The number of solution(s) the equation ${x^{{x^2}}} = {x^{4x + 5}}$ has:
(A) One solution only (B) Two solutions
(C) Three solutions (D) No solution
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
$\sin ({\pi ^x}) = {e^x} + {e^{ – x}}$Solve for real x
Using $A.M. \ge G.M.$,
$\frac{{{e^x} + {e^{ – x}}}}{2} \ge \sqrt {{e^x}.{e^{ – x}}} $
$ \Rightarrow {e^x} + {e^{ – x}} \ge 2$
$\therefore \sin ({\pi ^x}) \ge 2$ which is not possible for any real x.
So, no solution.
$\int\limits_0^1 {\frac{{{x^{e – 1}} – 1}}{{\ln x}}dx} = ?$
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 $\int\limits_0^1 {\frac{{{x^{e – 1}} – 1}}{{\ln x}}dx} = ?$
f is differentiable function such that$f(x + y) – f(x) = f(y) + 2xy$$x,y \in R$$f(x) = ?$
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
$f(x + y) – f(x) = f(y) + 2xy$
$x,y \in R$
$f(x) = ?$
Evaluate, $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos x + \cos 3x + \sin 2x)}^2}}}{{{{(\sin x – \sin 3x + 2\cos 2x)}^3}}}$
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,
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos x + \cos 3x + \sin 2x)}^2}}}{{{{(\sin x – \sin 3x + 2\cos 2x)}^3}}}$
$f(x + \sqrt x ) = x – \sqrt x $$f(380) = ?$
Substituting 361 in place of x,
$f(361 + \sqrt {361} ) = 361 – \sqrt {361} $
$ \Rightarrow f(361 + 19) = 361 – 19$
$ \Rightarrow f(380) = 342$