Tangents and its Equations Test 26

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Tangents and its Equations Test 26

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Weekly Quiz Competition

Question 1
1 / -0

The abscissa of the points, where the tangent to curve $$y={x}^{3} - 3{x}^{2} - 9x+5$$ is parallel to x-axis, are

A
$$x=0$$ and $$0$$

B
$$x=1$$ and $$-1$$

C
$$x=1$$ and $$-3$$

D
$$x=-1$$ and $$3$$

Solution

Given, $$y={ x }^{ 3 }-3{ x }^{ 2 }-9x+5$$
$$\Rightarrow \dfrac { dy }{ dx } =3{ x }^{ 2 }-6x-9$$
We know that, this equation gives the slope of the tangent to the curve. The tangent is parallel to x-axis,
$$\therefore \dfrac { dy }{ dx } =0$$
$$\Rightarrow 3{ x }^{ 2 }-6x-9=0$$
$$\Rightarrow x=-1,3$$
Question 2
1 / -0

Suppose that the equation $$f\left( x \right) ={ x }^{ 2 }+bx+c=0$$ has two distinct real roots $$\alpha $$ and $$\beta $$. The angle between the tangent to the curve $$y=f\left( x \right) $$ at the point $$\left( \dfrac { \alpha +\beta }{ 2 } ,f\left( \dfrac { \alpha +\beta }{ 2 } \right) \right) $$ and the positive direction of the $$x$$-axis is

A
$${ 0 }^{ }$$

B
$${ 30 }^{ }$$

C
$${ 60 }^{ }$$

D
$${ 90 }^{ }$$

Solution

Since, $$\alpha $$ and $$\beta $$ are the roots of
$$f\left( x \right) ={ x }^{ 2 }+bx+c$$
$$\therefore \alpha +\beta =-b$$ and $$\alpha \beta =c$$
$$\therefore f\left( \dfrac { \alpha +\beta }{ 2 } \right) ={ \left( \dfrac { \alpha +\beta }{ 2 } \right) }^{ 2 }+b\left( \dfrac { \alpha +\beta }{ 2 } \right) +c$$
$$={ \left( -\dfrac { b }{ 2 } \right) }^{ 2 }+b\left( -\dfrac { b }{ 2 } \right) +c$$
$$=\dfrac { { b }^{ 2 } }{ 4 } -\dfrac { { b }^{ 2 } }{ 2 } +c=-\dfrac { { b }^{ 2 } }{ 4 } +c$$

Now, $$\dfrac { dy }{ dx } =f^{ \prime }\left( x \right) =2x+b$$

At point, $$\left( \dfrac { \alpha +\beta }{ 2 } ,f\left( \dfrac { \alpha +\beta }{ 2 } \right) \right) $$, i.e., $$\left( -\dfrac { b }{ 2 } ,-\dfrac { { b }^{ 2 } }{ 4 } +c \right) $$,

$$\dfrac { dy }{ dx } =2\left( -\dfrac { b }{ 2 } \right) +b=0$$

Hence, the slope of the tangent to the curve and the positive direction of $$x$$-axis is $${ 0 }^{ }$$.
Question 3
1 / -0

The points on the curve $$9{y}^{2}={x}^{3}$$, where the normal to the curve makes equal intercepts with the axes are

A
$$\left( 4,\pm \cfrac { 8 }{ 3 } \right) $$

B
$$\left( 4,\cfrac { -8 }{ 3 } \right) $$

C
$$\left( 4,\pm \cfrac { 3 }{ 8 } \right) $$

D
$$\left( \pm 4,\cfrac { 3 }{ 8 } \right) $$

E
answer required

Solution

Let the point on curve be $$(h,k)$$
$$y^2= \dfrac{x^3}{9}\\
2y.\dfrac{dy}{dx}= \dfrac{1}{9} .3 x^2\\
\therefore \dfrac{dy}{dx}= \dfrac{x^2}{6y}$$
$$\dfrac{dy}{dx}$$ gives slope of curve at given point
$$\dfrac{dy}{dx}\;at\;(h,k) = \dfrac{h^2}{6k}$$
but this is the slope of tangent and we need slope of normal
As we know that slope of normal$$\times$$ slope of tangent is $$-1$$
$$\therefore$$ slope of normal is $$\dfrac{-6k}{h^2}$$
Given in question that normal makes equal intercepts on axes which means slope of normal is either $$1$$ or $$-1$$
$$-6k=h^2$$ and $$6k=h^2$$
Now go through the option you will that option A satisfy our equation
Question 4
1 / -0

The coordinates of the point P on the curve $$x = a(\theta + \sin \theta), y = a(1 - \cos \theta)$$ where the tangent is inclined at an angle $$\dfrac {\pi}{4}$$ to the x-axis, are

A
$$\left (a\left (\dfrac {\pi}{2} - 1\right ), a\right )$$

B
$$\left (a\left (\dfrac {\pi}{2} + 1\right ), a\right )$$

C
$$\left (a \dfrac {\pi}{2}, a\right )$$

D
$$(a, a)$$

Solution

$$x = a(\theta + \sin \theta), y = a(1 - \cos \theta)$$
$$\dfrac {dy}{d\theta} = a\sin \theta, \dfrac {dx}{d\theta} = a(1 + \cos \theta)$$
$$\dfrac {dy}{dx} = \dfrac {\sin \theta}{1 + \cos \theta} = \tan \dfrac {\theta}{2}$$
$$\Rightarrow \tan \dfrac {\theta}{2} = 1$$
$$\Rightarrow \dfrac {\theta}{2} = \dfrac {\pi}{4} \Rightarrow \theta = \dfrac {\pi}{2}$$
$$x = a(\theta + \sin \theta)$$
$$y = a(1 - \cos \theta)$$
$$x = a (\theta + \sin \theta)$$
$$y = a(1 - \cos \theta)$$
$$(x, y) \equiv \left (a \left (\dfrac {\pi}{2} + 1\right ), a\right )$$.
Question 5
1 / -0

Slope of Normal to the curve $$y = x^{2} - \dfrac {1}{x^{2}}$$ at $$(-1, 0)$$ is

A
$$\dfrac {1}{4}$$

B
$$-\dfrac {1}{4}$$

C
$$4$$

D
$$-4$$

Solution

We have: $$y=x^2-\dfrac{1}{x^2}=x^2-x^{-2}$$
Now differentiate,
$$\dfrac{dy}{dx}=2x+2x^{-3}$$
So slope of tangent at $$(-1,0)$$ is
$$\dfrac{dy}{dx}\bigg|_{x=-1}=2(-1)+2(-1)^3=-2-2=-4$$
Hence slope of normal at the same point is $$=\dfrac{1}{4}$$

Note: Normal and tangent are perpendicular to each other
Question 6
1 / -0

The slope of the tangent to the curve $$x=3t^2+1, y=t^3-1$$ at $$x=1$$ is

A
$$\dfrac{1}{2}$$

B
$$0$$

C
$$-2$$

D
$$\infty$$

Solution

Given, $$x=3t^2+1, y=t^3-1$$
Slope of the tangent to the given curve is $$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx}$$
$$= 3t^2 \times \dfrac{1}{6t}$$
$$=\dfrac{t}{2}$$
Since the slope has to be calculated at $$x = 1, $$ i.e. at $$3t^2 + 1 = 1$$, we get $$t = 0$$
Thus, the required slope is $$0.$$
Question 7
1 / -0

The equation of one of the curves whose slope at any point is equal to $$y+2x$$ is

A
$$y=2(e^x+x-1)$$

B
$$y=2(e^x-x-1)$$

C
$$y=2(e^x-x+1)$$

D
$$y=2(e^x+x+1)$$

Solution

Given, $$\dfrac{dy}{dx}=y+2x$$
Put $$y+2x=z$$
$$\Rightarrow\;\dfrac{dy}{dx}+2=\dfrac{dz}{dx}$$
$$\Rightarrow\;\dfrac{dy}{dx}=\dfrac{dz}{dx}-2\;\dots(ii)$$
From Eqs. (i) and (ii)
$$\dfrac{dz}{dx}-2=z$$
$$\Rightarrow\;\int\dfrac{dz}{.z+2}=\int\,dx$$
$$\Rightarrow\;log(z+2)=x+c$$
$$\Rightarrow\;log(y+2x+2)=x+c$$
$$\Rightarrow\;y+2x+2=e^{x+c}$$
$$\Rightarrow\;y+2x+2=e^x.e^c$$
$$\Rightarrow\;y=2[e^x-x-1]$$
Question 8
1 / -0

The slope of the tangent to the curve $$x={t}^{2}+3t-8$$, $$y=2{t}^{2}-2t-5$$ at the point $$(2,-1)$$ is

A
$$\cfrac{22}{7}$$

B
$$\cfrac{6}{7}$$

C
$$\cfrac{7}{6}$$

D
$$\cfrac{-6}{7}$$

Solution

We have $$x=t^2+3t-8$$ and $$y=2t^2-2t-5$$
$$\Rightarrow 2x-y=2(t^2+3t-8)-(2t^2-2t-5)=8t-11$$
Now substitute the point $$(2,-1)$$ to get value of $$t$$
$$\Rightarrow 2(2)-(-1)=8t-11\Rightarrow 5+11=8t\Rightarrow t=2$$
Now differentiate $$x$$ and $$y$$ w.r.t.
$$\Rightarrow \dfrac{dx}{dt}=2t+3$$ and $$\dfrac{dy}{dt}=4t-2$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}=\dfrac{4t-2}{2t+3}$$
$$\therefore \dfrac{dy}{dx}\bigg|_{t=2}=\dfrac{4(2)-2}{2(2)+3}=\dfrac{6}{7}$$
Question 9
1 / -0

If $$\Delta$$ is the area of the triangle formed by the positive x-axis and the normal and tangent to the circle $$x^{2} + y^{2} = 4$$ at $$(1, \sqrt {3})$$, then $$\Delta =$$

A
$$\dfrac {\sqrt {3}}{2}$$

B
$$\sqrt {3}$$

C
$$2\sqrt {3}$$

D
$$6$$

Solution

$$x^{2} + y^{2} = 4$$
Differentiating wrt x, we get
$$\dfrac {dy}{dx} = -\dfrac {x}{y}, m = \dfrac {-1}{\sqrt {3}}$$
Equation of tangent: $$ y - \sqrt {3} = \dfrac {-1}{\sqrt {3}} (x - 1)$$
$$\Rightarrow \sqrt {3}y - 3 = -x + 1$$
$$\Rightarrow x + \sqrt {3}y = 4$$

$$Area = \left |\dfrac {y_{1}^{2}(1 + m^{2})}{2m}\right | = 2\sqrt {3}$$
Question 10
1 / -0

If the tangent to the curve $$2y^{3} = ax^{2} + x^{3}$$ at the point $$(a, a)$$ cuts off intercepts $$\alpha$$ and $$\beta$$ on the coordinate axes where $$\alpha^{2} + \beta^{2} = 61$$ then the value of '$$a$$' is equal to

A
$$25$$

B
$$36$$

C
$$\pm 30$$

D
$$\pm 40$$

Solution

$$2y^{3} = ax^{2} + x^{3}$$

Diff. w.r.t $$x$$

$$6y^{2} \dfrac {dy}{dx} = 2ax + 3x^{2}$$

$$\dfrac {dy}{dx}|_{(a, a)} = \dfrac {2a^{2} + 3a^{2}}{6a^{2}} = \dfrac {5}{6}$$

Hence tangent at $$(a,a)$$ is,

$$y-a =\dfrac{5}{6}(x-a)\Rightarrow 6y-6a=5x-5a\Rightarrow 5x-6y+a=0$$

$$\therefore x - intercept = \dfrac {-a}{5} = \alpha$$

and $$y - intercept = \dfrac {a}{6} = \beta$$

$$\therefore \alpha^{2} + \beta^{2} = 61\Rightarrow \dfrac {a^{2}}{25} + \dfrac {a^{2}}{36} = 61$$

$$\dfrac{61a^2}{900}=61\Rightarrow a^2=900$$

$$\therefore a = \pm 30$$

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Tangents and its Equations Test 26

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SHARING IS CARING

Question 1

$$x=0$$ and $$0$$

$$x=1$$ and $$-1$$

$$x=1$$ and $$-3$$

$$x=-1$$ and $$3$$

Solution

Question 2

$${ 0 }^{ }$$

$${ 30 }^{ }$$

$${ 60 }^{ }$$

$${ 90 }^{ }$$

Solution

Question 3

$$\left( 4,\pm \cfrac { 8 }{ 3 } \right) $$

$$\left( 4,\cfrac { -8 }{ 3 } \right) $$

$$\left( 4,\pm \cfrac { 3 }{ 8 } \right) $$

$$\left( \pm 4,\cfrac { 3 }{ 8 } \right) $$

answer required

Solution

Question 4

$$\left (a\left (\dfrac {\pi}{2} - 1\right ), a\right )$$

$$\left (a\left (\dfrac {\pi}{2} + 1\right ), a\right )$$

$$\left (a \dfrac {\pi}{2}, a\right )$$

$$(a, a)$$

Solution

Question 5

$$\dfrac {1}{4}$$

$$-\dfrac {1}{4}$$

$$4$$

$$-4$$

Solution

Question 6

$$\dfrac{1}{2}$$

$$0$$

$$-2$$

$$\infty$$

Solution

Question 7

$$y=2(e^x+x-1)$$

$$y=2(e^x-x-1)$$

$$y=2(e^x-x+1)$$

$$y=2(e^x+x+1)$$

Solution

Question 8

$$\cfrac{22}{7}$$

$$\cfrac{6}{7}$$

$$\cfrac{7}{6}$$

$$\cfrac{-6}{7}$$

Solution

Question 9

$$\dfrac {\sqrt {3}}{2}$$

$$\sqrt {3}$$

$$2\sqrt {3}$$

$$6$$

Solution

Question 10

$$25$$

$$36$$

$$\pm 30$$

$$\pm 40$$

Solution

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