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 $\int\limits_0^1 {\frac{{{x^{e – 1}} – 1}}{{\ln x}}dx} = ?$ →
We have, $f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) – f(x)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2xh + f(h)}}{h}$
$ = 2x + \mathop {\lim }\limits_{h \to 0} \frac{{f(h)}}{h}$ Continue reading f is differentiable function such that
$f(x + y) – f(x) = f(y) + 2xy$
$x,y \in R$
$f(x) = ?$ →
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,
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos x + \cos 3x + \sin 2x)}^2}}}{{{{(\sin x – \sin 3x + 2\cos 2x)}^3}}}$ →
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 $\scriptscriptstyle \frac{1}{{102 \times 101 \times 100}} + \frac{1}{{101 \times 100 \times 99}} + ……… + \frac{1}{{3 \times 2 \times 1}} = ?$ →
Consider a particle in circular motion as shown in the figure. The position vector sweeps equal area in equal time. Select correct option(s).
(A) the speed is constant
(B) acceleration is constant
(C) $\left| {\overrightarrow {{a_r}} } \right|$ = constant
(D) $a_t = 0$ Continue reading Circular Motion Radius Area →
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}$
$\therefore{({a^2}{\sec ^2}\theta + {b^2}{\rm{cose}}{{\rm{c}}^2}\theta )_{\min }} = {a^2} + {b^2} + 2|ab|$
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 $
Physics, Math etc. by IIT Alumnus Online and Bhopal
