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 $y=4x^2+2x-8$ 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, $f(x)=sinx-ax+b$
$\Rightarrow f^{\prime }(x)=cosx-a$
For increasing function, $f^{\prime }(x)>0$
$\Rightarrow cosx>a$
Since, $cosx\in [-1,1]$
$\Rightarrow a<1$
$\Rightarrow a\in (-\infty ,1)$
(iii) We have, $f\left(x\right)=ax\frac{b}{x}$
$\Rightarrow f^{\prime }\left(x\right)=a-\frac{b}{x^2}$
Put $f^{\prime }\left(x\right)=0$
$\Rightarrow a=\frac{b}{x^2}$
$\Rightarrow x=\pm \sqrt{\frac{b}{a}}$
Now, $f^{\prime \prime }\left(x\right)=-b\cdot \frac{\left(-2\right)}{x^3}=\frac{2b}{x^3}$
At $x=\sqrt{\frac{b}{a}},f^{\prime \prime }\left(x\right)$
$=\frac{2b}{{\left(\frac{b}{a}\right)}^{3/2}}=\frac{2b\cdot a^{3/2}}{b^{3/2}}>0\left[\because a,b>0\right]$
$\therefore$ 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, $y=tanx$
$\Rightarrow \frac{dy}{dx}=se{c}^{2}x$
$\Rightarrow (\frac{dy}{dx}{)}_{(0,0)}=se{c}^{2}0=1$
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
$y-0=-1(x-0)$
$\Rightarrow y+x=0$