Tangents and its Equations Test 5

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Tangents and its Equations Test 5

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the normal to the curve ${ y }^{ 4 }=a{ x }^{ 3 }$ at $\left( a,a \right)$ is

A
$x+2y=3a$

B
$3x-4y+a=0$

C
$4x+3y=7a$

D
$4x-3y=0$

Solution

Given curve is ${ y }^{ 4 }=a{ x }^{ 3 }$
On differentiating with respect to $x$ , we get
$4{ y }^{ 3 }\dfrac { dy }{ dx } =3a{ x }^{ 2 }$
$\Rightarrow { \left( \dfrac { dy }{ dx } \right) }_{ \left( a,a \right) }=\dfrac { 3{ a }^{ 3 } }{ 4{ a }^{ 3 } } =\dfrac { 3 }{ 4 }$
$\therefore$ Equation of normal at point $\left( a,a \right)$ is
$y-a=-\dfrac { 4 }{ 3 } \left( x-a \right)$
$\Rightarrow 4x+3y=7a$

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Question 2
1 / -0

If tangent to the curve $\displaystyle x={ at }^{ 2 },y=2at$ is perpendicular to $x$ -axis, then its point of contact is:

A
$(a, a)$

B
$(0, a)$

C
$(0, 0)$

D
$(a, 0)$

Solution

We have, $x=at^2\Rightarrow \dfrac{dx}{dt}=2at$
And $y=2at\Rightarrow \dfrac{dy}{dt}=2a$
$\therefore \dfrac{dy}{dx}=\dfrac{1}{t}=\infty$ , for tangent to the curve perpendicular to $x-$ axis
$\Rightarrow t = 0$
so the point of contact is $(at^2,2at)=(0,0)$

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Question 3
1 / -0

The slope of the normal to the curve $y = 2x^2+ 3 \sin x$ at $x = 0$ is.

A
$3$

B
$\dfrac{1}{3}$

C
$-3$

D
$-\dfrac{1}{3}$

Solution

$y = 2x^2+ 3 \sin x$
$\cfrac{dy}{dx} = 4x+3\cos x$
Thus slope of tangent at $x=0$ is $4(0)+3\cos 0=3$
Hence slope of normal at the same point is $-\cfrac{1}{m}=-\cfrac{1}{3}$

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Question 4
1 / -0

The slope of the tangent to the curve $y=\displaystyle\int_{0}^{x}\dfrac{dt}{1+t^3}$ at the point where x=1 is

A
$\dfrac{1}{4}$

B
$\dfrac{1}{3}$

C
$\dfrac{1}{2}$

D
1

Solution

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

$m=\left ( \dfrac{dy}{dx} \right )x=1=\dfrac{1}{1+1}=\dfrac{1}{2}$

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Question 5
1 / -0

The equation of the tangent to the curve $x=\sqrt{t}, y=t-\dfrac{1}{\sqrt{t}}$ at $t=4$ is:

A
$x-4y=20$

B
$17x-4y=20$

C
$17x+4y=20$

D
$4x-17y=20$

Solution

Given, $x=\sqrt{t}$ .......(1), $y=t-\dfrac{1}{\sqrt{t}}$ ........(2).
When $t=4$ , $x=2,y=4-\dfrac{1}{2}=\dfrac{7}{2}$ .
Using $(1)$ in $(2)$ we get,
$y=x^2-\dfrac{1}{x}$ .
$\Rightarrow \dfrac{dy}{dx}=2x+\dfrac{1}{x^2}$
$\Rightarrow \dfrac{dy}{dx}\Big|_{x=2}=\dfrac{17}{4}$ .
Now, equation of tangent at $\left(2,\dfrac{7}{2}\right)$ is
$y-\dfrac{7}{2}=\dfrac{17}{4}(x-2)$
or, $17x-4y=20$ .

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Question 6
1 / -0

The chord of the curve $y = x^{2} + 2ax + b$ , joining the points where $x = \alpha$ and $x = \beta$ , is parallel to the tangent to the curve at abscissa $x =$

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

B
$\dfrac {2a + b}{3}$

C
$\dfrac {2\alpha + \beta}{3}$

D
$\dfrac {\alpha + \beta}{2}$

Solution

Given curve is $y=x^{2}+2ax+b$
Differentiate above equation w.r.t. $x$ we get
$\dfrac{dy}{dx}=2x+2a$ is the equation of tangent to the curve
But tangent to the curve is parallel to the chord of the curve which joins the points $x=\alpha$ and $x=\beta$
$\therefore$ Tangent to this curve is $=(\alpha+\beta)+2a$
$\therefore 2x + 2a = (\beta + \alpha) + 2a$
$\implies x = \dfrac {\alpha + \beta}{2}$

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Question 7
1 / -0

Consider the curve $y = e^{2x}$ .What is the slope of the tangent to the curve at (0, 1) ?

A
0

B
1

C
2

D
4

Solution

We know that slope of tangent to a curve $y=f(x)$ at a point $(x_1,y_1)$ equals $f^{'}(x_1)$
$\Rightarrow$ Slope of tangent to the curve $y=e^{2x}$ equals $\cfrac{d(e^{2x})}{dx} =2e^{2x}$
$\Rightarrow$ Slope of tangent at $(0,1)=f^{'}(0)=2$

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Question 8
1 / -0

If normal is drawn to ${ y }^{ 2 }=12x$ making an angle ${45}^{o}$ with the axis then the foot of the normal is

A
$\left( 3,8 \right)$

B
$\left( 3,-6 \right)$

C
$\left( 12,-12 \right)$

D
$\left( 8,-8 \right)$

Solution

Given that,
${y^2} = 12x - - \left( 1 \right)$
Diff. w.r.t. to $x$
$2y\dfrac{{dy}}{{dx}} = 12$
$y\dfrac{{dy}}{{dx}} = 6 - - - - \left( 2 \right)$
Slope tangent
$\dfrac{{dy}}{{dx}} = \tan 45$
$m = 1$
Slope at Normal from $(2)$
$\Rightarrow y\left( { - 1} \right) = 6$
$\Rightarrow y =- 6$
$\therefore x = 3$
Point $=(3,-6)$
Option $(A)$ is correct answer.

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Question 9
1 / -0

The equation of the normal to the curve $y=-\sqrt { x } +2$ at the point of its intersection with the bisector of the first quadrant is

A
$4x-y+16=0$

B
$4x-y=16$

C
$2x-y-1=0$

D
$2x-y+1=0$

Solution

Given
$y=-\sqrt { x } +2....(i)$
And bisector of first quadrant is
$y=x$ ..... $(ii)$
On solving Eqs(i) and (ii) we get
$x=1,4$
$\therefore$ From Eq (i)
Points are $(1,1)$ and $(4,0)$
But $(4,0)$ not satisfy Eq (ii)
$\therefore$ Point $(1,1)$ is only point of intersection of curve (i) and line (ii)
Now, slope of tangent of curve (i) is
$\cfrac { dy }{ dx } =\cfrac { -1 }{ 2\sqrt { x } }$
$\therefore$ Slope of tangent at point $(1,1)$ is
${ \left| \cfrac { dy }{ dx } \right| }_{ (1,1) }=\cfrac { -1 }{ 2 }$
and $\cfrac { -1 }{ { \left| \cfrac { dy }{ dx } \right| }_{ (1,1) } } =\cfrac { -1 }{ \cfrac { -1 }{ 2 } } =2\quad$
$\therefore$ Equation of the normal to the curve at point $(1,1)$ will be
$y-1=2(x-1)\Rightarrow 2x-y-1=0$

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Question 10
1 / -0

Equation of normal drawn to the graph of the function defined as $f\left( x \right) =\dfrac { \sin { { x }^{ 2 } } }{ x } ,x\neq 0$ and $f\left( 0 \right) =0$ at the origin is

A
$x+y=0$

B
$x-y=0$

C
$y=0$

D
$x=0$

Solution

Differentiating $f(x)$ w.r.t $x$ , we get
$f'(x)=\dfrac{x\cos(x^2).2x -\sin x^2.1}{x^2}$
$\Rightarrow f'(x)= \dfrac{2x^2\cos x^2- \sin x^2}{x^2}$

Find the value of $f'(x)$ at $x=0$ , we get
$f'(x)$ is in $\dfrac{0}{0}$ indeterminate form

Apply L'Hospital rule
$\lim_{x \to0}f'(x)= \dfrac{2x\cos x^2-4x^2\sin x^2}{2x}$
$\Rightarrow \lim_{x\to0} f'(x)=\dfrac{2x\times(\cos x^2-2x^2\sin x^2)}{2x}$
$\Rightarrow \lim_{x\to0} \cos x^2- 2x^2\sin x^2$
$\Rightarrow 1 - 0$
$\Rightarrow 1$

The slope of the tangent at $x=0$ on the curve $f(x)=\dfrac{sin x^2}{x}$ is $1$ .

We know that, product of slope of tangent and normal is $-1$ .
Therefore, $m_1\times m_2 =-1$
where, $m_1 =1$
Hence, $m_2 = -1$

Slope of normal $= -1$
$y =mx+c$
Here, $m=-1$

To find c: at $x=0$ we have $y=f(x)=0$
Substituting these values in above equation of a line, we get
$0 = -1 \times 0 +c$
$\Rightarrow c=0$

Therefore, equation of normal is $y=-x$ or $y+x=0$

Hence, correct answer id option (A)

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Tangents and its Equations Test 5

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Question 1

x+2y=3ax+2y=3ax+2y=3a

3x−4y+a=03x-4y+a=03x−4y+a=0

4x+3y=7a4x+3y=7a4x+3y=7a

4x−3y=04x-3y=04x−3y=0

Solution

Question 2

(a,a)(a, a)(a,a)

(0,a)(0, a)(0,a)

(0,0)(0, 0)(0,0)

(a,0)(a, 0)(a,0)

Solution

Question 3

333

13\dfrac{1}{3}31​

−3-3−3

−13-\dfrac{1}{3}−31​

Solution

Question 4

14\dfrac{1}{4}41​

13\dfrac{1}{3}31​

12\dfrac{1}{2}21​

1

Solution

Question 5

x−4y=20x-4y=20x−4y=20

17x−4y=2017x-4y=2017x−4y=20

17x+4y=2017x+4y=2017x+4y=20

4x−17y=204x-17y=204x−17y=20

Solution

Question 6

a+b2\dfrac {a + b}{2}2a+b​

2a+b3\dfrac {2a + b}{3}32a+b​

2α+β3\dfrac {2\alpha + \beta}{3}32α+β​

α+β2\dfrac {\alpha + \beta}{2}2α+β​

Solution

Question 7

0

1

2

4

Solution

Question 8

(3,8)\left( 3,8 \right) (3,8)

(3,−6)\left( 3,-6 \right) (3,−6)

(12,−12)\left( 12,-12 \right) (12,−12)

(8,−8)\left( 8,-8 \right) (8,−8)

Solution

Question 9

4x−y+16=04x-y+16=04x−y+16=0

4x−y=164x-y=164x−y=16

2x−y−1=02x-y-1=02x−y−1=0

2x−y+1=02x-y+1=02x−y+1=0

Solution

Question 10

x+y=0x+y=0x+y=0

x−y=0x-y=0x−y=0

y=0y=0y=0

x=0x=0x=0

Solution

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$x+2y=3a$

$3x-4y+a=0$

$4x+3y=7a$

$4x-3y=0$

$(a, a)$

$(0, a)$

$(0, 0)$

$(a, 0)$

$3$

$\dfrac{1}{3}$

$-3$

$-\dfrac{1}{3}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$x-4y=20$

$17x-4y=20$

$17x+4y=20$

$4x-17y=20$

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

$\dfrac {2a + b}{3}$

$\dfrac {2\alpha + \beta}{3}$

$\dfrac {\alpha + \beta}{2}$

$\left( 3,8 \right)$

$\left( 3,-6 \right)$

$\left( 12,-12 \right)$

$\left( 8,-8 \right)$

$4x-y+16=0$

$4x-y=16$

$2x-y-1=0$

$2x-y+1=0$

$x+y=0$

$x-y=0$

$y=0$

$x=0$