Tangents and its Equations Test 29

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Tangents and its Equations Test 29

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the tangent to the curves $$x=t\cos t$$ and $$y=t\sin t$$ at the origin is?

A
$$x=0$$

B
$$y=0$$

C
$$x+y=0$$

D
$$x-y=0$$

Solution

We have,
$$x=t\cos t $$ and $$y=t\sin t$$
$$\therefore \displaystyle\frac{dx}{dt}=\cos t -t \sin t$$
and $$\displaystyle\frac{dy}{dt}=\sin t +t\cos t$$
At the origin, we have
$$x=0, y=0$$
$$\Rightarrow t\cos t=0$$
and $$t\sin t=0$$
$$\Rightarrow t=0$$
The slope of the tangent at $$t=0$$ is

$$\displaystyle\frac{dy}{dx}=\left(\displaystyle\frac{dy/dt}{dx/dt}\right)_{t=0}=\left(\displaystyle\frac{\sin t+t \cos t}{\cos t - t\sin t}\right)_{t=0}=0$$
So, the equation of the tangent at the origin is $$y-0=0(x-0)$$
$$\Rightarrow y=0$$.
Question 2
1 / -0

Let $$f(x)=2{ x }^{ 3 }-5{ x }^{ 2 }-4x+3,\cfrac { 1 }{ 2 } \le x\le 3$$. The point at which the tangent to the curve is parallel to the X-axis is

A
$$(1,-4)$$

B
$$(2,-9)$$

C
$$(2,-4)$$

D
$$(2,-1)$$

E
$$(2,-5)$$

Solution

Given, $$f(x)=2{ x }^{ 3 }-5{ x }^{ 2 }-4x+3$$

$$f'(x)=6{x}^{2}-10x-4$$

If tangent is parallel to X-axis, then $$\cfrac { dy }{ dx } =0\quad $$

$$\Rightarrow 6{ x }^{ 2 }-10x-4=0\quad $$

$$\Rightarrow x=\cfrac { 5\pm \sqrt { 25+24 } }{ 6 } =\cfrac { 5\pm 7 }{ 6 } $$

$$\Rightarrow x=2,\cfrac { -1 }{ 3 } $$

$$\quad \therefore y=f(x)=2{ (2) }^{ 3 }-5{ (2) }^{ 2 }-4(2)+3=-9$$
Question 3
1 / -0

The equation of the tangent to the curve $$y={ x }^{ 3 }-6x+5$$ at $$(2,1)$$ is

A
$$6x-y-11=0$$

B
$$6x-y-13=0$$

C
$$6x+y+11=0$$

D
$$6x-y+11=0$$

Solution

The equation of the curve
$$y={ x }^{ 3 }-6x+5$$
$$\Rightarrow \cfrac { dy }{ dx } =3{ x }^{ 2 }-6$$
$$\Rightarrow { \left( \cfrac { dy }{ dx } \right) }_{ 2,1 }=6$$
Now, equation of the tangent at $$(2,1)$$ is
$$(y-1)=6(x-2)$$
$$\Rightarrow 6x-y-11=0$$
Question 4
1 / -0

The slope of the normal to the curve $$x=1-a\sin { \theta } $$, $$y=b\cos ^{ 2 }{ \theta }$$ at $$ \theta =\dfrac { \pi }{ 2 } $$ is

A
$$\dfrac { a }{ 2b } $$

B
$$\dfrac { 2a }{ b } $$

C
$$\dfrac { a }{ b } $$

D
$$\dfrac { -a }{ 2b } $$

Solution

Given, $$x=1-a\sin { \theta } $$ and $$y=b\cos ^{ 2 }{ \theta }$$
On differentiating with respect to $$\theta $$, we get
$$\dfrac { dx }{ d\theta } =-a\cos { \theta }$$
and $$ \dfrac { dy }{ d\theta } =2b\cos { \theta } \left( -\sin { \theta } \right)$$
Then, $$ \dfrac { dy }{ dx } =\dfrac { { dy }/{ d\theta } }{ { dx }/{ d\theta } } =\dfrac { 2b }{ a } \sin { \theta }$$
$$ \therefore$$ Slope of normal at the point $$ \theta =\dfrac { \pi }{ 2 }$$ is
$$ -\dfrac { dx }{ dy } =-\dfrac { 1 }{ { dy }/{ dx } } $$
$$=-\dfrac { 1 }{ \dfrac { 2b }{ a } \sin { \left( \dfrac { \pi }{ 2 } \right) } } =-\dfrac { a }{ 2b } $$
Question 5
1 / -0

The equation of the tangent to the curve $$y=4e^{-x/4}$$ at the point where the curve crosses Y-axis is equal to

A
$$3x+4y=16$$

B
$$4x+y=4$$

C
$$x+y=4$$

D
$$3x+4y=25$$

Solution

As the curve crosses Y-axis i.e. $$x=0$$
$$y=4e^{-0} \Rightarrow y = 4$$
Given, $$y=4e^{-x/4}$$
$$\Rightarrow \dfrac{dy}{dx} = 4e^{-x/4}\left(-\dfrac{1}{4}\right) = -e^{-x/4}$$
$$\Rightarrow \left(\dfrac{dy}{dx}\right)_{0,4} = -e^{-0}=-1$$
$$\therefore$$ Equation of tangent at $$(0, 4)$$ is $$y-4=-1(x-0)$$
$$\implies x+y=4$$
Question 6
1 / -0

The equation of tangent of the curve $$y = be^{-x/a}$$ at the point, where the curve meet y-axis is

A
$$bx + ay - ab = 0$$

B
$$ax - by - ab = 0$$

C
$$bx - ay - ab = 0$$

D
$$ax + by - ab = 0$$

Solution

We have
$$y = be^{-x/a}$$
$$\therefore \dfrac {dy}{dx} = \dfrac {-b}{a}e^{-x/a} .... (i)$$
Since, the curve meet y-axis. Hence, $$x = 0$$
$$\Rightarrow y = be^{-0/a} = b$$
Hence, point $$(0, b)$$ is on y-axis.
$$\therefore \left (\dfrac {dy}{dx}\right )_{(0, b)} = -\dfrac {b}{a}e^{-0/a} = -\dfrac {b}{a}$$
$$\therefore$$ Equation of tangent at $$(0, b)$$ is
$$y - b = -\dfrac {b}{a} (x - 0)$$
$$\Rightarrow ay - ab = -bx$$
$$\Rightarrow bx + ay - ab = 0$$.
Question 7
1 / -0

The tangents to curve $$y={ x }^{ 3 }-2{ x }^{ 2 }+x-2$$ which are parallel to straight line $$y=x$$, are

A
$$x+y=2$$ and $$x-y=\dfrac { 86 }{ 27 } $$

B
$$x-y=2$$ and $$x-y=\dfrac { 86 }{ 27 } $$

C
$$x-y=2$$ and $$x+y=\dfrac { 86 }{ 27 } $$

D
$$x+y=2$$ and $$x+y=\dfrac { 86 }{ 27 } $$

Solution

Given,
$$y={ x }^{ 3 }-2{ x }^{ 2 }+x-2$$

On differentiating both sides with respect to $$x$$, we get

$$\dfrac { dy }{ dx } =3{ x }^{ 2 }-4x+1$$

and $$y=x$$

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

$$\therefore$$ Slope of tangent will be $$ 3{ x }^{ 2 }-4x+1$$.

Since, the tangent is parallel to line $$y=x$$.

$$\therefore 3{ x }^{ 2 }-4x+1=1$$

$$\Rightarrow 3{ x }^{ 2 }-4x=0$$

$$\Rightarrow x\left( 3x-4 \right) =0$$

$$\Rightarrow x=0,\dfrac { 4 }{ 3 }$$

When $$x=0$$, then $$y=-2$$

When $$x=\dfrac { 4 }{ 3 } $$, then $$y=\dfrac { -50 }{ 27 }$$

Now, equation of tangents at point $$ \left( 0,-2 \right)$$ is

$$ y-{ y }_{ 1 }=\dfrac { dy }{ dx } \left( x-{ x }_{ 1 } \right)$$

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

$$ \Rightarrow y+2=x$$

$$\Rightarrow x-y=2$$ ...(i)

and equation of tangents at point $$\left( \dfrac { 4 }{ 3 } ,-\dfrac { 50 }{ 27 } \right) $$ is

$$y-{ y }_{ 1 }=\dfrac { dy }{ dx } \left( x-{ x }_{ 1 } \right) $$

$$y+\dfrac { 50 }{ 27 } =x-\dfrac { 4 }{ 3 }$$

$$ \Rightarrow x-y=\dfrac { 50 }{ 27 } +\dfrac { 4 }{ 3 }$$

$$ \Rightarrow x-y=\dfrac { 50+36 }{ 27 } $$

$$\Rightarrow x-y=\dfrac { 86 }{ 27 } $$ ....(ii)

Hence, equations (i) and (ii) are required equations of the tangents.
Question 8
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The slope of 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
$$-6$$

D
None of these

Solution

Given curves are $$ x=t^{2} + 3t - 8$$ ... $$(i)$$ and $$y = 2t^{2} - 2t - 5 $$ ... $$(ii)$$
At $$(2,-1),$$ From $$(i)$$
$$t^{2}+3t-10=0\implies t=2$$ or $$t=-5$$
From $$(ii)$$
$$2t^{2}-2t-4=0\implies t^{2}-t-2=0\implies t=2$$ or $$t=-1$$
From both the solutions, we get $$t=2$$

Differentiating both the equations w.r.t. $$t$$, we get
$$\dfrac{dx}{dt}=2t+3$$ ........ $$(iii)$$
$$\dfrac{dy}{dt}=4t-2$$ ....... $$(iv)$$

Now, $$ \dfrac{dy}{dx} = \dfrac {\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$$

$$ = \dfrac {4t-2}{2t+3}$$ .... From $$(iii)$$ and $$(iv)$$
$$\therefore \dfrac{dy}{dx} = \dfrac {4t-2}{2t+3}$$ is the slope of tangent to the given curve
$$\therefore \left|\dfrac{dy}{dx}\right|_{(2,-1)} = \left|\dfrac {4t-2}{2t+3}\right|_{t=2}=\dfrac{8-2}{4+3}=\dfrac{6}{7}$$ is the slope of tangent to the given curve at $$(2,-1)$$
Question 9
1 / -0

The points at which the tangent to the curve $$y = x^3 - 3x^2 - 9x + 7$$ is parallel to the x-axis are

A
$$(3, - 20)$$ and $$(- 1, 12)$$

B
$$(3, 20)$$ and $$(1, 12)$$

C
$$(1, -10)$$ and $$(2, 6)$$

D
None of these

Solution

Tangent to the curve is parallel to the axis is when slope of the tangent is 0.
$$\therefore$$ Equation of the curve is
$$y=x^3-3x^2-9x+7=0$$ ...... $$(i)$$
$$\therefore \dfrac{dy}{dx}=3x^2-6x-9$$

Now, the tangent is parallel to x-axis, then slope of the tangent is zero or we can say that $$\dfrac{dy}{dx}=0$$.
$$\Rightarrow 3x^2-6x-9=0$$
$$\Rightarrow 3(x^2-2x-3)=0$$
$$\Rightarrow (x-3)(x+1)=0$$
$$\Rightarrow x=3, -1$$

When $$x=3$$, then from Eq. $$(i),$$ we get
$$y=(3)^3-(3)\cdot (3)^2-9\cdot 3+7$$
$$=27-27-27+7=-20$$

When $$x =-1,$$ then from Eq. $$(i),$$ we get
$$y=(-1)^3-3(-1)^2-9(-1)+7$$
$$=-1-3+9+7=12$$
Hence, the points at which the tangent is parallel to x-axis are $$(3, -20)$$ and $$(-1, 12)$$.
Question 10
1 / -0

The slope of the tangent to the curve $$y=3{ x }^{ 2 }-5x+6$$ at $$\left( 1,4 \right) $$ is

A
$$-2$$

B
$$1$$

C
$$0$$

D
$$-1$$

E
$$2$$

Solution

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

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 29

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Tangents and its Equations Test 29

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SHARING IS CARING

Question 1

$$x=0$$

$$y=0$$

$$x+y=0$$

$$x-y=0$$

Solution

Question 2

$$(1,-4)$$

$$(2,-9)$$

$$(2,-4)$$

$$(2,-1)$$

$$(2,-5)$$

Solution

Question 3

$$6x-y-11=0$$

$$6x-y-13=0$$

$$6x+y+11=0$$

$$6x-y+11=0$$

Solution

Question 4

$$\dfrac { a }{ 2b } $$

$$\dfrac { 2a }{ b } $$

$$\dfrac { a }{ b } $$

$$\dfrac { -a }{ 2b } $$

Solution

Question 5

$$3x+4y=16$$

$$4x+y=4$$

$$x+y=4$$

$$3x+4y=25$$

Solution

Question 6

$$bx + ay - ab = 0$$

$$ax - by - ab = 0$$

$$bx - ay - ab = 0$$

$$ax + by - ab = 0$$

Solution

Question 7

$$x+y=2$$ and $$x-y=\dfrac { 86 }{ 27 } $$

$$x-y=2$$ and $$x-y=\dfrac { 86 }{ 27 } $$

$$x-y=2$$ and $$x+y=\dfrac { 86 }{ 27 } $$

$$x+y=2$$ and $$x+y=\dfrac { 86 }{ 27 } $$

Solution

Question 8

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

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

$$-6$$

None of these

Solution

Question 9

$$(3, - 20)$$ and $$(- 1, 12)$$

$$(3, 20)$$ and $$(1, 12)$$

$$(1, -10)$$ and $$(2, 6)$$

None of these

Solution

Question 10

$$-2$$

$$1$$

$$0$$

$$-1$$

$$2$$

Solution

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