Category Archives: IIT JEE
$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\sin x – 2\sin \frac{x}{2}} \right)}^2} + {{(1 – \cos x)}^3}}}{{\sin x\sin 2x – 8\cos x{{\sin }^2}\frac{x}{2} – \frac{4}{3}{{\sin }^4}x}} = ?$
The given limit $ = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {2\sin \frac{x}{2}\cos \frac{x}{2} – 2\sin \frac{x}{2}} \right)}^2} + {{\left( {2{{\sin }^2}\frac{x}{2}} \right)}^3}}}{{2{{\sin }^2}x\cos x – 8\cos x{{\sin }^2}\frac{x}{2} – \frac{4}{3}{{\sin }^4}x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\sin }^2}\frac{x}{2}{{\left( {\cos \frac{x}{2} – 1} \right)}^2} + 8{{\sin }^6}\frac{x}{2}}}{{8{{\sin }^2}\frac{x}{2}{{\cos }^2}\frac{x}{2}\cos x – 8\cos x{{\sin }^2}\frac{x}{2} – \frac{4}{3}.16{{\sin }^4}\frac{x}{2}{{\cos }^4}\frac{x}{2}}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\sin }^2}\frac{x}{2}.4{{\sin }^4}\frac{x}{4} + 8{{\sin }^6}\frac{x}{2}}}{{8{{\sin }^2}\frac{x}{2}\cos x\left( {{{\cos }^2}\frac{x}{2} – 1} \right) – \frac{{64}}{3}{{\sin }^4}\frac{x}{2}{{\cos }^4}\frac{x}{2}}}$ Continue reading $\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\sin x – 2\sin \frac{x}{2}} \right)}^2} + {{(1 – \cos x)}^3}}}{{\sin x\sin 2x – 8\cos x{{\sin }^2}\frac{x}{2} – \frac{4}{3}{{\sin }^4}x}} = ?$
The two blocks, each of mass M kg, are connected …
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\sin x}} – {e^{ – \sin x}} – 2\tan x}}{{\tan x – x}} = ?$
The given limit,
$ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 + \sin x + \frac{{{{\sin }^2}x}}{{2!}} + \frac{{{{\sin }^3}x}}{{3!}} + ……….} \right) – \left( {1 – \sin x + \frac{{{{\sin }^2}x}}{{2!}} – \frac{{{{\sin }^3}x}}{{3!}} + ……….} \right) – 2\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….} \right)}}{{\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….} \right) – x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sin x + \frac{{{{\sin }^3}x}}{{3!}} + \frac{{{{\sin }^5}x}}{{5!}} + ……….} \right) – 2\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….} \right)}}{{\frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sin x – x + \frac{{{{\sin }^3}x}}{{3!}} – \frac{{{x^3}}}{3} + \frac{{{{\sin }^5}x}}{{5!}} – \frac{2}{{15}}{x^5} + ……….} \right)}}{{\frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….}}$ Continue reading $\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\sin x}} – {e^{ – \sin x}} – 2\tan x}}{{\tan x – x}} = ?$
A block of mass $\sqrt 3$ kg is placed …
The string between mass m and 2m is inextensible and light …
In $\Delta ABC$, $\left| {\begin{array}{*{20}{c}}{\cos 3A}&{\cos 3B}&{\cos 3C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$Select the right option
(A) Triangle is isosceles
(B) Triangle is equilateral
(C) Triangle is right angled
(D) Triangle is scalene
We have, $\left| {\begin{array}{*{20}{c}}{4{{\cos }^3}A – 3\cos A}&{4{{\cos }^3}B – 3\cos B}&{4{{\cos }^3}C – 3\cos C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}{4{{\cos }^3}A}&{4{{\cos }^3}B}&{4{{\cos }^3}C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| – \left| {\begin{array}{*{20}{c}}{3\cos A}&{3\cos B}&{3\cos C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}{{{\cos }^3}A}&{{{\cos }^3}B}&{{{\cos }^3}C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$ as the 2nd determinant is 0.
$ \Rightarrow \cos A.\cos B.\cos C\left| {\begin{array}{*{20}{c}}{{{\cos }^2}A}&{{{\cos }^2}B}&{{{\cos }^2}C}\\1&1&1\\{\tan A}&{\tan B}&{\tan C}\end{array}} \right| = 0$ Continue reading In $\Delta ABC$,
$\left| {\begin{array}{*{20}{c}}{\cos 3A}&{\cos 3B}&{\cos 3C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$
Select the right option