Logarithm Examples

Here you will learn some logarithm examples for better understanding of logarithm concepts. Example 1 : 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 \(({x\over y})^{b-c}\) \(\times\) \(({y\over z})^{c-a}\) \(\times\) \(({z\over x})^{a-b}\). Solution : \(log_e

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Limits Examples

Here you will learn some limits examples for better understanding of limit concepts. Example 1 : If \(\displaystyle{\lim_{x \to \infty}}\)(\({x^3+1\over x^2+1}-(ax+b)\)) = 2, then find the value of a and b. 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 \(\implies\) \(\displaystyle{\lim_{x \to \infty}}\)\(x(1-a)-b-{a\over x}+{(1-b)\over x^2}\over 1+{1\over x^2}\) =

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Inverse Trignometric Function Examples

Here you will learn some inverse trignometric function examples for better understanding of inverse trigonometric function concepts. Example 1 : Find the value of \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\). 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

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Integration Examples

Here you will learn some integration examples for better understanding of integration concepts. Example 1 : 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 \(tan{x\over 2}\) =

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Function Examples

Here you will learn some function examples for better understanding of function concepts. Example 1 : Find the range of the given function \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\) Solution : Now 1 \(\le\) \(16sin^2x\) + 1) \(\le\) 17 \(\therefore\)   0 \(\le\) \(log_2(16sin^2x+1)\) \(\le\) \(log_217\) \(\therefore\)   2 – \(log_217\) \(\le\) 2 – \(log_2(16sin^2x+1)\) \(\le\) 2 Now consider 0

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