Solution : Here, tanx + secx = 2cosxย ย ย ย \(\implies\)ย ย ย sinx + 1 = \(2cos^2x\) \(\implies\) \(2sin^2x\) + sinx โ 1 = 0ย ย ย \(\implies\)ย ย sinx = \(1\over 2\), -1 But sinx = -1 \(\implies\) x = \(3\pi\over 2\) for which tanx + secx = 2cosxย is not defined. Thus, sinx = โฆ
Find the number of solutions of tanx + secx = 2cosx in [0, \(2\pi\)]. Read More ยป
Solution : Since, \({1\over 6}sin\theta\), \(cos\theta\) and \(tan\theta\) are in G.P. \(\implies\) \(cos^2\theta\) = \({1\over 6}sin\theta\).\(cos\theta\) \(\implies\) \(6cos^3\theta\) + \(cos^2\theta\) โ 1 = 0 \(\therefore\)ย ย (\(2cos\theta โ 1\))(\(3cos^2\theta\) + \(2cos\theta\) + 1) = 0 \(cos\theta\) = \(1\over 2\)ย ย ย ย (other values are imaginary) \(cos\theta\) = \(cos\pi\over 3\)ย ย \(\theta\) = \(2n\pi \pm {\pi\over 3}\),ย n โฆ
Solution : \(2log{2\over 5}\) + \(3log{25\over 8}\) + \(log{128\over 625}\) = \(log{2^2\over 5^2}\) + \(log({5^2\over 2^3})^3\) + \(log{2^7\over 5^4}\) = \(log({2^2\over 5^2}{5^6\over 2^9}{2^7\over 5^4})\) = log 1 = 0 Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Evaluate the given log : \(81^{l\over {log_5 3}}\) + \(27^{log_9 36}\) + \(3^{4\over โฆ
Find the value of \(2log{2\over 5}\) + \(3log{25\over 8}\) โ \(log{625\over 128}\). Read More ยป
Solution : We have \(2^{x + 2}\) > \(2^{2/x}\). Since the base 2 > 1, we have x + 2 > -\(2\over x\) (the sign of the inequality is retained) Now x + 2 + \(2\over x\)ย \(\implies\) \(x^2 + 2x + 2\over x\) > 0 \(\implies\) \((x + 1)^2 + 1\over x\) > 0ย โฆ
Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Read More ยป
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 \(2log{2\over 5}\) + \(3log{25\over 8}\) โ \(log{625\over 128}\). If \(log_a x\) โฆ
Solution : \(\displaystyle{\lim_{x \to \infty}}\)\(x^2 + x + 1\over {3x^2 + 2x โ 5}\) It is (\(\infty\over \infty\) form),ย ย Put x = \(1\over y\) = \(\displaystyle{\lim_{y \to 0}}\) \(1 + y + y^2\over {3 + 2y โ 5y^2}\) = \(1\over 3\) Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Evaluate : \(\displaystyle{\lim_{x \to โฆ
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 : \(\displaystyle{\lim_{x \to \infty}}\) \(x^2 + x + 1\over {3x^2 + 2x โ 5}\) Evaluate : \(\displaystyle{\lim_{x \to \infty}}\) \(({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}\) Evaluate โฆ
Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Read More ยป
Solution : We have, \(tan^{-1}(1)\) + \(cos^{-1}({-1\over 2})\) + \(sin^{-1}({-1\over 2})\) = \(\pi\over 4\) + \(2\pi\over 3\) โ \(\pi\over 6\) = \(3\pi\over 4\) 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 : \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) โฆ
Solution : We have, \(x^2\over 16\) โ \(y^2\over 9\) = 1 Compare given equation with \(x^2\over a^2\) โ \(y^2\over b^2\) = 1 a = 4 and b = 3 Since the normal to the given hyperbola whose slope is โmโ, isย y = mx \(\mp\) \({m(a^2+b^2)}\over \sqrt{a^2 โ m^2b^2}\) Hence, required equation of normal is โฆ
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 = 0ย i.e. x = -2. Hence, Domain(f) = R โ {-2} Similar Questions If y = 2[x] + 3 & y = โฆ
Find the domain of the function f(x) = \(1\over x + 2\). Read More ยป
