Application of Derivatives Test

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Application of Derivatives Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

If $$x+ 4y=14$$ is a normal to the curve $$y^2=\alpha x ^3-\beta $$ at $$(2,3)$$, then the value of $$\alpha+\beta$$ is

A
$$9$$

B
$$-5$$

C
$$7$$

D
$$-7$$

Solution

Given equation of curve is
$$y^2=\alpha x ^3-\beta $$
Since, it passes through (2,3)
$$4=8\alpha -\beta$$
$$\displaystyle \frac{dy}{dx}=\frac{3\alpha x^2}{2y}$$
Slope of tangent at (2,3) is $$2\alpha$$
Slope of normal at $$\displaystyle (2,3)= -\frac{1}{2\alpha}$$
Given equation of normal is $$x+4y=14$$
Slope of normal is $$-\cfrac{1}{4}$$
$$\Rightarrow \alpha =2$$
$$\Rightarrow \beta=7$$
Hence $$\alpha+\beta =9$$
Question 2
1 / -0

The slope of the tangent to the curve $$y= \sqrt {4-x^2}$$ at the point where the ordinate and the abscissa are equal is

A
-1

B
1

C
0

D
none of these

Solution

Given equation of curve is $$y= \sqrt {4-x^2}$$
$$y^{2}=4-x^{2}$$
$$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{x}{y}$$
Let $$P(x_1,y_1)$$ be the point on the curve such that $$x_1=y_1$$
So, slope of tangent to curve at P is $$-1$$.
Question 3
1 / -0

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

A
(0,1)

B
(1,0)

C
(1,1)

D
none of these

Solution

Given equation of curve is $$x+y=e^{xy}$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{y-1}{1-xe^{xy}}$$
Since, the tangent is parallel to y-axis,
$$\displaystyle\frac{dy}{dx}=\frac{1}{0}$$
$$\Rightarrow xe^{xy}=1$$
Also, $$y=0$$ on the line parallel to y-axis
$$\Rightarrow x=1$$
Hence, the required point is (1,0)
Question 4
1 / -0

The pressure P and volume V of a gas are connected by the relation $$PV^{1/4}=constant$$. The percentage increase in the pressure corresponding to a deminition of $$\dfrac12 \%$$ in the volume is

A
$$\dfrac {1}{2}$$ %

B
$$\dfrac {1}{4}$$ %

C
$$\dfrac {1}{8}$$ %

D
none of these

Solution

$$PV^{1/4}=constant$$
$$\displaystyle \Rightarrow P=\dfrac{k}{V^{{1}/{4}}}$$
$$\displaystyle \Rightarrow \dfrac{dP}{dV}=-\dfrac{k}{4}V^{-{5}/{4}}$$
Percentage error in V $$\displaystyle= -\dfrac{1}{2}\%$$
$$\Rightarrow\displaystyle \dfrac{\Delta V}{V} =-\dfrac{1}{200}$$
$$\Rightarrow\displaystyle {\Delta V}=-\dfrac{V}{200}$$
Approximate change in $$P\displaystyle=dP=(\dfrac{dP}{dV}){\Delta V}$$
$$\displaystyle =\dfrac{1}{800} {kV^{-1/4}}=\dfrac{1}{8}\%$$ of P
Percentage increase in $$V \ \displaystyle =\dfrac{1}{8}\%$$
Question 5
1 / -0

If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is

A
$$1\%$$

B
$$2\%$$

C
$$3\%$$

D
$$4\%$$

Solution

Time period $$\displaystyle T=2\pi \sqrt {\frac{l}{g}}$$
$$\displaystyle \frac { dT }{ dl } =\frac { \pi }{ \sqrt { gl } } $$
Percentage error in measuring l = 2%
$$\displaystyle \Rightarrow \frac { \Delta l }{ l } =\frac { 2 }{ 100 } $$
$$\displaystyle \Rightarrow \Delta l =\frac { 2l }{ 100 } $$
Approximate error in $$T \ \displaystyle= dT= \frac { dT }{ dl } \Delta l $$
$$\displaystyle =\frac{T}{100}=1\% of T$$
Percentage error in $$T \ =1 \%$$
Question 6
1 / -0

If there is an error of $$a \%$$ in measuring the edge of a cube, then the percentage error in its surface area is

A
$$2a$$

B
$$\dfrac {a}{2}$$

C
$$3a$$

D
None of the above

Solution

Surface area of cube $$S=6x^{2}$$
$$\Rightarrow log S = log 6 + 2log x $$
On differentiating we get,
$$ \dfrac{\Delta S}{S} = 2 \dfrac{\Delta x}{x}$$
$$ \dfrac{\Delta S}{S} = 2 a \%$$
Question 7
1 / -0

While measuring the side of an equilateral triangle an error of $$k \%$$ is marked, the percentage error in its area is

A
$$k \%$$

B
$$2k \%$$

C
$$\dfrac {k}{2}\%$$

D
$$3k \%$$

Solution

Area of equilateral triangle $$A=\dfrac { \sqrt { 3 } }{ 4 } { x }^{ 2 }$$
$$\displaystyle \dfrac{dA}{dx}=\dfrac { \sqrt { 3 } }{ 2 }x$$
Given, percentage error in measuring side $$=k\%$$
$$\Rightarrow \displaystyle \dfrac { \Delta x }{ x } =\dfrac { k }{ 100 } $$
$$\Rightarrow \displaystyle { \Delta x }=\dfrac {k x }{ 100 } $$
Now, approximate error in measuring A$$\displaystyle =dA= (\dfrac{dA}{dx}){ \Delta x}$$
$$\displaystyle = \dfrac{k }{100} \dfrac{\sqrt{3}}{2}x^{2} =2k\%$$ of A
Percentage error in measuring $$A =2k\%$$
Question 8
1 / -0

If $$y=x^n$$, then the ratio of relative errors in $$y$$ and $$x$$ is

A
$$1:1$$

B
$$2:1$$

C
$$1:n$$

D
$$n:1$$

Solution

$$y=x^{n}$$
$$\Rightarrow \displaystyle \dfrac{dy}{dx}=nx^{n-1}$$
Approximate error in y is $$\displaystyle dy=\left (\dfrac{dy}{dx}\right) \Delta x$$
$$=nx^{n-1} \Delta x$$
Relative error in y is $$\displaystyle \dfrac{dy}{y}=\dfrac{n}{x}\Delta x$$
Approximate error in x is $$\displaystyle dx=\left (\dfrac{dx}{dy}\right) \Delta y$$
$$\displaystyle=\dfrac{1}{nx^{n-1}} \Delta y$$
Relative error in x is $$\displaystyle \dfrac{dx}{x}=\dfrac{1}{nx^{n}}\Delta y$$
Required ratio $$\displaystyle = \dfrac{\dfrac{n}{x}\Delta x}{\dfrac{1}{nx^{n}}\Delta y}$$
$$\displaystyle =n^{2}x^{n-1} \dfrac{\Delta x}{\Delta y}$$
$$\displaystyle =\dfrac{n}{1}$$
So, the ratio of relative errors in y and x is $$ n:1$$.
Question 9
1 / -0

The circumference of a circle is measured as $$28 cm$$ with an error of $$0.01 cm$$. The percentage error in the area is

A
$$\dfrac {1}{14}$$

B
$$0.01$$

C
$$\dfrac {1}{7}$$

D
none of these

Solution

Circumference $$C=2\pi r$$
$$\Rightarrow\displaystyle r=\frac{14}{\pi}$$
Also, $$\displaystyle \frac{dC}{dr}=2\pi$$
Area of circle $$A=\pi r^{2} $$
$$\Rightarrow\displaystyle A=\frac{{14}^{2}}{\pi}$$
Also, $$\displaystyle \frac{dA}{dr}=2\pi r$$
$$\displaystyle \Rightarrow \frac{dA}{dC}=r=\frac{14}{pi}$$
Approximate error in $$A$$ is $$\displaystyle dA=( \frac{dA}{dC}) \Delta C$$
$$\displaystyle=\frac{14}{\pi}\frac{1}{100}$$
$$\displaystyle=\frac{1}{1400}$$ of A
Percentage error in $$A \ \displaystyle =\frac{1}{14}\%$$
Question 10
1 / -0

The height of a cylinder is equal to the radius. If an error of $$\alpha$$ % is made in the height, then percentage error in its volume is

A
$$\alpha$$ %

B
$$2\alpha$$ %

C
$$3\alpha$$ %

D
none of these

Solution

Volume of cylinder $$V= \pi { r }^{ 2 }h$$
Since, $$h=r$$
$$V=\pi {h}^{3}$$
$$\displaystyle \dfrac{dV}{dh}=3\pi h^{2}$$
Given, percentage error in measuring height $$=\alpha$$%
$$\Rightarrow \displaystyle \dfrac { \Delta h }{ h } =\dfrac { \alpha }{ 100 } $$
$$\Rightarrow \displaystyle { \Delta h }=\dfrac {\alpha h }{ 100 } $$
Now, approximate error in measuring V$$\displaystyle =dV= (\dfrac{dV}{dh}){ \Delta h}$$
$$\displaystyle = \dfrac{3\alpha }{100} {\pi h^{3}} =3\alpha$$% of V
Percentage error in measuring $$V =3\alpha$$%

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 22

Result

Application of Derivatives Test - 22

Score

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Rank

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SHARING IS CARING

Question 1

$$9$$

$$-5$$

$$7$$

$$-7$$

Solution

Question 2

-1

1

0

none of these

Solution

Question 3

(0,1)

(1,0)

(1,1)

none of these

Solution

Question 4

$$\dfrac {1}{2}$$ %

$$\dfrac {1}{4}$$ %

$$\dfrac {1}{8}$$ %

none of these

Solution

Question 5

$$1\%$$

$$2\%$$

$$3\%$$

$$4\%$$

Solution

Question 6

$$2a$$

$$\dfrac {a}{2}$$

$$3a$$

None of the above

Solution

Question 7

$$k \%$$

$$2k \%$$

$$\dfrac {k}{2}\%$$

$$3k \%$$

Solution

Question 8

$$1:1$$

$$2:1$$

$$1:n$$

$$n:1$$

Solution

Question 9

$$\dfrac {1}{14}$$

$$0.01$$

$$\dfrac {1}{7}$$

none of these

Solution

Question 10

$$\alpha$$ %

$$2\alpha$$ %

$$3\alpha$$ %

none of these

Solution

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