$x \to 1 – \frac{1}{x}$ yields,
$f\left( {1 – \frac{1}{x}} \right) + f\left[ {\frac{1}{{1 – \left( {1 – \frac{1}{x}} \right)}}} \right] = 1 – \frac{1}{x}$
$ \Rightarrow f\left( {\frac{{x – 1}}{x}} \right) + f(x) = \frac{{x – 1}}{x}……..(A)$ Continue reading $f(x) + f\left( {\frac{1}{{1 – x}}} \right) = x,\forall x \in \mathbb{R}$
$f(x) = ?$
In how many ways can 3 boys and 3 girls be seated at a round table such that exactly 2 boys sit together?
Solution
Circular permutations without any restriction = (6 – 1)! = 120
Circular permutations when no 2 boys are together which is B G B G B G situation is a circle = 2! $\times$ 3! = 12
Circular permutations when all the 3 boys are together = (4 – 1)! $\times $ 3! = 36
Circular permutations for exactly 2 boys sitting together = 120 – (12 + 36) = 72
Using summation formula of G.P. having 1st term as a and common ratio as r we have,
$\frac{{a({r^3} – 1)}}{{(r – 1)}} = 42$ ……..(A)
Squaring the terms of the G.P. yields another G.P. having 1st term as $a^2$ and common ratio as $r^2 $. Thus,
$\frac{{{a^2}({r^6} – 1)}}{{({r^2} – 1)}} = 1092$
$ \Rightarrow \frac{{{a^2}({r^3} – 1)({r^3} + 1)}}{{(r – 1)(r + 1)}} = 1092$ Continue reading Find G.P. such that
$\sum\limits_{r = 1}^3 {{t_r}} = 42$
$\sum\limits_{r = 1}^3 {{{({t_r})}^2}} = 1092$
-1 satisfies the given equation. Thus,
$7x^4 (x+1) – 7x^3 (x+1) + 7x^2 (x+1) -7x(x+1)+8(x+1)=0$
$\Rightarrow (x+1) (7x^4 – 7x^3 +7x^2 -7x +8) = 0$
Consider, $7x^4 – 7x^3 +7x^2 -7x +8 = 0$
Or, $7(x^4 – x^3 +x^2 -x) +8 = 0$ Continue reading Solve for $x \in \mathbb{R}$
$7x^5 + x + 8 = 0$
We have, ${2^{2x}} + \frac{{{2^{2x}}}}{2} = {3^x}\sqrt 3 + \frac{{{3^x}}}{{\sqrt 3 }}$
$ \Rightarrow \frac{3}{2}{.2^{2x}} = \frac{4}{{\sqrt 3 }}{.3^x}$
$ \Rightarrow \frac{{{2^{2x}}}}{8} = \frac{{{3^x}}}{{3\sqrt 3 }}$
$ \Rightarrow {2^{2x – 3}} = {3^{x – \frac{3}{2}}}$
$ \Rightarrow {2^{2\left( {x – \frac{3}{2}} \right)}} = {3^{x – \frac{3}{2}}}$
$ \Rightarrow {4^{x – \frac{3}{2}}} = {3^{x – \frac{3}{2}}}$
The above is only possible if $x – \frac{3}{2} = 0$ or $x=\frac {3}{2}$.
Using the substitution $x + \frac{1}{x} + 4 = t$, we have
$\begin{array}{l}x + \frac{1}{x} = t – 4\\ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = {(t – 4)^2}\\ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 16 = {(t – 4)^2} + 14\end{array}$
Now, $f(t) = {(t – 4)^2} + 14$
$\therefore f(17) = {(17 – 4)^2} + 14 = 169 + 14 = 183$
We have, $\sqrt 13 > \sqrt 7$ & $\sqrt 12 > \sqrt 6$
$\therefore \sqrt 13 + \sqrt 12 > \sqrt 7 + \sqrt 6 $
$\Rightarrow \frac {1}{\sqrt 13 + \sqrt 12} < \frac {1}{\sqrt 7 + \sqrt 6}$
$\Rightarrow \frac {\sqrt 13 – \sqrt 12}{(\sqrt 13 + \sqrt 12)(\sqrt 13 – \sqrt 12)} < \frac {\sqrt 7 – \sqrt 6}{(\sqrt 7 + \sqrt 6)(\sqrt 7 – \sqrt 6)}$
$\therefore \sqrt 13 – \sqrt 12 < \sqrt 7 – \sqrt 6 $
