Category Archives: IIT JEE

IIT JEE

$\int\limits_0^1 {\frac{{{x^{e – 1}} – 1}}{{\ln x}}dx} = ?$

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Using the result,

For $I(\lambda ) = \int\limits_a^b {f(x,\lambda )dx} $, $\frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_a^b {\left[ {\frac{\partial }{{\partial \lambda }}f(x,\lambda )} \right]dx} $

Let, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } – 1}}{{\ln x}}dx} $

$\frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_0^1 {\left[ {\frac{\partial }{{\partial \lambda }}\left( {\frac{{{x^\lambda } – 1}}{{\ln x}}} \right)} \right]dx} $ Continue reading $\int\limits_0^1 {\frac{{{x^{e – 1}} – 1}}{{\ln x}}dx} = ?$ →

IIT JEE

Evaluate,

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos x + \cos 3x + \sin 2x)}^2}}}{{{{(\sin x – \sin 3x + 2\cos 2x)}^3}}}$

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The given limit,

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(2\cos 2x\cos x + 2\sin x\cos x)}^2}}}{{{{( – 2\cos 2x\sin x + 2\cos 2x)}^3}}}$

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\cos }^2}x{{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{( – \sin x + 1)}^3}}}$

$\mathop { = \lim }\limits_{x \to \frac{\pi }{2}} \frac{{(1 + \sin x)(1 – \sin x){{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{( – \sin x + 1)}^3}}}$ Continue reading Evaluate,

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos x + \cos 3x + \sin 2x)}^2}}}{{{{(\sin x – \sin 3x + 2\cos 2x)}^3}}}$ →

IIT JEE

$\frac{1}{{102 \times 101 \times 100}} + \frac{1}{{101 \times 100 \times 99}} + ……… + \frac{1}{{3 \times 2 \times 1}} = ?$

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On rewriting we have the series as,

$\frac{1}{{1 \times 2 \times 3}} + \frac{1}{{2 \times 3 \times 4}} + …………….. + \frac{1}{{100 \times 101 \times 102}}$

$ = \sum\limits_{r = 1}^{100} {\frac{1}{{r(r + 1)(r + 2)}}} $

$ = \frac{1}{2}\sum\limits_{r = 1}^{100} {\left[ {\frac{1}{{r(r + 1)}} – \frac{1}{{(r + 1)(r + 2)}}} \right]} $ Continue reading $\frac{1}{{102 \times 101 \times 100}} + \frac{1}{{101 \times 100 \times 99}} + ……… + \frac{1}{{3 \times 2 \times 1}} = ?$ →

IIT JEE

Find the least value of

${a^2}{\sec ^2}\theta + {b^2}{\rm{cose}}{{\rm{c}}^2}\theta $

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The given expression can be written as,

$({a^2}{\sec ^2}\theta – {a^2}) + {a^2} + ({b^2}{\rm{cose}}{{\rm{c}}^2}\theta – {b^2}) + {b^2}$

$ = {a^2}{\tan ^2}\theta + {b^2}{\cot ^2}\theta + {a^2} + {b^2}$

Using A.M. $\geq $ G.M.,

$\frac{{{a^2}{{\tan }^2}\theta + {b^2}{{\cot }^2}\theta }}{2} \ge \sqrt {{a^2}{{\tan }^2}\theta .{b^2}{{\cot }^2}\theta } $

$ \Rightarrow {a^2}{\tan ^2}\theta + {b^2}{\cot ^2}\theta \ge 2|ab|$

$ \Rightarrow {a^2}{\tan ^2}\theta + {b^2}{\cot ^2}\theta + {a^2} + {b^2} \ge 2|ab| + {a^2} + {b^2}$

$\therefore{({a^2}{\sec ^2}\theta + {b^2}{\rm{cose}}{{\rm{c}}^2}\theta )_{\min }} = {a^2} + {b^2} + 2|ab|$

IIT JEE

$\mathop {\lim }\limits_{x \to 0} {e^{\frac{{\tan x – \sin x}}{{{{\sin }^3}x}}}} = ?$

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The given limit = ${e^{\mathop {\lim }\limits_{x \to 0} \frac{{\tan x – \sin x}}{{{{\sin }^3}x}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\cos x}} – 1}}{{{{\sin }^2}x}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{\cos x.{{\sin }^2}x}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{\cos x.(1 – {{\cos }^2}x)}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x.(1 + \cos x)}}}}$

$ = {e^{1/2}} = \sqrt e $