We have, $f(x) = – \frac{1}{{{x^2}}}f\left( {\frac{1}{x}} \right)$
$\therefore I = \int\limits_{\tan \alpha }^{\cot \alpha } { – \frac{1}{{{x^2}}}f\left( {\frac{1}{x}} \right)dx} $
Let, $\frac{1}{x} = t$
$ \Rightarrow – \frac{1}{{{x^2}}}dx = dt$
$\therefore I = \int\limits_{\cot \alpha }^{\tan \alpha } {f(t)dt} =-I $
$ \Rightarrow I = 0$
The required area = shaded area in the figure = $\frac{1}{2}({5^2} – {3^2})\left( {\frac{\pi }{4} – \frac{\pi }{8}} \right)$
$ = \frac{1}{2} \times 16 \times \frac{\pi }{8} = \pi $ square unit.
$x \to \frac{1}{x}$ in the original equation yields,
$f\left( {\frac{1}{x}} \right) + f(x) + 3f\left( { – \frac{1}{x}} \right) = \frac{1}{x}$ ……..(A)
Original equation – (A) yields,
$3f( – x) – 3f\left( { – \frac{1}{x}} \right) = x – \frac{1}{x}$ Continue reading Find $f(x)$ such that,
$f(x) + f\left( {\frac{1}{x}} \right) + 3f( – x) = x$
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, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}(\pi /2 – x)}}} $
$ \Rightarrow I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\cot }^{\sqrt {\tan \alpha } }}x}}} = \int\limits_0^{\pi /2} {\frac{{{{\sin }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\sin }^{\sqrt {\tan \alpha } }}x + {{\cos }^{\sqrt {\tan \alpha } }}x}}} $ ……..(A)
Also, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}x}} = } \int\limits_0^{\pi /2} {\frac{{{{\cos }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\cos }^{\sqrt {\tan \alpha } }}x + {{\sin }^{\sqrt {\tan \alpha } }}x}}}$ ……..(B)
(A) + (B) gives, $2I = \int\limits_0^{\pi /2} {dx} = \frac{\pi }{2}$
$ \Rightarrow I = \frac{\pi }{4}$
$\therefore \frac{{dI}}{{d\alpha }} = 0$
Assuming $\beta > \alpha $ which of the following option(s) is/are correct?
(A) $y \le \alpha $
(B) $y \ge \beta $
(C) $\alpha \le y \le \beta $
(D) $y \in \mathbb{R}$ Continue reading $y = \frac{{{x^2} – \alpha \beta }}{{2x – \alpha – \beta }}$
$x \in \mathbb{R} – \left\{ {\frac{{\alpha + \beta }}{2}} \right\}$
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 $
