Assuming the existence of the compound shown that is given trivial name, “Ravana” what would you call it in IUPAC?
$y = f(x) = \frac{x}{{x + 5}}$ $f(5x) = g(y)$ $g(y) = ?$
$f(5x) = \frac{{5x}}{{5x + 5}} = \frac{x}{{x + 1}}$
Given, $y = \frac{x}{{x + 5}}$
$\therefore y.x + 5y = x$ Continue reading $y = f(x) = \frac{x}{{x + 5}}$ $f(5x) = g(y)$ $g(y) = ?$
f'(0) = ? for an even function f that is differentiable at 0
Since f is differentiable at 0 we have,
$\mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) – f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 – h) – f(0)}}{{ – h}}$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f(h) – f(0)}}{h} = – \mathop {\lim }\limits_{h \to 0} \frac{{f( – h) – f(0)}}{h}$
Since f is an even function, f(-h) = f(h) Continue reading f'(0) = ? for an even function f that is differentiable at 0
$3{\cos ^2}x + 2\sin x + 1$
${(3{\cos ^2}x + 2\sin x + 1)_{\max }} = ?$
The given expression = $- 3{\sin ^2}x + 2\sin x + 4$
= – $\left[ {{{(\sqrt 3 \sin x)}^2} – 2.\frac{1}{{\sqrt 3 }}.\sqrt 3 \sin x + \frac{1}{3} – \frac{1}{3}} \right]$ + 4
$ = – \left[ {{{\left( {\sqrt 3 \sin x – \frac{1}{{\sqrt 3 }}} \right)}^2} – \frac{1}{3}} \right] + 4$ Continue reading $3{\cos ^2}x + 2\sin x + 1$
