comparing this with \(x^2\over a^2\) + \(y^2\over b^2\) = …
Solution : \(\because\) Equation of ellipse is \(9x^2 + 16y^2\) = 144 or \(x^2\over 16\) + \({(y-3)}^2\over 9\) = 1 comparing this with \(x^2\over a^2\) + \(y^2\over b^2\) = 1 then we get \(a^2\) = 16 and \(b^2\) = 9 and comparing the line y = x + k with y = mx + c …
For what value of k does the line y = x + k touches the ellipse \(9x^2 + 16y^2\) = 144. Read More »
Solution : Here S = (2, 3) & S’ is (-2, 3) and b = \(\sqrt{5}\) \(\implies\) SS’ = 4 = 2ae \(\implies\) ae = 2 but \(b^2\) = \(a^2(1-e^2)\) \(\implies\) 5 = \(a^2\) – 4 \(\implies\) a = 3 Hence the equation to major axis is y = 3. Centre of ellipse is midpoint …
Solution : Family of circles is \(x^2 + y^2 – 2x – 4y + 1\) + \(\lambda\)(\(x^2 + y^2 – 1\)) = 0 (1 + \(\lambda\))\(x^2\) + (1 + \(\lambda\))\(y^2\) – 2x – 4y + (1 – \(\lambda\))) = 0 \(x^2 + y^2 – {2\over {1 + \lambda}} x – {4\over {1 + \lambda}}y + …
Solution : Pair of normals are (x + 2y)(x + 3) = 0 \(\therefore\) Normals are x + 2y = 0, x + 3 = 0 Point of intersection of normals is the center of the required circle i.e. \(C_1\)(-3,3/2) and center of the given circle is \(C_2\)(2,3/2) and radius \(r^2\) = \(\sqrt{4 + {9\over …
Solution : Since the normal to the circle always passes through the center, so the equation of the normal will be the line passing through (5,6) & (\(5\over 2\), -1) i.e. y + 1 = \(7\over {5/2}\)(x – \(5\over 2\)) \(\implies\) 5y + 5 = 14x – 35 \(\implies\) 14x – 5y – 40 = …
Solution : 5 different mangoes can be distributed by following ways among 3 children such that each gets at least 1 : Total number of ways : (\(5!\over 3!1!1!2!\) + \(5!\over 2!2!2!\)) \(\times\) 3! Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children = \(3^7\) (as …
Solution : There are 4 odd digits (1, 1, 3, 3) and 4 odd places(first, third, fifth and seventh). At these places the odd digits can be arranged in \(4!\over 2!2!\) = 6 Then at the remaining 3 places, the remaining three digits(2, 2, 4) can be arranged in \(3!\over 2!\) = 3 ways Therefore, …
