Tangents and its Equations Test 16

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Tangents and its Equations Test 16

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Weekly Quiz Competition

Question 1
1 / -0

The slope of the tangent to the locus $$y=\cos^{-1}\left ( \cos x \right )$$ at $$x=\displaystyle \frac{\pi }{4}$$ is

A
$$1$$

B
$$0$$

C
$$2$$

D
$$-1$$

Solution

For $$x=\cfrac { \pi }{ 4 } \quad y=\cos^{ -1 }\cos\left( \cfrac { \pi }{ 4 } \right) =\cfrac { \pi }{ 4 } $$
Therefore point is $$\left( \cfrac { \pi }{ 4 } ,\cfrac { \pi }{ 4 } \right) $$
Now $$y=\cos^{ -1 }\cos\left( x \right) \Rightarrow \cos\left( y \right) =\cos\left( x \right) $$
Slope is $$\cfrac { dy }{ dx } =\cfrac { -\sin\left( x \right) }{ -\sin\left( y \right) } =1$$
Question 2
1 / -0

$$P(2, 2)$$ and $$Q\left ( \displaystyle \frac{1}{2}, -1 \right )$$ are two points on the parabola $$y^{2}=2x$$. The coordinates of the point $$R$$ on the parabola, where tangent to the curve is parallel to the chord $$PQ$$ is

A
$$\left ( \displaystyle \frac{5}{4}, \sqrt{\frac{5}{2}} \right )$$

B
$$(2, -1)$$

C
$$\left ( \displaystyle \frac{1}{8}, \frac{1}{2} \right )$$

D
none of these

Solution

Let point $$\left( 2{ t }^{ 2 },2t \right) $$ lies on curve $${ y }^{ 2 }=2x$$
Then slope is $$\displaystyle\dfrac { dy }{ dx } =\dfrac { 2 }{ 2y } =\dfrac { 1 }{ 2t } $$
And this slope is equal to the the slope of line $$PQ =\displaystyle\dfrac { -1-2 }{ \dfrac { 1 }{ 2 } -2 } =2$$
Equating slope we get $$\displaystyle t=\dfrac { 1 }{ 4 } $$
Hence point is $$\displaystyle\left( \dfrac { 1 }{ 8 } ,\dfrac { 1 }{ 2 } \right) $$
Question 3
1 / -0

The number of tangents to the curve $$y^{2}-2x^{3}-4y+8=0$$ that pass through $$(1, 0)$$ is

A
$$3$$

B
$$1$$

C
$$2$$

D
$$6$$

Solution

Point $$(0,1)$$ doesn't lie on curve, so let $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ is point of contact
Therefore slope of $${ y }^{ 2 }-2{ x }^{ 3 }-4y+8=0$$ is
$$\cfrac { dy }{ dx } =\cfrac { 6{ x }^{ 2 } }{ 2y-4 } =\cfrac { 6{ x }_{ 1 }^{ 2 } }{ 2{ y }_{ 1 }-2 } $$
And slope of line trough $$\left( 1,0 \right) $$ and $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ is $$\\ \cfrac { 0-{ y }_{ 1 } }{ 1-{ x }_{ 1 } } $$
Equating slopes we get $$6{ x }_{ 1 }^{ 2 }-6{ x }_{ 1 }^{ 3 }=-2{ y }_{ 1 }^{ 2 }-4{ y }_{ 1 }$$
And on solving it with equation of curve we get a quadratic equation. That give two values of slope of tangent.
Hence 2 tangents is possible.
Question 4
1 / -0

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

A
$$(0, 1)$$

B
$$(1, 0)$$

C
$$(1, 1)$$

D
none of these

Solution

We have, $$x+y=e^{xy}\Rightarrow (1)$$
Differentiate both sides w.r.t. $$x$$ we get,
$$1+\dfrac{dy}{dx}=e^{xy}(y+x\dfrac{dy}{dx})$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$$
Now for tangent to be y-axis, $$\left|\dfrac{dy}{dx}\right|\to \infty$$
$$\Rightarrow 1-xe^{xy}=0\Rightarrow (2)$$
Solving (1) and (2) we get, $$(x,y)=(1,0)$$, which is the required point
Question 5
1 / -0

Let $$\displaystyle f(x)=e\:^{x}\sin x$$ be the equation of a curve. If at $$\displaystyle x=a,0\leq a\leq 2\pi$$, the slope of the tangent is the maximum then the value of $$a$$ is

A
$$\displaystyle \pi /2$$

B
$$\displaystyle 3\pi /2$$

C
$$\displaystyle \pi $$

D
$$\displaystyle \pi /4$$

Solution

Let $$y={ e }^{ x }sin\left( x \right) $$
Slope is $$\cfrac { dy }{ dx } ={ e }^{ x }\left( sin\left( x \right) +cos\left( x \right) \right) $$
To maximize slope
$$\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ 2e }^{ x }cos\left( x \right) =0$$
Critical point is $$x=\cfrac { \pi }{ 2 } $$
And at $$x=\cfrac { \pi }{ 2 } $$ is the point of maxima
Question 6
1 / -0

The equation of tangent to the curve $$y=e^{-\left | x \right |}$$ at the point where the curve cuts the line $$x=1$$ is

A
$$x+y=e$$

B
$$e(x+y)=1$$

C
$$y+ex=1$$

D
none of these

Solution

Substituting $$x=1$$ in $$y={ e }^{ -\left| x \right| }$$, we get $$y=\cfrac { 1 }{ e } $$
Slope at point $$\left( 1,\cfrac { 1 }{ e } \right) $$ is
$$\cfrac { dy }{ dx } =-\cfrac { 1 }{ e } $$
Hence equation of tangent is
$$y-\cfrac { 1 }{ e } =-\cfrac { 1 }{ e } (x-1)\\ ey+x=2$$
Question 7
1 / -0

The equation of the curve is given by $$x=e^{t}\sin t$$, $$y=e^{t}\cos t$$. The inclination of the tangent to the curve at the point $$t=\displaystyle \frac{\pi }{4}$$ is

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

B
$$\displaystyle \frac{\pi }{3}$$

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

D
$$0$$

Solution

$$dx=e^{t}(\cos t+\sin t).dt$$
$$dy=e^{t}[\cos t-\sin t).dt$$
Hence
$$\dfrac{dy}{dx}_{t=\tfrac{\pi}{4}}$$

$$=\dfrac{\cos t-\sin t}{\cos t+\sin t}_{t=\tfrac{\pi}{4}}$$

$$=0$$
Question 8
1 / -0

If the curve $$y=ax^3+bx^2+cx$$ is inclined at $$45^0$$ to the x-axis at $$(0,0)$$ but it touches the x-axis at $$(1,0)$$ then the value of $$a,b$$ and $$c$$ are given by

A
$$a=-1,b=2,c=1$$

B
$$a=1,b=-2,c=1$$

C
$$a=1,b=1,c=-2$$

D
$$a=-2,b=1,c=1$$

Solution

$$\displaystyle \frac { dy }{ dx } =3a{ x }^{ 2 }+2bx+c$$ (given)
At $$\displaystyle \left( 0,0 \right) ,\frac { dy }{ dx } =c=\tan { { 45 }^{ 0 } } =1\Rightarrow c=1$$
At $$\displaystyle \left( 1,0 \right) ,\frac { dy }{ dx } =3a+2b+c$$
Tangent at $$(1,0)$$ is $$y-0=(3a+2b+c).(x-1)$$
This is x-axis if $$3a+2b+c=0$$
which is true if $$a=1,b=-2$$
Hence $$a=1,b=-2,c=1$$
Question 9
1 / -0

The point on the curve $$\sqrt{x}+\sqrt{y}=2a^{2}$$, where the tangent is equally inclined to the axes, is

A
$$\left ( a^{4}, a^{4} \right )$$

B
$$\left ( 0, 4a^{4} \right )$$

C
$$\left ( 4a^{4}, 0 \right )$$

D
none of these

Solution

$$y=x$$ is the equation of line as it is equally inclined to axes
So let point $$(k,k)$$ is the point of tangency
Therefore, $$\sqrt { k } +\sqrt { k } =2{ a }^{ 2 }\\ k={ a }^{ 4 }$$
Hence point is $$\left( { a }^{ 4 },{ a }^{ 4 } \right) $$
Question 10
1 / -0

The angle between two tangents to the ellipse $$\displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{9}=1$$ at the points where the line $$y=1$$ cuts the curve is

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

B
$$\tan^{-1}\displaystyle \frac{6\sqrt{2}}{7}$$

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

D
none of these

Solution

Substituting $$y=1$$ in $$\cfrac { { x }^{ 2 } }{ 16 } +\cfrac { { y }^{ 2 } }{ 9 } =1$$
We get $$x=\pm \cfrac { 8\sqrt { 2 } }{ 3 } $$
And slope of tangents $$\cfrac { dy }{ dx } =\cfrac { 9x }{ 16y } =m_{1}, m_{2} =\pm \cfrac { 3\sqrt { 2 } }{ 2 } $$
Therefore $$tan\left( \theta \right) =\dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\left| -\cfrac { 6\sqrt { 2 } }{ 7 } \right| \Rightarrow \theta ={ tan }^{ -1 }\left( \cfrac { 6\sqrt { 2 } }{ 7 } \right) $$

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 16

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Tangents and its Equations Test 16

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SHARING IS CARING

Question 1

$$1$$

$$0$$

$$2$$

$$-1$$

Solution

Question 2

$$\left ( \displaystyle \frac{5}{4}, \sqrt{\frac{5}{2}} \right )$$

$$(2, -1)$$

$$\left ( \displaystyle \frac{1}{8}, \frac{1}{2} \right )$$

none of these

Solution

Question 3

$$3$$

$$1$$

$$2$$

$$6$$

Solution

Question 4

$$(0, 1)$$

$$(1, 0)$$

$$(1, 1)$$

none of these

Solution

Question 5

$$\displaystyle \pi /2$$

$$\displaystyle 3\pi /2$$

$$\displaystyle \pi $$

$$\displaystyle \pi /4$$

Solution

Question 6

$$x+y=e$$

$$e(x+y)=1$$

$$y+ex=1$$

none of these

Solution

Question 7

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

$$\displaystyle \frac{\pi }{3}$$

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

$$0$$

Solution

Question 8

$$a=-1,b=2,c=1$$

$$a=1,b=-2,c=1$$

$$a=1,b=1,c=-2$$

$$a=-2,b=1,c=1$$

Solution

Question 9

$$\left ( a^{4}, a^{4} \right )$$

$$\left ( 0, 4a^{4} \right )$$

$$\left ( 4a^{4}, 0 \right )$$

none of these

Solution

Question 10

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

$$\tan^{-1}\displaystyle \frac{6\sqrt{2}}{7}$$

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

none of these

Solution

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