Application of Derivatives Test - 30

Given, $$r=9 cm, \Delta r=0.03cm$$

Given, $$y= ax^3+bx^2+cx+5$$

The slope of normal $$\displaystyle x=a\left ( \theta -\sin \theta

 \right );\: \: \: \: y=a\left ( 1-\cos \theta  \right )\: \: \: at\: \:

\: \theta =\dfrac{\pi }{2}$$
$$\displaystyle \left ( \dfrac{dx}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\left ( 1-\cos \theta  \right )=a $$
$$\displaystyle \left ( \dfrac{dy}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\sin \theta =a $$
$$\therefore \left ( \dfrac{dy}{dx} \right )=1$$
Therefore, slope of normal at the given point is, $$m= \left ( -\dfrac{dx}{dy} \right )=-1$$
Hence, option 'C' is correct.

Given, $$\displaystyle g(x)=\ln(h(x))$$
$$\Rightarrow g'(x)=\cfrac{h'(x)}{h(x)}$$
$$\Rightarrow g''(x)=\cfrac{h(x).h''(x)-(h'(x))^2}{(h(x))^2}$$
Using given condition, Clearly $$g''(x)< 0, x  \epsilon J$$
Hence $$g(x)$$ is concave down on $$J.$$

Since the same line is tangent at one point $$x=a$$ and normal at other point $$x=b$$

$$y^2=4ax$$
$$\Rightarrow 2y\cfrac{dy}{dx}=4a\Rightarrow \cfrac{dy}{dx}=\cfrac{2a}{y}$$
Therefore, slope of normal to the given curve is, $$=\left(-\cfrac{dx}{dy}\right)_{(at^2,2at)}=-\cfrac{2at}{2a}=-t$$
Hence, option 'C' is correct.

Given  curve is $$xy + ax - by = 0$$
Slope of tangent at $$(1, 1)$$ is $$\displaystyle x.\frac{dy}{dx}+y+a-b.\frac{dy}{dx}=0 $$
$$\displaystyle

\left ( \frac{dy}{dx} \right )_{\left ( 1,1 \right )}=-\left [

\frac{\left ( a+y \right )}{x-b} \right ]_{\left ( 1,1 \right

)}=-\frac{\left ( a+1 \right )}{1-b} $$
$$\displaystyle \therefore -\frac{\left ( a+1 \right )}{1-b}=2 $$
So, $$2b - a = 3  $$                                        ..........(1)
$$\displaystyle \because (1, 1)$$ lies on curve $$xy + ax - by = 0$$
$$\displaystyle \therefore a-b=-1 $$         .........(2)
From (1) and (2), we get
$$a = 1, b = 2$$
Hence, option 'A' is correct.

$$\displaystyle y=\cos \left ( x+y \right )$$           $$\displaystyle \left ( -2\pi \leq x\leq 2\pi  \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\sin \left ( x+y \right )\left ( 1+\frac{dy}{dx} \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}\left ( 1+\sin \left ( x+y \right ) \right )=-\sin \left ( x+y \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}$$
Given tangent is parallel to the line, $$x + 2y = 0$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}=-\frac{1}{2}$$
$$\displaystyle 2\sin \left ( x+y \right )=1+\sin \left ( x+y \right )$$
$$\displaystyle \sin \left ( x+y \right )=1\Rightarrow \cos(x+y)=0\Rightarrow x+y=\cfrac{\pi}{2}$$
Also the point lies on the given curve,
 $$y=0\Rightarrow x= \dfrac{\pi }{2}$$
Thus the point is, $$\displaystyle \left ( \frac{\pi }{2},0 \right )$$
Hence, option 'A' is correct.

Let point be $$\displaystyle \left ( x_{1},y_{1} \right )$$
$$y = \displaystyle x^{3}+5$$
$$\Rightarrow \displaystyle \dfrac{dy}{dx}=3x^{2}$$
Now, the slope of tangent at this point is $$=\displaystyle \left ( \frac{dy}{dx} \right )_{x_{1}y_{1}}=3x_{1}^{2}$$
It is $$\displaystyle \perp $$ to $$x + 3y - 2 = 0$$
So $$\displaystyle 3x_{1}^{2}\times -\frac{1}{3}=-1\Rightarrow x_1=\pm 1 $$
Thus,
at $$\displaystyle x_{1}=1$$,             $$\Rightarrow$$                $$\displaystyle y_{1}=6$$
and at $$\displaystyle x_{1}=-1$$      $$\Rightarrow$$              $$\displaystyle y_{1}=4$$
Thus, the points are $$(1, 6)$$ and $$(-1, 4)$$
Hence, option 'D' is correct.

Let the point is $$P(a,b)$$
Now the given curve is $$y=2x^2-x+1$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}=4x-1$$
Thus slope of tangent at P is $$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=4a-1$$
But given the tangent is parallel to line $$y=3x+4$$
$$\Rightarrow 4a-1=3\Rightarrow a=1$$
Also the point P lies on the given curve,
$$b=2a^2-a+1=2$$
Therefore, the point P is $$(1,2)$$
Hence, option 'B' is correct.