Application of Derivatives Test - 12

$$y=3\sin { \theta  } \cos { \theta  } $$

$$x^2y^2-2x=4-4y$$
$$2xy^2+2y\cdot x^2\cdot\displaystyle\frac{dy}{dx}-2=-4\cdot\frac{dy}{dx}$$
$$\displaystyle\frac{dy}{dx}(2y\cdot x^2+4)=2-2x\cdot y^2$$
$$\left.\displaystyle\frac{dy}{dx}\right|_{2, -2}=\displaystyle\frac{2-2\times 2\times 4}{2(-2)\times 4+4}=\frac{+14}{+12}=\frac{7}{6}$$
 eq of tangent :$$(y+2)=\displaystyle\frac{7}{6}(x-2)\Rightarrow 7x-6y=26$$.

Given equation of curve is $$x^{2}=y-6$$

Given equation of conic is
$$y - 6 = x^2$$ 
$$\dfrac{dy}{dx}=2x$$
Slope of tangent at $$(2,10)$$ is $$4.$$

Equation of tangent to conic is 
$$y-10=4(x-2)$$
$$\Rightarrow y-10=4x-8$$
$$\Rightarrow y=4x+2$$          .....(1)

Given equation of circle is 
$$x^{ 2 }+y^{ 2 }+8x-2y=k$$
$$\Rightarrow (x+4)^2+(y-1)^2=k+17$$

Given $$(\alpha, \beta)$$ is a point of tangency for fixed $$k$$
Radius = Length of tangent from center $$(-4,1)$$
$$\Rightarrow \sqrt{k+17}=\dfrac{15}{\sqrt{17}}$$
$$\Rightarrow k+17=\dfrac{225}{17}$$

So, equation of circle at $$(\alpha, \beta)$$ is 
$$(\alpha+4)^2+(\beta-1)^2=\dfrac{225}{17}$$

Now, eqn (1) is tangent to the circle for fixed $$k$$ at $$(\alpha,\beta)$$
$$\Rightarrow (\alpha+4)^2+(4\alpha+1)^2=\dfrac{225}{17}$$
$$\Rightarrow 17\alpha^2+16\alpha+17=\dfrac{225}{17}$$
$$\Rightarrow 289\alpha^2+272\alpha+64=0$$
$$\Rightarrow \alpha =\dfrac{-8}{17}$$  (Here $$D=0$$)

So, by (1), $$\beta =\dfrac{2}{17}$$

$$f(x) = (a^{2} =- 3a + 2)(\cos^{2}x / 4 - \sin^{2}x / 4) + (a - 1)x + \sin 1$$
$$\Rightarrow f(x) = (a - 1)(a - 2)\cos x/2 + (a - 1) x + \sin 1$$
$$\Rightarrow f'(x) = -\dfrac {1}{2} (a - 1)(a - 2) \sin \dfrac {x}{2} + (a - 1)$$
$$\Rightarrow f'(x) = (a - 1) \left [1 - \dfrac {(a - 2)}{2}\sin \dfrac {x}{2}\right ]$$
If $$f(x)$$ does not possess critical points, then $$f'(x)\neq 0$$ for any $$x \epsilon R$$
$$\Rightarrow (a - 1)\left [1 -\dfrac {(a - 2)}{2} \sin \dfrac {x}{2}\right ]\neq 0$$ for any $$x\epsilon R$$
$$\Rightarrow a\neq 1$$ and $$1 - \left (\dfrac {a - 2}{2}\right )\sin \dfrac {x}{2} = 0$$
must not have any solution in $$R$$.
$$\Rightarrow a\neq 1$$ and $$\sin \dfrac {x}{2} = \dfrac {2}{a - 2}$$ is not solvable in $$R$$.
$$\Rightarrow a\neq 1$$ and $$\left |\dfrac {2}{a - 2}\right | > 1$$ [For $$a = 2, f(x) = x + \sin 1\therefore f'(x) = 1 \neq 0]$$
$$\Rightarrow a \neq 1$$ and $$|a - 2| < 2\Rightarrow a\neq 1$$ and $$-2 < a - 2 < 2$$
$$\Rightarrow a\neq 1$$ and $$0 < a < 4\Rightarrow a\epsilon (0, 1)\cup (1, 4)$$.

Given, $$f(x) = \sqrt {x}(7x - 6) = 7x^{3/2} - 6x^{1/2}$$
Thus $$f'(x) = 7\times \dfrac {3}{2}x^{1/2} - 6\times \dfrac {1}{2} x^{-1/2}$$
When tangent is parallel to $$x$$ axis $$f'(x) = 0$$.
Therefore, $$\dfrac {21}{2}x^{1/2} - 3x^{-1/2} = 0$$
$$\Rightarrow \dfrac {21}{2}\sqrt {x} = \dfrac {3}{\sqrt {x}}$$
$$\Rightarrow 7x = 2$$

Given the function $$f(x) = x^{3} - 1, \quad x\epsilon [-1, 1]$$

Since y-axis is major axis.
$$\Rightarrow f(4a) < f(a^2-5)$$
$$\Rightarrow 4a > a^2-5$$                 (Q $$f$$ is decreasing)
$$\Rightarrow  a^2-4a - 5 < 0$$
$$\Rightarrow  a \in (-1, 5)$$

Stationary means $$f'(x)=0$$

$$f'(x)=cos  x$$
$$f'(x)=0$$  when $$x=\dfrac {\pi}{2}, \dfrac {3\pi}{2}$$