Application of Derivatives Test - 27

Slope of normal
$$=\dfrac{-1}{f'(x)}$$
$$=tan(\dfrac{3\pi}{4})$$
$$=-1$$
Hence
$$f'(x)=1$$
Or 
$$\dfrac{dy}{dx}=1$$
Or 
$$y=x+c$$
Hence
$$y=f(x)$$ is an equation of a straight line parallel to $$y=x$$.
Hence
$$f'(x)$$ is independent of x and its value is 1.

Hence slope of tangents
$$=tan60^{0}$$
$$=\sqrt{3}$$
$$=\dfrac{dy}{dx}$$
Hence
$$\dfrac{dy}{dx}$$
$$=3x^{2}$$
$$=\sqrt{3}$$
Or 
$$x^{2}=\dfrac{1}{\sqrt{3}}$$
Hence
$$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}$$.
Thus y
$$=\dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$$.

$$\displaystyle f\left ( x \right )= \left ( k^{2}-3x+2 \right )\cos \frac{x}{2}+\left ( k-1 \right )x+\sin 1$$
$$\displaystyle

{f}'\left ( x \right )= \left ( k-1 \right )\left ( k-2 \right )\left (

-\frac{1}{2}\sin \frac{x}{2} \right )+\left ( k-1 \right )$$
$$=\displaystyle \left ( k-1 \right )\left [ 1-\frac{k-2}{2}\sin \frac{x}{2} \right ]$$
Since f(x) does not possess critical points therefore f'(x) is not equal to zero.
i.e.,

$$\displaystyle k\neq 1$$ or $$\displaystyle 1-\frac{k-2}{2}\sin

\frac{x}{2}= 0$$ does not posses a solution or $$\displaystyle \sin

\frac{x}{2}= \frac{2}{k-2}$$ does not have a solution.
Hence we must

have $$\displaystyle \left | \frac{2}{k-2} \right |> 1$$ as

$$\displaystyle \left | \sin \frac{x}{2} \right |< 1.$$
Above implies that $$\displaystyle \left | k-2 \right |^{2}\leq 4$$
or $$\displaystyle -2< \left ( k-2 \right )< 2$$
$$\displaystyle \because x^{2}< a^{2}\Rightarrow \left ( x^{2}-a^{2} \right )= -ive$$ or $$\displaystyle -a< x< a$$
$$\displaystyle \therefore 0 < k< 4$$. Also $$\displaystyle k\neq 1.$$
$$\displaystyle \therefore k \epsilon \left ( 0,1 \right )\cup \left ( 1,4 \right )$$

$$y=[x+\sqrt{x+\sqrt{x}}]^{\frac{1}{2}}$$

$$y^{2}=x+\sqrt{x+\sqrt{x}}$$

$$y^{2}-x=\sqrt{x+\sqrt{x}}$$

$$y^{4}-2xy^{2}+x^{2}=x+\sqrt{x}$$

Differentiating with respect to x gives us

$$4y^{3}y'-2y^{2}-4xyy'+2x=1+\dfrac{1}{2\sqrt{x}}$$

At $$x=1$$

$$4y^{3}_{x=1}y'-2y^{2}_{x=1}-4yy'_{x=1}+2=1+\dfrac{1}{2}$$

$$4y^{3}_{x=1}y'-2y^{2}_{x=1}-4yy'_{x=1}=\dfrac{-1}{2}$$

Now at $$x=1$$
$$y=\sqrt{1+\sqrt{1+\sqrt{1}}}$$

$$=\sqrt{1+\sqrt{2}}$$

Substituting in the above equation gives us

$$4y^{3}_{x=1}y'-2y^{2}_{x=1}-4yy'_{x=1}=\dfrac{-1}{2}$$

$$4(\sqrt{1+\sqrt{2}})(1+\sqrt{2})y'-2(1+\sqrt{2})-4(\sqrt{1+\sqrt{2}})y'=\dfrac{-1}{2}$$

$$4(\sqrt{1+\sqrt{2}})y'(1+\sqrt{2}-1)=\dfrac{-1}{2}+2(1+\sqrt{2})$$

$$4\sqrt{2}(\sqrt{1+\sqrt{2}})y'=\dfrac{3}{2}+2\sqrt{2}$$

$$8\sqrt{2}(\sqrt{1+\sqrt{2}})y'=3+4\sqrt{2}$$

$$y'=\dfrac{3+4\sqrt{2}}{8\sqrt{2}(\sqrt{1+\sqrt{2}})}$$
Hence

$$\dfrac{dy}{dx}_{x=1}=\dfrac{3+4\sqrt{2}}{8\sqrt{2}(\sqrt{1+\sqrt{2}})}$$

Given equation of curve $$\displaystyle y=4-x^{2}$$     ...(i)
$$\dfrac{dy}{dx}=-2x$$

Given points $$A(-2,0)$$ and $$(1,3)$$
Slope of AB $$=\dfrac{3}{3}=1$$

$$\Rightarrow -2x=1$$
$$\Rightarrow x=\dfrac{-1}{2}$$

So, by equation (i), we get
$$y=\dfrac{15}{4}$$
Hence, the point is $$(-\dfrac{1}{2},\dfrac{15}{4})$$