$x=-i$ satisfies the given equation,
$\therefore x^2(x+i)-xi(x+i)+2(x+i)=0$
$\Rightarrow (x+i)(x^2-xi+2)=0$
Now consider, $x^2-xi+2=0$
$x=-i$ again satisfies the above equation,
$\therefore x(x+i)-2i(x+i)=0$
$\Rightarrow (x+i)(x-2i)=0$
$x=2i$ is another solution,
So, $x=-i$, $x=-i$, $x=2i$ are the three solutions.
Using Sine Rule in the left triangle,
$\frac {sin \alpha}{a} = \frac {sin \beta}{b}$
Using Sine Rule in the right triangle,
$\frac {sin 20^\circ}{a} = \frac {sin 140^\circ}{b}$
Dividing the above equations we have,
$\frac {sin \alpha}{sin 20^\circ}=\frac {sin \beta}{sin 140^\circ}$ Continue reading $\alpha = ?$ in the Triangle
(A) 1 (B) 2 (C) 3 (D) None of the options given
We have, ${x^{\left( {{x^x}} \right)}} = x$
Taking log, ${x^x}\ln |x| = \ln |x|$
$ \Rightarrow \ln |x|\left( {{x^x} – 1} \right) = 0$
$\ln |x| = 0,x = \pm 1$ Or ${x^x} = 1$
Taking log, $x\ln |x| = 0$
x = 0 Or $\ln |x| = 0,x = \pm 1$
x = 0 is rejected as it leads to $0^0$ situation.
Hence, two solutions or option (B).
We have, $\frac{{({x^2} + 5x + 6)({x^2} + 9x + 20)}}{{({x^2} – 5x + 6)({x^2} – 9x + 20)}} = 1$
$ \Rightarrow \frac{{{x^2} + 9x + 20}}{{{x^2} – 5x + 6}} = \frac{{{x^2} – 9x + 20}}{{{x^2} + 5x + 6}} = \frac{{({x^2} + 9x + 20) – ({x^2} – 9x + 20)}}{{({x^2} – 5x + 6) – ({x^2} + 5x + 6)}} = \frac{{18x}}{{ – 10x}} = – \frac{{9}}{{5}}$
(assuming $x \ne 0$)
$ \Rightarrow \frac{{{x^2} + 9x + 20}}{{{x^2} – 5x + 6}} – 1 = – \frac{9}{5} – 1$
$ \Rightarrow \frac{{14x + 14}}{{{x^2} – 5x + 6}} = – \frac{{14}}{5}$
$ \Rightarrow \frac{{x + 1}}{{{x^2} – 5x + 6}} = – \frac{1}{5}$
$ \Rightarrow {x^2} – 5x + 6 + 5x + 5 = 0$ Or ${x^2} + 11 = 0$
So, no real solution if $x \neq 0$.
If x = 0, LHS of the original equation = $\frac{{2 \times 3 \times 4 \times 5}}{{( – 2) \times ( – 3) \times ( – 4) \times ( – 5)}}$ = 1 = RHS
Thus, x = 0 is the only real solution.
The given integral can be split as,
$ \int\limits_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 – x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}} dx + \int\limits_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{1 + \cos 2x}}} dx$
$\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 – x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}$ is an odd function. Thus, the given integral
$ = 0 + \int\limits_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{2{{\cos }^2}x}}} dx = \frac{1}{2}\int\limits_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} {{{\sec }^2}xdx} = \frac{1}{2}\left. {\tan x} \right|_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} = \frac{1}{{\sqrt 3 }}$
Let $a+b+c=k$. So,
$k(a+b)=15$
$k(b+c)=6$
$k(c+a)=-3$
Adding all of the above, $k.2k=18$
$\Rightarrow k=\pm 3$ Continue reading $(a+b)(a+b+c)=15$$(b+c)(a+b+c)=6$$(c+a)(a+b+c)=-3$
$\{ a,b,c\} \equiv ?$
A disc or radius r on top of another disc of radius 4r is made to go around it via pure rolling motion. How many maximum full rotations will the small disc make in this process? Continue reading Disc Rolling on Disc
