Tangents and its Equations Test 37

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Tangents and its Equations Test 37

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Weekly Quiz Competition

Question 1
1 / -0

Tangents are drawn from origin to the curve $$y=\sin{x}+\cos{x}$$. Then their points of contact lie on the curve

A
$$\dfrac{1}{x^{2}}+\dfrac{2}{y^{2}}=1$$

B
$$\dfrac{1}{x^{2}}-\dfrac{2}{y^{2}}=-1$$

C
$$\dfrac{2}{x^{2}}+\dfrac{2}{y^{2}}=1$$

D
$$\dfrac{2}{x^{2}}-\dfrac{2}{y^{2}}=1$$

Solution
Question 2
1 / -0

The slope of normal to the curve y= log (logx) at x = e is

A
e

B
-e

C
$$\frac{1}{e}$$

D
-$$\frac{1}{e}$$

Solution

$$\begin{array}{l} \frac { d }{ { dx } } \log \left( { \log x } \right) \\ Substitute\, u=\log x \\ =\dfrac { d }{ { du } } \log u\frac { d }{ { dx } } \log x \\ =\dfrac { 1 }{ u } \left( { \frac { 1 }{ x } } \right) \\ =\dfrac { 1 }{ { \log x } } \left( { \frac { 1 }{ x } } \right) \\ Substitute\, x=e \\ =\dfrac { 1 }{ { \log e } } \left( { \dfrac { 1 }{ e } } \right) \\ =\dfrac { 1 }{ e } \end{array}$$
Question 3
1 / -0

The inclination of the tangent at $$\theta = \dfrac {\pi}{3}$$ on the curve $$x = a(\theta + \sin \theta), y = a(1 + \cos \theta)$$ is

A
$$\dfrac {\pi}{3}$$

B
$$\dfrac {\pi}{6}$$

C
$$\dfrac {2\pi}{3}$$

D
$$\dfrac {5\pi}{6}$$

Solution

For the curve ,

$$\displaystyle \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{\dfrac{d(a(1+cos\theta))}{d\theta}}{\dfrac{d(a(\theta+sin\theta))}{d\theta}} = \dfrac{-asin\theta}{a+acos\theta} = \dfrac{-sin\theta}{1+cos\theta}$$

So, If the inclination of tangent at $$\theta = \dfrac{\pi}{3}$$ be $$y$$ ,

Then, $$\tan(y) =$$ $$\dfrac{-sin(\dfrac{\pi}{3})}{1+cos\dfrac{\pi}{3}} = \dfrac{-1}{\sqrt{3}}$$

Thus,$$ y = \pi-\dfrac{\pi}{6}= \dfrac{5\pi}{6} $$
Question 4
1 / -0

Let f be a real-valued differentiable function on R (the set of all real numbers) such that $$f(1)=1$$. If the y-intercept of the tangent at any point P(x, y) on the curve $$y=f(x)$$ is equal to the cube of the abscissa of P, then the value of $$f(-3)$$ is equal to?

A
$$-3$$

B
$$3$$

C
$$9$$

D
$$-9$$

Solution

$$P\equiv (x,y)$$
Let slope by $$\cfrac { dy }{ dx } $$
So, equation of tangent will be
$$(X-m)\cfrac { dy }{ dx } =(Y-y)$$
for Y- intercept , $$x=0$$
So, $$Y=y-x\cfrac { dy }{ dx } $$
Now according to question ,
$${ x }^{ 3 }=y-x\cfrac { dy }{ dx } \\ \cfrac { dy }{ dx } -\cfrac { y }{ x } ={ x }^{ 2 }\\ \cfrac { xdy-ydx }{ { x }^{ 2 } } =-xdx\\ d(\cfrac { y }{ x } )=-xdx$$
On integrating , we get
$$\cfrac { y }{ x } =\cfrac { -{ x }^{ 2 } }{ 2 } +C$$
Now, $$y(1)=1$$
So, we get $$\cfrac { y }{ x } =\cfrac { -{ x }^{ 2 } }{ 2 } +\cfrac { 3 }{ 2 } $$
So, $$y(-3)=f(-3)=9$$
Question 5
1 / -0

The greatest slope among the lines represented by the equation $$4x^2 - y^2 + 2y - 1 = 0 $$ is -

A
$$-3$$

B
$$-2$$

C
$$2$$

D
$$3$$

Solution

$$4{x}^{2} - {y}^{2} + 2y - 1 = 0$$
$${y}^{2} - 2y + 1 = 4{x}^{2}$$
$${\left( y - 1 \right)}^{2} = {\left( 2x \right)}^{2}$$
$$y - 1 = \pm 2x$$
$$y = 2x + 1 \text{ or } y = -2x + 1$$
Comparing the above equations with $$y = mx + c$$, we have
Slope $$\left( m \right) = 2, -2$$
Hence the greatest slope will be $$2$$.
Question 6
1 / -0

If the tangent at $$(x_{1}, y_{1})$$ to the curve $$x^{3}+y^{3}=a^{3}$$ meets the curve again at $$(x_{2}, y_{2})$$ then

A
$$\dfrac{x_{2}}{x_{1}}+\dfrac{y_{2}}{y_{1}}=-1$$

B
$$\dfrac{x_{2}}{y_{1}}+\dfrac{x_{1}}{y_{2}}=-1$$

C
$$\dfrac{x_{1}}{x_{2}}+\dfrac{y_{1}}{y_{2}}=-1$$

D
$$\dfrac{x_{2}}{x_{1}}+\dfrac{y_{2}}{y_{1}}=1$$

Solution

Given $$x^3+y^3=a^3$$
The derivative is
$$\dfrac{dy}{dx}=-\dfrac{x^2}{y^2}.....(1)$$
Therefore, slope of tangent at $$(x_1,y_1)$$ is
$$-\dfrac{x_1^2}{y_1^2}.........(2)$$
The tangent passes through $$(x_2,y_2)$$, therefore the slope of tangent is also given by
$$\dfrac{y_2-y_1}{x_2-x_1}.......(3)$$
Comparing the two slope equations we get
$$\dfrac{y_2-y_1}{x_2-x_1}=-\dfrac{x_1^2}{y_1^2}$$

$$\dfrac{y_2^3-y_1^3}{x_2^3-x_1^3}\times\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}=-\dfrac{x_1^2}{y_1^2}$$

$$-\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}=-\dfrac{x_1^2}{y_1^2}$$

$$x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_2^2$$

$$x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1y_2+x_1^2y_2^2$$

$$x_1^2y_2^2-x_2^2y_1^2=x_1x_2y_1^2-x_1^2y_1y_2$$

$$(x_1y_2-x_2y_1)(x_1y_2+x_2y_1)=x_1y_1(x_2y_1-x_1y_2)$$

$$x_1y_2+x_2y_1=-x_1y_1$$

$$\dfrac{x_2}{x_1}+\dfrac{y_2}{y_1}=-1$$
Question 7
1 / -0

The ordinate of all points on the curve $$y=\dfrac{1}{2\sin^{2}x+3\cos^{2}x}$$ where the tangent is horizontal, is

A
Always equal to $$\dfrac{1}{2}$$

B
Always equal to $$\dfrac{1}{3}$$

C
$$\dfrac{1}{2}$$ or $$\dfrac{1}{3}$$ according as $$n$$ is an even or an odd integer.

D
$$\dfrac{1}{2}$$ or $$\dfrac{1}{3}$$ according as $$n$$ is an even integer

Solution

$$y=\dfrac{1}{2sin^{2}x+3cos^{2}x}=\dfrac{1}{2+cos^{2}x}$$
$$tan\theta =\dfrac{dy}{dx}= \dfrac{-2sinx\, cos x}{(2+cos^{2}x)^{2}}=0$$
$$\therefore sin\, x=0$$ or $$cos\, x= 0$$
If $$sin\, x=0, y= \dfrac{1}{3}$$, if $$cos\, x=0, y= \dfrac{1}{2}$$
Question 8
1 / -0

The curve given by $$x + y = {e^{xy}}$$ has an tangents parallel to the y-axis at the point

A
$$(0,1)$$

B
$$(1,0)$$

C
$$(1,1)$$

D
$$(0,0)$$

Solution

$$\dfrac{dy}{dx}$$(differentiate the terms)
$$1+\dfrac{dy}{dx}=e^{xy}\left(x\dfrac{dy}{dx}+y\right)$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$$
Check by option
$$\dfrac{dy}{dx}\left|_{(0, 1)}\right.=\dfrac{1-1}{1-0}=\dfrac{0}{1}=0$$ Not coming infinity
$$\displaystyle\dfrac{dy}{dx}\left|_{(1, 0)}\right.=\dfrac{-1}{0}=\infty$$
This point satisfy
$$\therefore$$ option B$$(1, 0)$$.
Question 9
1 / -0

A curve C has the property that if the tangent drawn at any point 'P' on C meets the coordinate axes at A and B, and P is midpoint of AB. If the curve passes through the point $$(1, 1)$$ then the equation of the curve is?

A
$$xy=2$$

B
$$xy=3$$

C
$$xy=1$$

D
$$xy=4$$

Solution

By option verification passing through $$(1,1)$$
$$xy=1$$
Question 10
1 / -0

Number of possible tangents to the curve $$y = \cos \left( {x + y} \right), - 3\pi \leqslant x \leqslant 3\pi $$, that are parallel to the line $$x + 2y = 0$$, is

A
1

B
2

C
3

D
4

Solution

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Tangents and its Equations Test 37

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Question 1

$$\dfrac{1}{x^{2}}+\dfrac{2}{y^{2}}=1$$

$$\dfrac{1}{x^{2}}-\dfrac{2}{y^{2}}=-1$$

$$\dfrac{2}{x^{2}}+\dfrac{2}{y^{2}}=1$$

$$\dfrac{2}{x^{2}}-\dfrac{2}{y^{2}}=1$$

Solution

Question 2

e

-e

$$\frac{1}{e}$$

-$$\frac{1}{e}$$

Solution

Question 3

$$\dfrac {\pi}{3}$$

$$\dfrac {\pi}{6}$$

$$\dfrac {2\pi}{3}$$

$$\dfrac {5\pi}{6}$$

Solution

Question 4

$$-3$$

$$3$$

$$9$$

$$-9$$

Solution

Question 5

$$-3$$

$$-2$$

$$2$$

$$3$$

Solution

Question 6

$$\dfrac{x_{2}}{x_{1}}+\dfrac{y_{2}}{y_{1}}=-1$$

$$\dfrac{x_{2}}{y_{1}}+\dfrac{x_{1}}{y_{2}}=-1$$

$$\dfrac{x_{1}}{x_{2}}+\dfrac{y_{1}}{y_{2}}=-1$$

$$\dfrac{x_{2}}{x_{1}}+\dfrac{y_{2}}{y_{1}}=1$$

Solution

Question 7

Always equal to $$\dfrac{1}{2}$$

Always equal to $$\dfrac{1}{3}$$

$$\dfrac{1}{2}$$ or $$\dfrac{1}{3}$$ according as $$n$$ is an even or an odd integer.

$$\dfrac{1}{2}$$ or $$\dfrac{1}{3}$$ according as $$n$$ is an even integer

Solution

Question 8

$$(0,1)$$

$$(1,0)$$

$$(1,1)$$

$$(0,0)$$

Solution

Question 9

$$xy=2$$

$$xy=3$$

$$xy=1$$

$$xy=4$$

Solution

Question 10

1

2

3

4

Solution

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