Application of Derivatives Test

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Application of Derivatives Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

The graph of the function $$f(x) = 2x^3 - 7$$ goes :

A
up to the right and down to the left

B
down to the right and up to the left

C
up to the right and up to the left

D
down to the right and down to the left

E
none of these ways.

Solution

Given $$f(x) =2x^{3}-7$$

$$\Rightarrow f^{'} (x) =6x^{2}>0\Rightarrow$$ strictly increasing function

$$\therefore $$Function is strictly increasing function

And $$f(0)=0$$ when $$x=\sqrt[3]{\dfrac{7}{2} }$$

$$\Rightarrow $$Graph will go to infinity to the right as $$x\rightarrow \infty $$

$$\Rightarrow $$Graph will go to minus infinity to the left as $$x\rightarrow - \infty $$

$$\therefore $$Graph goes up to the right and down to the left

Option 'A' is correct
Question 2
1 / -0

The length of the segment of the tangent line to the curve $$x=a\cos ^{ 3 }{ t } ,y=\sin ^{ 3 }{ t } $$, at any point on the curve cut off by the coordinate axes is

A
$$4a$$

B
$$a$$

C
$${a}^{2}$$

D
$$2a$$
Question 3
1 / -0

Find the approximate error in the volume of a cube with edge $$x$$ cm, when the edge is increased by $$2\%$$

A
$$4\%$$

B
$$2\%$$

C
$$6\%$$

D
$$8\%$$

Solution

in a mutiplication while multiplying we have to add the percentage errors

so if the edge of the cube is $$x$$cm then the volume will be $${ x }^{ 3 }$$

but for errors we have to add the percentage

therefore the total error percentage now becomes $$6\%$$

therefore the error is increased by $$4\%$$
Question 4
1 / -0

Which one of the following be the gradient of the hyperbola $$xy=1$$ at the point $$\left(t,\dfrac{1}{t}\right)$$

A
$$-\dfrac{1}{t}$$

B
$$-\dfrac{1}{t^2}$$

C
$$\dfrac{1}{t}$$

D
$$-\dfrac{2}{t^2}$$

Solution

Given the hyperbola
$$xy=1$$.
Now differentiate both sides with respect to $$x$$ we get,
$$y+x\dfrac{dy}{dx}=0$$
or, $$\dfrac{dy}{dx}=-\dfrac{y}{x}$$
Now the gradient of the given hyperbola at $$\left(t,\dfrac{1}{t}\right)$$ is
$$\left.\dfrac{dy}{dx}\right|_{\left(t,\dfrac{1}{t}\right)}=-\dfrac{1}{t^2}$$
Question 5
1 / -0

The local maximum value of $$x{(1-x)}^{2},0\le x\le 2$$ is

A
$$2$$

B
$$\dfrac {4}{27}$$

C
$$5$$

D
$$2,\dfrac {4}{27}$$

Solution

$$\begin{array}{l} Let\, f\left( x \right) =x{ \left( { 1-x } \right) ^{ 2 } } \\ f'\left( x \right) -1{ \left( { 1-x } \right) ^{ 2 } }+x\times 2\left( { 1-x } \right) \times -1 \\ ={ \left( { 1-x } \right) ^{ 2 } }-2x\left( { 1-x } \right) \end{array}$$

$$\begin{array}{l} f''\left( x \right) =-2\left( { 1-x } \right) -2\left[ { 1-x+x\times -1 } \right] \\ =-2+2x-2\left[ { 1-x-x } \right] \\ =-2+2x-2+4x \\ =6x-4 \end{array}$$

For critical points, $$F'\left( x \right) = 0$$
$$\begin{array}{l} \Rightarrow { \left( { 1-x } \right) ^{ 2 } }-2x\left( { 1-x } \right) =0 \\ \Rightarrow \left( { 1-x } \right) \left( { 1-x-2x } \right) =0 \\ \Rightarrow \left( { 1-x } \right) \left( { 1-3x } \right) =0 \\ \Rightarrow \left( { x-1 } \right) \left( { 3x-1 } \right) =0 \\ x=1 ,\quad \frac { 1 }{ 3 } \end{array}$$

$$f'\left( 1 \right) =6\left( 1 \right) -4=6-4=2>0\, \min imum$$
$$f''\left( { \dfrac { 1 }{ 3 } } \right) =6\times \dfrac { 1 }{ 3 } -4=2-4=-2<0\, { { miximum } }$$

$$f\left( { \dfrac { 1 }{ 3 } } \right) =\dfrac { 1 }{ 3 } { \left( { 1-\dfrac { 1 }{ 3 } } \right) ^{ 2 } }=\dfrac { 1 }{ 3 } \times \dfrac { 4 }{ 9 } =\dfrac { 4 }{ { 27 } } $$

So, $$x = \dfrac{1}{3}$$ is the point of local Maximum and the local maximum value is $$\dfrac{4}{{27}}$$
Question 6
1 / -0

Function $$f(x)=x-\ell nx$$ is decreasing, when

A
$$x \in (0,1)$$

B
$$x \in (-1,1)$$

C
$$x \in (1,\infty)$$

D
$$None\ of\ these$$

Solution

$$f\left( x \right) = x - \ln x$$
$$f'\left( x \right) = 1 - \frac{1}{x}$$
$$ = \frac{{x - 1}}{x}$$
$$f'\left( x \right)\,is\,not\,define\,at\,x = 0$$
$$For\,critical\,po{\mathop{\rm int}} s\,$$
$$f'\left( x \right) = 0$$
$$ = \frac{{x - 1}}{x} = 0$$
$$ = x - 1 = 0$$
$$x = 1$$
$$So,\,critical\,po{\mathop{\rm int}} s\,are\,x = 1\,and\,x = 0$$
$$f'\left( x \right) = \frac{{x - 1}}{x} > 0\ for\ all x \in \left( {1,\infty } \right)$$
$$and\,f'\left( x \right) = \frac{{x - 1}}{x} < 0 \ for\ all \left( {0,1} \right)$$
$$so,\,f\left( x \right)\,is\,decreasing\,\forall x \in \left( {0,1} \right)$$
Question 7
1 / -0

The slope of the tangent to the curve $$xy+ax-by=0$$ at the point $$(1,1)$$ is $$2$$, then value of $$a$$ and $$b$$ are respectively:

A
$$1,2$$

B
$$2,1$$

C
$$3,5$$

D
None of these

Solution

$$xy+ax-by=0$$
$$(1,1)$$ lies on curve
$$1+a-b=0$$
$$a-b=-1$$ ....... $$(i)$$
$$y+x\dfrac { dy }{ dx } +a-b\dfrac { dy }{ dx } =0$$
$$1+2+a-2b=0$$
$$a-2b=-3$$ ........ $$(ii)$$
Solving $$(i)$$ and $$(ii)$$, we get
$$a=1,b=2$$
Question 8
1 / -0

The interval in which the function $$f(x) = {x^3}$$ increases less rapidly than $$\,g(x) = 6{x^2} + 15x + 5$$ is :

A
$$( - \infty , - 1)\,\,\,\,$$

B
$$( - 5,1)\,\,\,\,$$

C
$$( - 1,5)$$

D
$$(5,\infty )$$

Solution

$${\text{Increasing}}\;{\text{rate = }}\dfrac{d}{{dx}}\left( {{f_1}{x_1}} \right)$$
$${\text{f}}\left( x \right) = 6{x^2} + 15x + 5$$
$${\text{f}}\left( x \right) = 12x + 15$$
$${\text{ }}\dfrac{d}{{dx}}{\left( x \right)^3} < \;\dfrac{d}{{dx}}\left( {6{x^2} + 15x + 5} \right)$$
$${\text{ 3}}{{\text{x}}^2}{\text{ < 12x + 15}}$$
$${\text{3}}{{\text{x}}^2} - 12x - 15 < 0$$
$${x^2} - 4x - 5 < 0$$
$${x^2} - 5x + x - 5 < 0$$
$$\left( {x + 1} \right)\left( {x - 5} \right) < 0$$
$$x \in \left( { - 1,5} \right)$$
Question 9
1 / -0

The values of $$x$$ for which the tangents to the curves $$y=x\cos{x},y=\cfrac{\sin{x}}{x}$$ are parallel to the axis of $$x$$ are roots of (respectively)

A
$$\sin{x}=x,\tan{x}=x$$

B
$$\cot{x}=x,\sec{x}=x$$

C
$$\cot{x}=x,\tan{x}=x$$

D
$$\tan{x}=x,\cot{x}=x$$

Solution

$$y=x\cos x$$

$$\dfrac{dy}{dx}=\cos x-x\sin x=0$$

$$\implies x=\cot x$$

$$y=\dfrac{\sin x}{x}$$

$$\dfrac{dy}{dx}=\dfrac{x\cos x-\sin x}{x^{2}}=0$$

$$\implies x=\tan x$$
Question 10
1 / -0

A curve with equation of the form $$y=a{x}^{4}+b{x}^{3}+cx+d$$ has zero gradient at the point $$(0,1)$$ and also touches the x-axis at the point $$(-1,0)$$ then

A
$$a=3$$

B
$$b=4$$

C
$$c+d=1$$

D
for $$x< -1$$ the curve has a negative gradient

Solution

given $$y=ax^4+bx^3+cx+d$$
$$y_{}{'}=4ax^3+3bx^2+c=0$$ at (0,1)
so $$c=0$$
curve passes through point (-1,0)
so $$0=a-b-c+d=>a+d=b$$
here $$y_{}{'}<0$$ for $$x<-1$$

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Application of Derivatives Test - 16

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Application of Derivatives Test - 16

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SHARING IS CARING

Question 1

up to the right and down to the left

down to the right and up to the left

up to the right and up to the left

down to the right and down to the left

none of these ways.

Solution

Question 2

$$4a$$

$$a$$

$${a}^{2}$$

$$2a$$

Question 3

$$4\%$$

$$2\%$$

$$6\%$$

$$8\%$$

Solution

Question 4

$$-\dfrac{1}{t}$$

$$-\dfrac{1}{t^2}$$

$$\dfrac{1}{t}$$

$$-\dfrac{2}{t^2}$$

Solution

Question 5

$$2$$

$$\dfrac {4}{27}$$

$$5$$

$$2,\dfrac {4}{27}$$

Solution

Question 6

$$x \in (0,1)$$

$$x \in (-1,1)$$

$$x \in (1,\infty)$$

$$None\ of\ these$$

Solution

Question 7

$$1,2$$

$$2,1$$

$$3,5$$

None of these

Solution

Question 8

$$( - \infty , - 1)\,\,\,\,$$

$$( - 5,1)\,\,\,\,$$

$$( - 1,5)$$

$$(5,\infty )$$

Solution

Question 9

$$\sin{x}=x,\tan{x}=x$$

$$\cot{x}=x,\sec{x}=x$$

$$\cot{x}=x,\tan{x}=x$$

$$\tan{x}=x,\cot{x}=x$$

Solution

Question 10

$$a=3$$

$$b=4$$

$$c+d=1$$

for $$x< -1$$ the curve has a negative gradient

Solution

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