Application of Derivatives Test

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Application of Derivatives Test - 26

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Weekly Quiz Competition

Question 1
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 2
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 3
1 / -0

The curve $$\displaystyle \frac{x^{n}}{a^{n}}+\frac{y^{n}}{b^{n}}=2$$ touches the line $$\displaystyle \frac{x}{a}+\frac{y}{b}=2$$ at the point

A
$$(b, a)$$

B
$$(a, b)$$

C
$$(1, 1)$$

D
$$\displaystyle \left ( \frac{1}{a}, \frac{1}{b} \right )$$

Solution

Slope of curve$$\cfrac { { x }^{ n } }{ { a }^{ n } } +\cfrac { { y }^{ n } }{ { b }^{ n } } =2$$ is
$$\cfrac { dy }{ dx } =-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } } $$
Slope of tangent $$\cfrac { { x } }{ { a } } +\cfrac { { y } }{ { b } } =2$$ is
$$\cfrac { dy }{ dx } =-\cfrac { { b } }{ { a } } $$
Equating both slopes, we get
$$-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } } =-\cfrac { { b } }{ { a } } $$
Only $$x=a\quad y=b$$ satisfy the above equation.
Hence, option 'B' is correct.
Question 4
1 / -0

If the line joining the points $$(0, 3)$$ and $$(5, -2)$$ is the tangent to the curve $$\displaystyle y=\frac{c}{x+1}$$ then the value of $$c$$ is

A
$$1$$

B
$$-2$$

C
$$4$$

D
none of these

Solution

Slop of line joining the points $$(0,3)\quad and\quad (5,2)$$ is given by
$$\dfrac { 3+2 }{ 0-5 } =-1$$

This is equal to the slop of tangent on the curve and that is given by

$$\dfrac { dy }{ dx } =\dfrac { -c }{ { \left( x+1 \right) }^{ 2 } } \\ \dfrac { dy }{ dx } =-1\\ \Rightarrow c={ \left( x+1 \right) }^{ 2 }...................(1)$$\

Equation of line joining the points $$(0,3)\quad and\quad (5,2)$$ is given by

$$(y-3)=-1(x-0)$$

Solving equation of tangent and the curve for point of intersection

$$\dfrac { c }{ x+1 } +x=3$$.............(2)

Solving (1) and (2)

$$x=1$$

Putting this in (2), we get c=4
Question 5
1 / -0

If error in measuring the edge of a cube is $$k$$% then the percentage error in estimating its volume is

A
$$k$$

B
$$3k$$

C
$$\displaystyle \frac{k}{3}$$

D
none of these

Solution

Let the actual length of the cube be a.
Therefore the measured length of the cube will be
$$=a(1\pm0.0k)$$
$$=a(1\pm\dfrac{k}{100})$$
Considering positive error,
$$a'=a(1+\dfrac{k}{100})$$
$$V'=a^{3}(1+\dfrac{k}{100})^{3}$$

$$=a^{3}(1+3(\dfrac{k}{100})+3(\dfrac{k}{100})^{2}+(\dfrac{k}{100})^{3})$$

Since $$\dfrac{k}{100}<<1$$, hence we neglect the higher order terms.
Thus
$$V'=a^{3}(1+3(\dfrac{k}{100}))$$

Actual volume V
$$V=a^{3}$$
Therefore
$$V'-V=a^{3}(1+\dfrac{3k}{100})-a^{3}$$

$$=a^{3}(\dfrac{3k}{100})$$

$$\dfrac{V'-V}{V}=\dfrac{a^{3}\dfrac{3k}{100}}{a^{3}}$$

$$=\dfrac{3k}{100}$$

$$=\dfrac{3k}{100}$$

$$\dfrac{V'-V}{V}\times 100=3k$$
Therefore percentage error in volume is $$3k$$.
Question 6
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 7
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 $$tan\left( \theta \right) =\dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\left| -\cfrac { 6\sqrt { 2 } }{ 7 } \right| \Rightarrow \theta ={ tan }^{ -1 }\left( \cfrac { 6\sqrt { 2 } }{ 7 } \right) $$
Question 8
1 / -0

Angle between the tangents to the curve $$y= x^{2}-5x+6$$ at the points $$(2,0)$$ and $$\left ( 3,0 \right )$$ is

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

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

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

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

Solution

Given equation of curve $$\displaystyle y=x^{2}-5x+6$$

$$\displaystyle \Rightarrow \frac{dy}{dx}=2x-5$$

Slope of tangent to the curve at $$(2,0)$$ is

$$\displaystyle \left(\frac{dy}{dx}\right)_{(2,0)}=2(2)-5=-1=m_{1}$$

Slope of tangent to the curve at $$(3,0)$$ is

$$\displaystyle \left(\frac{dy}{dx}\right)_{(3,0)}=2(3)-5=1=m_{2}$$

Since $$\displaystyle m_{1}m_{2}=-1$$

$$\therefore $$ Angle between the tangents to the curve at $$(2,0)$$ and $$(3, 0)$$ is $$\displaystyle \dfrac{\pi}{2}$$
Question 9
1 / -0

The number of tangents to the curve $$\displaystyle y= e^{\left | x \right |}$$ at the point $$(0,1)$$ is

A
2

B
1

C
4

D
0

Solution

For $$x>0$$,
$$y=e^{x}$$.
$$\dfrac{dy}{dx}=e^{x}$$.
Now
$$\dfrac{dy}{dx}_{x=0}=1$$.
Hence
$$y-1=1(x-0)$$
Or
$$x-y+1=0$$ is the required equation of tangent.
Similarly for $$x<0$$
$$y=e^{-x}$$.
$$\dfrac{dy}{dx}=-e^{-x}$$.
Now
$$\dfrac{dy}{dx}_{x=0}=-1$$.
Hence
$$y-1=-1(x-0)$$
Or
$$x+y-1=0$$ is the required equation of tangent.
Hence at $$x=0$$ we will have 2 tangents to the curve $$y=e^{|x|}$$ which are mutually perpendicular.
Question 10
1 / -0

The curve $$y+e^{xy}+x= 0$$ has a tangent parellel to y-axis at a point

A
$$\left ( -1,\:0 \right )$$

B
$$\left ( 1,\:0 \right )$$

C
$$\left ( 1,\:1 \right )$$

D
$$\left ( 0,\:0 \right )$$

Solution

$$\dfrac{dy}{dx}+e^{xy}[x\dfrac{dy}{dx}+y]+1=0$$
Or
$$\dfrac{dy}{dx}[1+x.e^{xy}]+1+y.e^{xy}=0$$
Or
$$\dfrac{dy}{dx}=\dfrac{-(1+y.e^{xy})}{1+x.e^{xy}}$$
Now
$$1+xe^{xy}=0$$ since the slope of the tangent is $$90^{0}$$.
Hence
$$xe^{xy}=-1$$
Or
$$x(-x-y)=-1$$
Or
$$x^{2}+xy=1$$
Or
$$y=\dfrac{1-x^{2}}{x}$$
Or
$$y=\dfrac{1}{x}-x$$ ...(i)
Considering $$y=0$$,
$$x^{2}=1$$
$$x=\pm1$$.
Hence we get 2 points $$(1,0)$$ and $$(-1,0)$$.
Out of these 2, only $$(-1,0)$$ lies on the curve.
Hence the required point is $$(-1,0)$$.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 26

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Application of Derivatives Test - 26

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SHARING IS CARING

Question 1

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

$$(2, -1)$$

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

none of these

Solution

Question 2

$$3$$

$$1$$

$$2$$

$$6$$

Solution

Question 3

$$(b, a)$$

$$(a, b)$$

$$(1, 1)$$

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

Solution

Question 4

$$1$$

$$-2$$

$$4$$

none of these

Solution

Question 5

$$k$$

$$3k$$

$$\displaystyle \frac{k}{3}$$

none of these

Solution

Question 6

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

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

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

none of these

Solution

Question 7

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

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

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

none of these

Solution

Question 8

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

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

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

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

Solution

Question 9

2

1

4

0

Solution

Question 10

$$\left ( -1,\:0 \right )$$

$$\left ( 1,\:0 \right )$$

$$\left ( 1,\:1 \right )$$

$$\left ( 0,\:0 \right )$$

Solution

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