Question

Fill in the blanks.
(i) The curves $y = 4x^2 + 2x - 8$ and $y = x^3 - x + 10$ touch each other at the point P.
(ii) The values of a for which the function $f(x) = sinx - ax + b$ increases on $R$ are Q.
(iii) The least value of the function $f(x) = ax + \frac{b}{x}(a > 0, b > 0, x > 0)$ is R.
(iv) The equation of normal to the curve $y = tanx$ at $(0,0)$ is S.

	P	Q	R	S
(a)	$\left(-\frac{1}{3}, - \frac{74}{9}\right)\,\,\,\,$	$\left(-\infty, 1\right)\,\,\,\,$	$\sqrt{ab}\,\,\,\,$	$y - 2x = 0$
(b)	$(3, 34)\,\,\,\,$	$\left(-\infty, 1\right)\,\,\,\,$	$2\sqrt{ab}\,\,\,\,$	$y + x = 0$
(c)	$(3, 34)\,\,\,\,$	$\left(-\infty, -1\right)\,\,\,\,$	$\sqrt{ab}\,\,\,\,$	$y = 0$
(d)	$\left(-\frac{1}{3}, - \frac{74}{9}\right)\,\,\,\,$	$\left(-\infty, 1\right)\,\,\,\,$	$2\sqrt{ab}\,\,\,\,$	$y - 2x = 0$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

(i) Given, Curves are $y = 4x^2 + 2 x - 8$ and $y = x^3 - x+ 10$
$\Rightarrow \frac{dy}{dx} = 8x + 2$
and $\frac{dy}{dx} = 3x^{2} - 1$
Since, the slope of both curves should be same.
$\therefore 8x + 2 = 3x^{2} - 1$
$\Rightarrow 3x^{2} - 8x - 3 = 0$
$\Rightarrow \left(3x + 1\right)\left(x - 3\right) = 0$
$\therefore x = -\frac{1}{3}$ and $x = 3$
For $x = -\frac{1}{3}$,
$y = 4\left(-\frac{1}{3}\right)^{2} + 2\left(\frac{-1}{3}\right) - 8$
$= -\frac{74}{9}$
and for $x = 3$, $y = 4\left(3\right)^2 + 2\left(3\right) - 8$
$ = 36+ 6 - 8 = 34$
Hence, the required points are $\left(3, 34\right)$ and $\left(-\frac{1}{3}, \frac{-74}{9}\right)$.
But only point $\left(3,34\right)$ satisfies both equations.
(ii) We have, $f(x) = sin \,x - ax + b$
$\Rightarrow f'(x ) = cosx - a$
For increasing function, $f'(x) > 0$
$ \Rightarrow cosx > a$
Since, $cosx \, \in [-1,1]$
$ \Rightarrow a < 1$
$\Rightarrow a \in (-\infty, 1)$
(iii) We have, $f\left(x\right) = ax \frac{b}{x}$
$\Rightarrow f'\left(x\right) = a- \frac{b}{x^{2}}$
Put $f'\left(x\right) = 0$
$\Rightarrow a = \frac{b}{x^{2}}$
$\Rightarrow x = \pm \sqrt{\frac{b}{a}}$
Now, $f''\left(x\right) =-b\cdot \frac{\left(-2\right)}{x^{3}} = \frac{2b}{x^{3}}$
At $x = \sqrt{\frac{b}{a}}, f''\left(x\right)$
$ =\frac{2b}{\left(\frac{b}{a}\right)^{3/2}} = \frac{2b\cdot a^{3/2}}{b^{3/2}} > 0 \,\,\left[\because a, b > 0\right]$
$\therefore $ Least value of $f\left(x\right)$ i.e. $f\left(\sqrt{\frac{b}{a}}\right) = a\cdot\sqrt{\frac{b}{a}} + \frac{b}{\sqrt{\frac{b}{a}}}$
$= a \cdot a^{-1/2 }\cdot b^{1/2} + b \cdot b^{-1/2} \cdot a^{1/2}$
$= \sqrt{ab} + \sqrt{ab} = 2\sqrt{ab}$.
(iv) We have, $y = tan\,x$
$\Rightarrow \frac{dy}{dx} = sec^2\,x$
$\Rightarrow (\frac{dy}{dx})_{(0, 0)} = sec^2\,0 = 1$
and slope of normal $= - \frac{1}{\left(\frac{dy}{dx}\right)} = -\frac{1}{1}$
$\therefore $ Equation of normal to the curve $y = tanx$ at $(0,0)$ is
$y - 0 = - 1( x - 0 )$
$ \Rightarrow y + x = 0$