Maths Questions – Page 33

Evaluate : \(\int\) \(dx\over {3sinx + 4cosx}\)

Solution : I = \(\int\) \(dx\over {3sinx + 4cosx}\) = \(\int\) \(dx\over {3[{2tan{x\over 2}\over {1+tan^2{x\over 2}}}] + 4[{1-tan^2{x\over 2}\over {1+tan^2{x\over 2}}}]}\) = \(\int\) \(sec^2{x\over 2}dx\over {4+6tan{x\over 2}-4tan^2{x\over 2}}\) let \(tan{x\over 2}\) = t, \(\therefore\) \({1\over 2}sec^2{x\over 2}\)dx = dt so I = \(\int\) \(2dt\over {4+6t-4t^2}\) = \(1\over 2\) \(\int\) \(dt\over {1-(t^2-{3\over 2}t})\) = \(1\over 2\) […]

Evaluate : \(\int\) \(dx\over {3sinx + 4cosx}\) Read More »

Find the asymptotes of the hyperbola \(2x^2 + 5xy + 2y^2 + 4x + 5y\) = 0. Find also the general equation of all the hyperbolas having the same set of asymptotes.

Solution : Let \(2x^2 + 5xy + 2y^2 + 4x + 5y + k\) = 0 be asymptotes. This will represent two straight line so \(abc + 2fgh – af^2 – bg^2 – ch^2\) = 0 \(\implies\) 4k + 25 – \(25\over 2\) – 8 – \(25\over 4\)k = 0 \(\implies\) k = 2 \(\implies\)

Find the asymptotes of the hyperbola \(2x^2 + 5xy + 2y^2 + 4x + 5y\) = 0. Find also the general equation of all the hyperbolas having the same set of asymptotes. Read More »

Find the equation of the tangent to the hyperbola \(x^2 – 4y^2\) = 36 which is perpendicular to the line x – y + 4 = 0

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Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line x – y = 0 \(\therefore\) m\(\times\)1 = -1 \(\implies\) m = -1 Since \(x^2-4y^2\) = 36 or \(x^2\over 36\) – \(y^2\over 9\) = 1 Comparing this with \(x^2\over a^2\) – \(y^2\over b^2\) = 1 \(\therefore\); \(a^2\)

Find the equation of the tangent to the hyperbola \(x^2 – 4y^2\) = 36 which is perpendicular to the line x – y + 4 = 0 Read More »

The eccentricity of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is

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Solution : Equation of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is \(-x^2-3y^2\) = 1 \(\implies\) \(-x^2\over 1\) + \(y^2\over {1/3}\) = 1 Here \(a^2\) = 1, \(b^2\) = \(1\over 3\) \(\therefore\) eccentricity e = \(\sqrt{1 + a^2/b^2}\) = \(\sqrt{1+3}\) = 2 Similar Questions Angle between asymptotes of hyperbola xy=8 is Find the

The eccentricity of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is Read More »

If the foci of a hyperbola are foci of the ellipse \(x^2\over 25\) + \(y^2\over 9\) = 1. If the eccentricity of the hyperbola be 2, then its equation is :

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Solution : For ellipse e = \(4\over 5\), so foci = (\(\pm\)4, 0) for hyperbola e = 2, so a = \(ae\over e\) = \(4\over 2\) = 2, b = \(2\sqrt{4-1}\) = 2\(\sqrt{3}\) Hence the equation of the hyperbola is \(x^2\over 4\) – \(y^2\over 12\) = 1 Similar Questions Find the equation of the ellipse

If the foci of a hyperbola are foci of the ellipse \(x^2\over 25\) + \(y^2\over 9\) = 1. If the eccentricity of the hyperbola be 2, then its equation is : Read More »

Find the period of the function f(x) = \(e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}\)

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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

Find the period of the function f(x) = \(e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}\) Read More »

Find the inverse of the function f(x) = \(log_a(x + \sqrt{(x^2+1)})\); a > 1 and assuming it to be an onto function.

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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 \(\sqrt{(y^2+1)}\) –

Find the inverse of the function f(x) = \(log_a(x + \sqrt{(x^2+1)})\); a > 1 and assuming it to be an onto function. Read More »

Find the range of the function \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\)

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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 »

Find the equation of the tangents to the ellipse \(3x^2+4y^2\) = 12 which are perpendicular to the line y + 2x = 4.

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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 \(x^2\over a^2\) + \(y^2\over b^2\) =

Find the equation of the tangents to the ellipse \(3x^2+4y^2\) = 12 which are perpendicular to the line y + 2x = 4. Read More »

For what value of k does the line y = x + k touches the ellipse \(9x^2 + 16y^2\) = 144.

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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 »