Application of Derivatives Test - 38

Equation of tangent is $$y-2x+1=0$$

It is tangent at $$x=1$$, so for $$x=1$$

$$y-2(1)+1=0\\ \Rightarrow y=1$$

So, its is tangent to the curve at $$(1,1)$$

Slope of tangent $$= -\left (\dfrac{-2}{1}\right)=2$$

$$xy+ax+by=0\\ y+x\dfrac { dy }{ dx } +a+b\dfrac { dy }{ dx } =0$$

Now $$\dfrac { dy }{ dx } =2$$ at $$(1,1)$$

$$1+1(2)+a+b(2)=0\\ \Rightarrow a+2b=-3$$     .....(i)

$$(1,1)$$ also lies on the curve $$xy+ax+by=0$$

$$\Rightarrow 1(1)+a(1)+b(1)=0\\ \Rightarrow a+b=-1$$     .......(ii)

Solving (i) and (ii), we get

$$\Rightarrow a=1,b=-2$$

So, option E is correct.

Given curve is
$$y = 5 + x - x^{2}$$     ..... (i)
On differentiating w.r.t. $$x$$, we get
$$\dfrac {dy}{dx} = 1 - 2x$$
Slope of normal $$= \dfrac {-1}{(dy/dx)} = \dfrac {-1}{1 - 2x}$$
$$= \dfrac {1}{2x - 1}$$     .... (ii)
Since, normal makes equal intercepts.
$$\therefore \theta = 135^{\circ}$$
From Eq. (ii), we have
$$\dfrac {1}{2x - 1} = \tan 135^{\circ} = -1$$
$$\Rightarrow -2x + 1 = 1\Rightarrow x = 0$$
Then, from Eq. (i), $$y = 5$$
So, the required point is $$(0, 5)$$.

Given curves are $$ y = 2^x $$ and $$ y =3^x $$
The point of intersection is $$ 3^x = 2^x \Rightarrow x = 0 $$
On differentiating w.r.t. $$x,$$ we get 
$$ \dfrac {dy}{dx} = 2^x \log 2 = m_1 $$
and $$ \dfrac {dy}{dx} = 3^x \log 3 = m_2 $$
Therefore, $$  \tan \alpha = \dfrac {m_2 - m_1}{1 + m_1m_2} $$
$$ = \dfrac {3^x \log 3 - 2^x \log 2 }{1+3^x \times 2^x \log 3 \times \log 2} $$
At $$x=0$$, 

Given curve is
$$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1 $$    ... (i)
On differentiating w.r.t. to $$x$$, we get
$$\dfrac {1}{\sqrt {a}} \cdot \dfrac {1}{2\sqrt {x}} + \dfrac {1}{\sqrt {b}} \cdot \dfrac {1}{2\sqrt {y}} \dfrac {dy}{dx} = 0$$
$$\Rightarrow \dfrac {dy}{dx} = -\dfrac {\sqrt {b}\sqrt {y}}{\sqrt {a}\sqrt {x}} \Rightarrow \left [\dfrac {dy}{dx}\right ]_{(x_{1}, y_{1})} = \dfrac {-\sqrt {b}\sqrt {y_{1}}}{\sqrt {a}\sqrt {x_{1}}}$$
Equation of tangent passing through the point $$(x_{1}, y_{1})$$ is
$$(y - y_{1}) = \dfrac {-\sqrt {b}\sqrt {y_{1}}}{\sqrt {a}\sqrt {x_{1}}} (x - x_{1})$$
$$\dfrac {y}{\sqrt {by_{1}}} - \dfrac {y_{1}}{\sqrt {by_{1}}} = -\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {x_{1}}{\sqrt {ax_{1}}}$$
$$\Rightarrow \dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = \dfrac {x_{1}}{\sqrt {ax_{1}}} + \dfrac {y_{1}}{\sqrt {by_{1}}} = \sqrt {\dfrac {x_{1}}{a}} + \sqrt {\dfrac {y_{1}}{b}}$$
$$\Rightarrow \dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = 1$$ [from Eq. (i)]
$$\left [\because at (x_{1}, y_{1}), \sqrt {\dfrac {x_{1}}{a}} + \sqrt {\dfrac {y_{1}}{b}} = 1\right ]$$
But $$\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = k$$ (given)
Therefore, $$k = 1$$

Given curve is $$y = x^2 - \dfrac{1}{x^2}$$
On differentiating both sides w.r.t. $$x$$, we get
$$\dfrac{dy}{dx} = 2x + \dfrac{2}{x^3}$$
At point $$(-1, 0)$$.
$$\dfrac{dy}{dx} = 2(-1) + \dfrac{2}{(-1)^3} = -4$$
Therefore, slope of normal to the curve
$$= - \dfrac{1}{dy/dx}$$
$$= - \dfrac{1}{-4} = \dfrac{1}{4}$$

Given curve is
$$y=3{ x }^{ 2 }-5x+6\Rightarrow \dfrac { dy }{ dx } =6x-5$$
$$\therefore $$ Slope of tangent at 
$$\left( 1,4 \right) =6\times 1-5=1$$

$$f(x)=2x^3-15x^2+36x+6$$

Let the normal $$P(at_1,2at_1)$$ by $$y= -t_1x + 2at_1+at_1^3$$

$$(2,14)$$ lies on the curve $$y=ax^3+bx+4$$

$$f(x)+2f(1-x)={ x }^{ 2 }+2...(i)$$
$$\Rightarrow f(1-x)+2f(x)={ (1-x) }^{ 2 }+2...(ii)$$
replacing $$x\rightarrow 1-x\quad $$
multiplying (ii) by $$2$$ then subtracting (1) we get
$$3f(x)={ x }^{ 2 }-4x+4={ (x-2) }$$
$$\Rightarrow f(x)=\cfrac { { (x-2) }^{ 2 } }{ 3 } $$
differentiating w.r.to $$x$$ both sides we get
$$f'(x)=\cfrac { 2(x-2) }{ 3 } >0\forall \left( 2,\infty  \right) $$