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If \(log_e x\) – \(log_e y\) = a, \(log_e y\) – \(log_e z\) = b & \(log_e z\) – \(log_e x\) = c, then find the value of \(({x\over y})^{b-c}\) \(\times\) \(({y\over z})^{c-a}\) \(\times\) \(({z\over x})^{a-b}\).

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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 \(\implies\) \(z\over x\) = …

If \(log_e x\) – \(log_e y\) = a, \(log_e y\) – \(log_e z\) = b & \(log_e z\) – \(log_e x\) = c, then find the value of \(({x\over y})^{b-c}\) \(\times\) \(({y\over z})^{c-a}\) \(\times\) \(({z\over x})^{a-b}\). Read More »

Evaluate : \(\displaystyle{\lim_{x \to \infty}}\) \(({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}\)

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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 : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) …

Evaluate : \(\displaystyle{\lim_{x \to \infty}}\) \(({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}\) Read More »

Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(xln(1+2tanx)\over 1-cosx\)

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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 »

Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \((2+x)sin(2+x)-2sin2\over x\)

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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 »

If \(\displaystyle{\lim_{x \to \infty}}\)(\({x^3+1\over x^2+1}-(ax+b)\)) = 2, then find the value of a and b.

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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 …

If \(\displaystyle{\lim_{x \to \infty}}\)(\({x^3+1\over x^2+1}-(ax+b)\)) = 2, then find the value of a and b. Read More »

Solve the equation : 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\)

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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 »

Prove that : \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over 3}\) + \(tan^{-1}{63\over 16}\) = \(\pi\)

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Solution : We have, \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over 3}\) + \(tan^{-1}{63\over 16}\) = \(tan^{-1}{12\over 5}\) + \(tan^{-1}{3\over 4}\) + \(tan^{-1}{63\over 16}\) = \(\pi\) + \(tan^{-1}({{{12\over 5} + {3\over 4}}\over {1 – {12\over 5} \times {3\over 4}}})\) + \(tan^{-1}{63\over 16}\) = \(\pi\) + \(tan^{-1}{63\over (-16)}\) + \(tan^{-1}{63\over 16}\) = \(\pi\) – \(tan^{-1}{63\over 16}\) + \(tan^{-1}{63\over 16}\) …

Prove that : \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over 3}\) + \(tan^{-1}{63\over 16}\) = \(\pi\) Read More »

Evaluate \(sin^{-1}(sin10)\)

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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 \(\therefore\)  \(sin^{-1}(sin10)\) …

Evaluate \(sin^{-1}(sin10)\) Read More »

Prove that : \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(sin^{-1}{56\over 65}\)

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Solution : We have, L.H.S. = \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(tan^{-1}{5\over 12}\) + \(tan^{-1}{3\over 4}\) \(\because\) [ \(cos^{-1}{12\over 13}\) = \(tan^{-1}{5\over 12}\) & \(sin^{-1}{3\over 5}\) = \(tan^{-1}{3\over 4}\) ] L.H.S. = \(tan^{-1}({{{5\over 12} + {3\over 4}}\over {1 – {5\over 12}.{3\over 4}}})\) = \(tan^{-1}{56\over 33}\) R.H.S. = \(sin^{-1}{56\over 65}\) = \(tan^{-1}{56\over 33}\) L.H.S = …

Prove that : \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(sin^{-1}{56\over 65}\) Read More »

Find the value of \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\).

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Solution : \(sin^{-1}({-\sqrt{3}\over 2})\) = – \(sin^{-1}({\sqrt{3}\over 2})\) = \(-\pi\over 3\) \(cos^{-1}(cos({7\pi\over 6}))\) = \(cos^{-1}(cos({2\pi – {5\pi\over 6}}))\) = \(cos^{-1}(cos({5\pi\over 6}))\) = \(5\pi\over 6\) Hence \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\) = \(-\pi\over 3\) + \(5\pi\over 6\) = \(\pi\over 2\) Similar Questions Solve the equation : 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\) Prove that : \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over …

Find the value of \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\). Read More »

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