Tangents and its Equations Test 42

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Tangents and its Equations Test 42

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Weekly Quiz Competition

Question 1
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 curves are,
$$x=3t^2+1$$ ---- ( 1 )
$$y=t^3-1$$ ---- ( 2 )
Substituting $$x=1$$ in ( 1 ) we get,
$$\Rightarrow$$ $$3t^2+1=1$$
$$\Rightarrow$$ $$3t^2=0$$
$$\Rightarrow$$ $$t=0$$
Differentiate ( 1 ) w.r.t. $$t,$$ we get
$$\Rightarrow$$ $$\dfrac{dx}{dt}=6t$$
Differentiate ( 2 ) w.r.t. $$t,$ we get
$$\Rightarrow$$ $$\dfrac{dy}{dt}=3t^2$$

$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3t^2}{6t}$$

$$\therefore$$ $$\dfrac{dy}{dx}=\dfrac{t}{2}$$

Slope of the tangent $$=\left(\dfrac{dy}{dx}\right)_{t=0}=\dfrac{0}{2}=0$$
Question 2
1 / -0

The point on the curve $$y=12x-x^2$$, where the slope of the tangent is zero will be

A
$$(0, 0)$$

B
$$(2, 16)$$

C
$$(3, 9)$$

D
$$(6, 36)$$

Solution

Let the point be $$P(x, y)$$
$$y=12x-x^2$$
$$\dfrac{dy}{dx}=12-2x$$
$$\left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=12-2x_1$$

since slope of tangent is zero
so $$\left(\dfrac{dy}{dx}\right)_{x_1, y_1}=0$$
$$12-2x_1=0$$
$$2x_1=12$$
$$x_1=6$$

Also curve passing through tangent
$$y_1=12x_1-x^2_1$$
$$y_1=12\times 6-36$$
$$y_1=72-36$$
$$y_1=36$$
The points are $$(6, 36)$$.
Question 3
1 / -0

The point on the curve $$y=x^2-3x+2$$ where tangent is perpendicular to $$y=x$$ is

A
$$(0, 2)$$

B
$$(1, 0)$$

C
$$(-1, 6)$$

D
$$(2, -2)$$

Solution

$$y=x^2-3x+2$$ [ Given ]

$$y=x$$ [ Given ]

Differentiating both sides w.r.t. $$x,$$ we get

$$\Rightarrow$$ $$\dfrac{dy}{dx}=1$$

Let $$(x_1,y_1)$$ be the required point.

It is given that point lies on the curve.

$$\therefore$$ $$y_1=x_1^2-3x_1+2$$

$$\Rightarrow$$ $$y=x^2-3x+2$$

Differentiating both sides w.r.t. $$x,$$ we get

$$\therefore$$ $$\dfrac{dy}{dx}=2x-3$$

Slope of the tangent $$=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=2x_1-3$$

The tangent is perpendicular to the line.

$$\therefore$$ Slope of the tangent $$=\dfrac{-1}{slope\,of\,the\,line}=\dfrac{-1}{1}=-1$$

Now,

$$2x_1-3=-1$$

$$\Rightarrow$$ $$2x_1=2$$

$$\Rightarrow$$ $$x_1=1$$

$$\Rightarrow$$ $$y_1=x_1^2-3x_1+2$$

$$\Rightarrow$$ $$y_1=(1)^2-3(1)+2$$

$$\Rightarrow$$ $$y_1=1-3+2$$

$$\therefore$$ $$y_1=0$$

$$\therefore$$ $$(x_1,y_1)=(1,0)$$
Question 4
1 / -0

Any tangent to the curve $$y=2x^7+3x+5$$

A
Is parallel to $$x$$-axis

B
Is parallel to $$y$$-axis

C
Makes an acute angle with $$x$$-axis

D
Makes an obtuse angle with $$x$$-axis

Solution

We have, $$y=2x^7+3x+5$$

$$\dfrac{dy}{dx}=14x^6+3$$

$$\Rightarrow \dfrac{dy}{dx} > 3$$ ($$\because x^6$$ is always positive for any real value)

$$\Rightarrow \dfrac{dy}{dx} > 0$$ so, $$\tan \theta > 0$$

Hence, $$\theta$$ lies in first quadrant.

Thus, the tangent to curve makes an acute angle.
Question 5
1 / -0

The normal to the curve $$x^2=4y$$ passing through $$(1, 2)$$ is

A
$$x+y=3$$

B
$$x-y=3$$

C
$$x+y=1$$

D
$$x-y=1$$

Solution

Curve is $$x^2=4y$$

Diff wrt to x
$$2x=\dfrac{4dy}{dx}\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{2}$$

Slope of normal $$=\dfrac{-1}{dy/dx}=\dfrac{-2}{x}$$
Let (h, k) be the point where normal and curve intersects
$$\therefore$$ Slope of normal at (h, k)$$=-2/h$$

Equation of normal passes through (h, k)
$$y-y_1=m(x-x_1)$$
$$y-k=\dfrac{-2}{h}(x-h)$$

Since normal passes through $$(1, 2)$$
$$2-k=\dfrac{-2}{h}(1-h)$$
$$k=2+\dfrac{2}{h}(1-h)$$ ..........$$(1)$$

since (h, k) lies on the curve $$x^2=4y$$
$$h^2=4k$$
$$k=\dfrac{h^2}{4}$$ ....... $$(2)$$

using $$(1)$$ and $$(2)$$
$$2+\dfrac{2}{h}(1-h)=\dfrac{h^2}{4}$$
$$\dfrac{2}{h}=\dfrac{h^2}{4}$$
$$h=2$$

Putting $$h=2$$ in $$(2)$$
$$k=\dfrac{h^2}{4}, k=1$$
$$h=2$$ and $$k=1$$ putting in equation of normal

$$\Rightarrow y-k=\dfrac{-2(k-h)}{h}$$

$$\Rightarrow y-1=\dfrac{-2(x-2)}{2}=y-1=-1(x-2)$$

$$\Rightarrow y-1=-x+2\\$$
$$\Rightarrow x+y=2+1\\$$
$$\Rightarrow x+y=3$$.
Question 6
1 / -0

Mark the correct alternative of the following.
The point on the curve $$9y^2=x^3$$, where the normal to the curve makes equal intercepts with the axes is?

A
$$(4, \pm 8/3)$$

B
$$(-4, 8/3)$$

C
$$(-4, -8/3)$$

D
$$(8/3, 4)$$

Solution

Let the required point be $$(x_1, y_1)$$

equation of the curve is $$9y^2=x^2$$

since $$(x_1, y_1)$$ lies on the curve, therefore
$$9y^2_1=x^3_1$$ .......$$(1)$$

Now $$9y^2=x^3$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{x^2}{6y}$$

$$\Rightarrow \left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=\dfrac{x^2}{6y^1}$$

since normal to the curve at $$(x_1, y_1)$$ makes equal intersepts with the coordinate axis, therefore slope of the normal $$=\pm 1$$

$$\Rightarrow \dfrac{1}{-(dy/dx)_{(x_1, y_1)}}=\pm 1$$

$$\Rightarrow (dy/dx)_{(x_1, y_1)}=\pm 1$$

$$\rightarrow \dfrac{x^2_1}{6y^1}=\pm 1$$

$$\Rightarrow x^4_1=36y^2_1=36\left(\dfrac{x^3_1}{9}\right)$$ (using $$1$$)

$$\Rightarrow x^4_1=4x^3_1\Rightarrow x^3_1(x_1-4)=0$$

$$\Rightarrow x_1=0, 4$$

Putting $$x_1=0$$ in $$(1)$$ we get
$$9y^2_1=0\Rightarrow y_1=0$$

Putting $$x_1=4$$ in $$(1)$$ we get
$$9y^2_1=(4)^3\Rightarrow y_1=\pm \dfrac{8}{3}$$

But the line making equal intersepts with the coordinate axes cannot pass through origin.

Hence, the required points are $$\left(4, \dfrac{8}{3}\right)$$ and $$\left(4, \dfrac{-8}{3}\right)$$.
Question 7
1 / -0

Mark the correct alternative of the following.
The line $$y=mx+1$$ is a tangent to the curve $$y^2=4x$$, if the value of m is?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$1/2$$

Solution

Given equation of the tangent to the given curve
$$y=mx+1$$
Now substituting the value of y in
$$y^2=4x$$, we get
$$\Rightarrow (mx+1)^2=4x$$
$$\Rightarrow m^2x^2+1+2mx-4x=0$$
$$\Rightarrow m^2x^2+x(2m-4)+1=0$$ ..........$$(1)$$
Since, a tangent touches the curve at one point, the root of equation $$(1)$$ must be equal.
Thus, we get
Discriminant, $$D=b^2-4ac=0$$
$$(2m-4)^2-4(m^2)(1)=0$$
$$\Rightarrow 4m^2-16m+16-4m^2=0$$
$$\Rightarrow -16m+16=0$$
$$\Rightarrow m=1$$.
Question 8
1 / -0

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

A
$$\dfrac {22}{7}$$

B
$$\dfrac {6}{7}$$

C
$$\dfrac {7}{6}$$

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

Solution

Given curve

$$x = t^2 + 3t - 8$$ and $$y = 2t^2 - 2t - 5$$

slope of tangent to the curve $$= \dfrac{dy}{dx} = \dfrac{dy/dt}{dx / dt}$$

$$\therefore \dfrac{dx}{dt} = 2t + 3 , \, \dfrac{dy}{dt} = 4t - 2$$

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{4t - 2}{2t + 3}$$

$$\left(\dfrac{dy}{dx} \right)_{t = 2} $$ $$\dfrac{4(2) - 2}{2(2) + 3} = \dfrac{6}{7}$$
Question 9
1 / -0

Consider the equation $$x^y=e^{x-y}$$
What is $$\dfrac{d^2y}{dx^2}$$ at $$x=1$$ equal to ?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

$${x}^{y} = {e}^{x - y}$$
Taking $$\log$$ both sides, we have

$$y \log{x} = \left( x - y \right) \log{e}$$

$$y \log{x} = x - y ..... \left( 1 \right)$$

$$\dfrac{y}{x}=1+\log x$$

At $$x = 1 \Rightarrow y = 1$$

Differentiating equation $$\left( 1 \right)$$ w.r.t. $$x$$, we have

$$\cfrac{dy}{dx} \log{x} + \cfrac{y}{x} = 1 - \cfrac{dy}{dx}$$

$$\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}$$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}$$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{y \log{x}}{x \left( 1 + \log{x} \right)}$$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{\log{x}}{{\left( 1 + \log{x} \right)}^{2}}$$

Again differentiating above equation w.r.t. $$x$$, we have

$$\cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{{\left( 1 + \log{x} \right)}^{2} \cdot \frac{1}{x} - \log{x} \cdot 2 \left( 1 + \log{x} \right) \frac{1}{x}}{{\left( 1 + \log{x} \right)}^{4}}$$

$$\Rightarrow \cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{\frac{1}{x} \left( 1 - \log{x} \right)}{{\left( 1 + \log{x} \right)}^{3}}$$

At $$x = 1$$,
$$\cfrac{{d}^{2}y}{d{x}^{2}} = 1$$
Question 10
1 / -0

Consider the equation $$x^y=e^{x-y}$$
What is $$\dfrac{dy}{dx}$$ at $$x=1$$ equal to ?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

$${x}^{y} = {e}^{x - y}$$

Taking $$\log$$ both sides, we have

$$y \log{x} = \left( x - y \right) \log{e}$$

$$y \log{x} = x - y ..... \left( 1 \right)$$

At $$x = 1 \Rightarrow y = 1$$

Differentiating equation $$\left( 1 \right)$$ w.r.t. $$x$$, we have

$$\cfrac{dy}{dx} \log{x} + \cfrac{y}{x} = 1 - \cfrac{dy}{dx}$$

$$\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}$$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}$$

At $$x = 1, \; y = 1$$

$$\cfrac{dy}{dx} = 0$$

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 42

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Tangents and its Equations Test 42

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SHARING IS CARING

Question 1

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

$$0$$

$$-2$$

$$\infty$$

Solution

Question 2

$$(0, 0)$$

$$(2, 16)$$

$$(3, 9)$$

$$(6, 36)$$

Solution

Question 3

$$(0, 2)$$

$$(1, 0)$$

$$(-1, 6)$$

$$(2, -2)$$

Solution

Question 4

Is parallel to $$x$$-axis

Is parallel to $$y$$-axis

Makes an acute angle with $$x$$-axis

Makes an obtuse angle with $$x$$-axis

Solution

Question 5

$$x+y=3$$

$$x-y=3$$

$$x+y=1$$

$$x-y=1$$

Solution

Question 6

$$(4, \pm 8/3)$$

$$(-4, 8/3)$$

$$(-4, -8/3)$$

$$(8/3, 4)$$

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$1/2$$

Solution

Question 8

$$\dfrac {22}{7}$$

$$\dfrac {6}{7}$$

$$\dfrac {7}{6}$$

$$-\dfrac {6}{7}$$

Solution

Question 9

$$0$$

$$1$$

$$2$$

$$4$$

Solution

Question 10

$$0$$

$$1$$

$$2$$

$$4$$

Solution

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