Solution : We have \(2^{x+2}\) > \(2^{-2/x}\) Since the base 2 > 1, we have x + 2 > \(-2\over x\) (the sign of inequality is retained) Now, x + 2 + \(-2\over x\) > 0 \(\implies\) \({x^2 + 2x + 2}\over x\) > 0 \(\implies\) \(({x+1})^2 + 1\over x\) > 0 \(\implies\) x \(\in\) …
Solve for x : \(2^{x+2}\) > \(({1\over 4})^{1/x}\) Read More »
Solution : \(log_a x\) = p \(\implies\) \(a^p\) = x \(\implies\) a = \(x^{1/p}\) Similarly \(b^q\) = \(x^2\) \(\implies\) b = \(x^{2/q}\) Now, \(log_x \sqrt{ab}\) = \(log_x \sqrt{x^{1/p}x^{2/q}}\) = \(log_x x^{({1\over p}+{2\over q}){1\over 2}}\) = \(1\over {2p}\) + \(1\over q\). Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Evaluate the …
If \(log_a x\) = p and \(log_b {x^2}\) = q then \(log_x \sqrt{ab}\) is equal to Read More »
Solution : \(log_e x\) – \(log_e y\) = a \(\implies\) \(log_e {x\over y}\) = a \(\implies\) \(x\over y\) = \(e^a\) \(log_e y\) – \(log_e z\) = b \(\implies\) \(log_e {y\over z}\) = b \(\implies\) \(y\over z\) = \(e^b\) \(log_e z\) – \(log_e x\) = c \(\implies\) \(log_e {z\over x}\) = c \(\implies\) \(z\over x\) = …
Solution : Here f(x) = \({7x^2+1\over 5x^2-1}\) \(\phi\)(x) = \({x^5\over {1-x^3}}\) = \(x^2x^3\over 1-x^3\) = \(x^2\over {1\over x^3}-1\) \(\therefore\) \(\displaystyle{\lim_{x \to \infty}}\) f(x) = \(7\over 5\) & \(\displaystyle{\lim_{x \to \infty}}\) \(\phi\)(x) \(\rightarrow\) – \(\infty\) \(\implies\) \(\displaystyle{\lim_{x \to \infty}}\) \((f(x))^{\phi (x)}\) = \(({7\over 5})^{-\infty}\) = 0 Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) …
Solution : \(\displaystyle{\lim_{x \to 0}}\) \(xln(1+2tanx)\over 1-cosx\) = \(\displaystyle{\lim_{x \to 0}}\) \(xln(1+2tanx)\over {1-cosx\over x^2}.x^2\).\(2tanx\over 2tanx\) = 4 Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Evaluate : \(\displaystyle{\lim_{x \to \infty}}\) \(({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}\) Evaluate the limit : \(\displaystyle{\lim_{x \to \infty}}\) \(x^2 + x + 1\over {3x^2 + 2x – 5}\) Evaluate : \(\displaystyle{\lim_{x …
Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(xln(1+2tanx)\over 1-cosx\) Read More »
Solution : \(\displaystyle{\lim_{x \to 0}}\) \(2(sin(2+x)-sin2)+xsin(2+x)\over x\) = \(\displaystyle{\lim_{x \to 0}}\)(\(2.2.cos(2+{x\over 2})sin{x\over 2}\over x\) + sin(2+x)) = \(\displaystyle{\lim_{x \to 0}}\)\(2cos(2+{x\over 2})sin{x\over 2}\over {x\over 2}\) + \(\displaystyle{\lim_{x \to 0}}\)sin(2+x) = 2cos2 + sin2 Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Evaluate : \(\displaystyle{\lim_{x \to \infty}}\) \(({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}\) Evaluate : \(\displaystyle{\lim_{x \to 0}}\) …
Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \((2+x)sin(2+x)-2sin2\over x\) Read More »
Solution : \(\displaystyle{\lim_{x \to \infty}}\)(\({x^3+1\over x^2+1}-(ax+b)\)) = 2 \(\implies\) \(\displaystyle{\lim_{x \to \infty}}\)\(x^3(1-a)-bx^2-ax+(1-b)\over x^2+1\) = 2 \(\implies\) \(\displaystyle{\lim_{x \to \infty}}\)\(x(1-a)-b-{a\over x}+{(1-b)\over x^2}\over 1+{1\over x^2}\) = 2 \(\implies\) 1 – a = 0, -b = 2 \(\implies\) a = 1, b = -2 Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Evaluate : \(\displaystyle{\lim_{x \to …
Solution : Here, 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\) cos(2\(tan^{-1}({2x+1})\)) = x { We Know cos2x = \({1-tan^2x\over {1+tan^2x}}\)} \(\therefore\) \({{1-{(2x+1)}^2}\over {1-{(2x+1)}^2}}\) = x \(\implies\) (1 – 2x – 1)(1 + 2x + 1) = x(\(4x^2 + 4x + 2\)) \(\implies\) -2x.2(x + 1) = 2x(\(2x^2 + 2x + 1\)) \(\implies\) 2x(\(2x^2 + 2x + 1 + 2x …
Solve the equation : 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\) Read More »
Solution : We have, \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over 3}\) + \(tan^{-1}{63\over 16}\) = \(tan^{-1}{12\over 5}\) + \(tan^{-1}{3\over 4}\) + \(tan^{-1}{63\over 16}\) = \(\pi\) + \(tan^{-1}({{{12\over 5} + {3\over 4}}\over {1 – {12\over 5} \times {3\over 4}}})\) + \(tan^{-1}{63\over 16}\) = \(\pi\) + \(tan^{-1}{63\over (-16)}\) + \(tan^{-1}{63\over 16}\) = \(\pi\) – \(tan^{-1}{63\over 16}\) + \(tan^{-1}{63\over 16}\) …
Solution : We know that \(sin^{-1}(sinx)\) = x, if \(-\pi\over 2\) \(\le\) x \(\le\) \(\pi\over 2\) Here, x = 10 radians which does not lie between -\(\pi\over 2\) and \(\pi\over 2\) But, \(3\pi\) – x i.e. \(3\pi\) – 10 lie between -\(\pi\over 2\) and \(\pi\over 2\) Also, sin(\(3\pi\) – 10) = sin 10 \(\therefore\) \(sin^{-1}(sin10)\) …
