Tangents and its Equations Test 13

Result

Tangents and its Equations Test 13

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The equations of the tangents to the curve $$y = x^4$$ from the point (2, 0) not on the curve, are given by

A
$$y=0$$

B
$$y-1=5(x-1)$$

C
$$\displaystyle y - \frac{{4098}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)$$

D
$$\displaystyle y - \frac{{32}}{{243}} = \frac{{80}}{{81}}\left( {x - \frac{2}{3}} \right)$$

Solution

$$y=4\left ( \frac{8}{3} \right )^3 (x-2)$$

$$y-0 = m(x-2)$$
$$y=m(x-2)$$
$$m(x_1 -2) = x_1^4$$
$$4x_1^4 - 8 x_1^3 = x_1^4$$ ( $$m=4x_1^3$$)
$$3x_1^4 = 8x_1^3$$
$$x_1 =\frac{8}{3}$$
$$\therefore \left ( y -\frac{4098}{81} \right ) = \frac{2048}{27} \left ( x- \frac{8}{3} \right )$$
Question 2
1 / -0

For the curve $$ y=3\sin \theta\cos\theta, x=e^{\theta}\sin \theta, 0\leq \theta\leq\pi$$; the tangent is parallel to $$x$$ -axis when $$\theta$$ is

A
$$0$$

B
$$\displaystyle \frac{\pi}{2}$$

C
$$\displaystyle \frac{\pi}{4}$$

D
$$\displaystyle \frac{\pi}{6}$$

Solution

$$\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{d\theta}=d\dfrac{3}{2}(sin2\theta)$$
$$\dfrac{dy}{d\theta}=d\dfrac{3}{4}(cos2\theta)$$ ---- (1)
$$\dfrac{dy}{d\theta}=e^{\theta}sin\theta+e^{\theta}cos\theta$$ ---- (2)
$$\dfrac{dx}{dy}=\dfrac{3}{4}\dfrac{cos2\theta}{e^{\theta}(sin\ \theta+ cos\ \theta)}=0$$
$$cos2\theta=0$$
$$2\theta={\pi}/{2}$$
$$\theta={\pi}/{4}$$
Question 3
1 / -0

The curve given by the equation $$y-e^{xy}+x=0$$ has a vertical tangent at the point

A
$$(0, 1)$$

B
$$(1, 1)$$

C
$$(- 1, 1)$$

D
$$(1, 0)$$

Solution

$$y-e^{xy}+x=0$$

$$\therefore \dfrac{dy}{dx}-e^{xy}\left ( x\dfrac{dy}{dx}+y \right )+1=0$$

$$\Rightarrow \dfrac{dy}{dx}=\dfrac{-1+ye^{xy}}{1-xe^{xy}}$$

$$\therefore $$ if the tangent is vertical at $$(x,y),xe^{xy}=1$$ 1 which is possible only if $$x=1,y=0.$$
Question 4
1 / -0

The sum of the intercepts made on the axes of coordinates by any tangent to the curve $$\sqrt{x}+\sqrt{y}=2$$ is equal to

A
$$4$$

B
$$2$$

C
$$8$$

D
None of these

Solution

Curve equation $$\sqrt{x}+\sqrt{y}=2 \Rightarrow x > 0, y>0$$
$$\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}=-\sqrt{\dfrac {y}{x}}$$
and since in question its asking in general not for a particular point on curve. Hence it we find it for any point on curve then it will be true for any point.
So as use can see (1, 1) lies on curve
So, $$\dfrac{dy}{dx} \int_{1, 1}=-1$$
So tangent equation at (1, 1) is $$y= (-1)x+c$$
and it will pass through (1, 1)
So, $$c=2$$
tangent equation : $$y= -x +2$$
$$x+y=2$$
$$\dfrac{x}{2}+\dfrac{y}{2} =1$$
So x - intercept $$=2$$
y- intercept $$=2$$
So sum $$=4$$
Question 5
1 / -0

If the circle $$x^{2}+y^{2}+2gx+2fy+c=0$$ is touched by $$y=x$$ at $$P$$ such that
$$OP=6\sqrt{2},$$ then the value of $$c$$ is

A
36

B
144

C
72

D
None of these

Solution

Let the point of contact be $$x_{1},y_{1}$$
However it lies on the line $$y=x$$
Hence
$$x_{1}=y_{1}$$
Applying distance formula, we get
$$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$$
$$2x_{1}^2=72$$
$$x_{1}=6$$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.
$$2x+2yy'+2g+2fy'=0$$
Now $$y'=1$$ since $$y'$$ is the slope of the line $$y=x$$.
By substituting, we get
$$x+y=-(g+f)$$
Now $$x_{1}=y_{1}=6$$
Hence
$$12=-(g+f)$$ ...(i)
Substituting $$x=y=6$$ in the equation of the circle, we get
$$36+36+2(6)(g+f)+c=0$$
$$72+12(-12)+c=0$$
$$c=144-72$$
$$c=72$$
Question 6
1 / -0

The point on the curve $$3y=6x-5x^3 $$, the normal at which passes through the origin is

A
$$\left( 1,\displaystyle \frac { 1 }{ 3 } \right) $$

B
$$\left( \displaystyle \frac { 1 }{ 3 } ,1 \right) $$

C
$$\left( 2,\displaystyle \frac { -28 }{ 3 } \right) $$

D
none of these

Solution

Let $$P(x_1,y_1)$$ be the required point .
Given equation of curve is
$$3y=6x-5x^{3}$$
$$\dfrac{dy}{dx}=2-5x^{2}$$
Slope of tangent at P is $$2-5{x_{1}}^{2}$$
Slope of normal at P is $$\displaystyle\dfrac{1}{5{x_{1}}^{2}-2}$$
Equation of normal at P is
$$y-y_{1}=\dfrac{1}{5{x_{1}}^{2}-2}(x-x_1)$$
Since it passes through origin,
$$\displaystyle\Rightarrow y_{1}=\dfrac{x_1}{5{x_{1}}^{2}-2}$$
Here , we can solve by check options.
So ,option A satisfies above equation.
Hence $$P(1,\dfrac{1}{3})$$ is the required point.
Question 7
1 / -0

The number of tangents to the curve $$x^{3/2} + y^{3/2}= 2a^{3/2}$$, $$a>0$$, which are equally inclined to the axes, is

A
$$2$$

B
$$1$$

C
$$0$$

D
$$4$$

Solution

For a tangent to be equally inclined to the axes, the slope of the tangent should be 1.
Thus
$$y'=1$$.
Now
$$x^{\frac{3}{2}}+y^{\frac{3}{2}}=2a^{\frac{3}{2}}$$.

$$\dfrac{3}{2}(\sqrt{x})+\dfrac{3}{2}(\sqrt{y})=0$$

$$\sqrt{x}=-\sqrt{y}$$

$$x=y$$
$$x_{1}=y_{1}$$

Substituting int he above equation, we get
$$2x^{\frac{3}{2}}=2a^{\frac{3}{2}}$$

$$x=y=a$$.
Hence the tangent touches the curve at $$(a,a)$$.
Thus we get only one tangent, satisfying that criteria.
Question 8
1 / -0

If $$m$$ is the slope of a tangent to the curve $$e^y= 1+x^2$$, then

A
$$\left | m \right | > 1$$

B
$$m>1$$

C
$$m>-1$$

D
$$\left | m \right | \leq 1$$

Solution

$$e^{y}=1+x^{2}$$
$$\dfrac{dy}{dx}=\dfrac{2x}{1+x^{2}}$$
So $$|m|=|\dfrac{2x}{x^{2}+1}| $$
Now, since $$x^{2}\ge 0$$
$$\Rightarrow x^{2}+1 \ge 1$$
$$\Rightarrow \dfrac{1}{ x^{2}+1} \le 1$$
$$\Rightarrow \dfrac{2x}{ x^{2}+1} \le 1$$
$$ |m| \le 1$$
Question 9
1 / -0

If at each point of the curve $$y=x^3-ax^2 +x+1$$, the tangent is inclined at an acute angle with the positive direction of the $$x$$-axis, then

A
$$a>0$$

B
$$a\leq \sqrt 3$$

C
$$-\sqrt 3 \leq a \leq \sqrt 3 $$

D
none of these

Solution

$$y=x^3-ax^2 +x+1$$
$$\dfrac{dy}{dx}=3x^{2}-2ax+1$$
Since, the tangent is inclined at an acute angle with the positive direction of x-axis.
$$\dfrac{dy}{dx}>0$$
$$\Rightarrow 3x^{2}-2ax+1>0$$
$$\Rightarrow D<0$$
$$\Rightarrow 4(a^2-3)<0$$
$$\Rightarrow (a-\sqrt{3})(a+\sqrt{3})<0$$
$$\Rightarrow -\sqrt{3}<a<\sqrt{3}$$
Question 10
1 / -0

The distance between the origin and the tangent to the curve $$y = e^{2x} + x^{2}$$ drawn at the point $$x = 0$$ is

A
$$\dfrac {1}{\sqrt 5}$$

B
$$\dfrac {2}{\sqrt 5}$$

C
$$\dfrac {-1}{\sqrt 5}$$

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

Solution

$$y=e^{2x} + x^2$$
At $$x=0 \Rightarrow y=1$$
$$\displaystyle \frac{dy}{dx}=2e^{2x}+2x$$
Slope of tangent at (0,1) $$=2$$
Equation of tangent through (0,1) is
$$y-1=2(x-0)$$
$$\Rightarrow y-2x-1=0$$
Distance of tangent from origin $$\displaystyle=|\frac{-1}{\sqrt{1+4}}|=\frac{1}{\sqrt{5}} $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 13

Result

Tangents and its Equations Test 13

Score

-

Rank

-

SHARING IS CARING

Question 1

$$y=0$$

$$y-1=5(x-1)$$

$$\displaystyle y - \frac{{4098}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)$$

$$\displaystyle y - \frac{{32}}{{243}} = \frac{{80}}{{81}}\left( {x - \frac{2}{3}} \right)$$

Solution

Question 2

$$0$$

$$\displaystyle \frac{\pi}{2}$$

$$\displaystyle \frac{\pi}{4}$$

$$\displaystyle \frac{\pi}{6}$$

Solution

Question 3

$$(0, 1)$$

$$(1, 1)$$

$$(- 1, 1)$$

$$(1, 0)$$

Solution

Question 4

$$4$$

$$2$$

$$8$$

None of these

Solution

Question 5

36

144

72

None of these

Solution

Question 6

$$\left( 1,\displaystyle \frac { 1 }{ 3 } \right) $$

$$\left( \displaystyle \frac { 1 }{ 3 } ,1 \right) $$

$$\left( 2,\displaystyle \frac { -28 }{ 3 } \right) $$

none of these

Solution

Question 7

$$2$$

$$1$$

$$0$$

$$4$$

Solution

Question 8

$$\left | m \right | > 1$$

$$m>1$$

$$m>-1$$

$$\left | m \right | \leq 1$$

Solution

Question 9

$$a>0$$

$$a\leq \sqrt 3$$

$$-\sqrt 3 \leq a \leq \sqrt 3 $$

none of these

Solution

Question 10

$$\dfrac {1}{\sqrt 5}$$

$$\dfrac {2}{\sqrt 5}$$

$$\dfrac {-1}{\sqrt 5}$$

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

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.