Application of Derivatives Test

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Application of Derivatives Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

Let $$f'(\sin { x } )<0$$ and $$f''(\sin { x } )>0>\forall x\in \left( 0,\cfrac { \pi }{ 2 } \right) $$ and $$g(x)=f'(\sin { x } )+f'(\cos { x } ) $$, then $$g(x)$$ is decreasing in

A
$$\left( \cfrac { \pi }{ 4 } ,\cfrac { \pi }{ 2 } \right) $$

B
$$\left( 0,\cfrac { \pi }{ 4 } \right) $$

C
$$\left( 0,\cfrac { \pi }{ 2 } \right) $$

D
$$\left( \cfrac { \pi }{ 6 } ,\cfrac { \pi }{ 2 } \right) $$
Question 2
1 / -0

$$f(x)=e^{3x}-sinx+x^{2}$$, Find $$f '(x)$$

A
$$3e^{3x}-cosx+2x$$

B
$$e^{3x}+cosx+2x$$

C
$$3e^{3x}+cosx+x$$

D
$$3e^{3x}+sinx+2x$$
Question 3
1 / -0

Given that f(x) is a differentiable function of x and that $$f(x).f(y)=f(x)-4-f(y)+f(xy)-2$$ and that $$f(2)=5$$. Then $$f'(3)$$ is equal to

A
2

B
24

C
15

D
19
Question 4
1 / -0

The curve that passes through the point $$(2,3)$$ and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by

A
$${ \left( \cfrac { x }{ 2 } \right) }^{ 2 }+{ \left( \cfrac { y }{ 3 } \right) }^{ 2 }=2\quad $$

B
$$2y-3x=0$$

C
$$y=\cfrac { 6 }{ x } $$

D
$${ x }^{ 2 }+{ y }^{ 2 }=13$$

Solution

Let $$B$$ be the pt. of contact $$y=f(x)$$ be the curve
then $$AB=BC$$ (given condition)
$$C\equiv (2x,0)$$ $$A\equiv (0,2y)$$
Slope $$m$$ of line $$AC$$ is
$$m=\dfrac{dy}{dx}=\dfrac{0.2y}{2x-0}$$
$$\dfrac{dy}{dx}=\dfrac{-y}{x}$$
$$\dfrac{dy}{y}=\dfrac{-dx}{x}$$
$$\ln y=-\ln x+c$$
$$yx=c$$
as $$(2, 3)$$ satisfies the curve
$$\therefore 2\times 3=c$$
$$\therefore y=\dfrac{6}{x}$$ is the curve
$$\therefore $$ Option $$C$$ is correct
Question 5
1 / -0

If the line $$y=4x-5$$ touches to the curve $${ y }^{ 2 }=a{ x }^{ 3 }+b$$ at the point $$(2,3)$$ then $$7a+2b=$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

The slope of given line $$y=4x-5$$ is $$m=4$$.

Now, since the given line touches the curve at $$(2,3)$$, the slope of curve $$\dfrac{dy}{dx}$$ at the given point is equal to the slope $$m$$ of the line.

Hence, for slope of curve, differentiating both sides w.r.t x:-

$$2y\dfrac{dy}{dx}=3ax^2$$

Putting the value $$x=2 $$ and $$y=3$$

$$6\dfrac{dy}{dx}=3a\times 4 \implies 6\times4=12a$$

$$\implies a=2$$

Now, the given point lies on the curve, so it must satisfy the equation of the curve:-

$$\implies (3)^2=2\times (2)^3+b$$

$$\implies b=-7$$

$$7a+2b=(7\times 2-2\times 7)=0$$

Hence, answer is option-(A).
Question 6
1 / -0

If $$f(x)$$ is an even function, where $$f(x)\ne 0$$, then which one of the following is correct?

A
$$f'(x)$$ is an even function

B
$$f'(x)$$ is an odd function

C
$$f'(x)$$ may be an even or odd function depending on the type of function

D
$$f'(x)$$ is a constant function

Solution

Let $$f(x)$$ be an even function

$$\Rightarrow f(x) = f(-x) $$

Differentiate both sides, we get

$$\dfrac{df}{dx} = \dfrac{d(f(-x))}{dx} \quad ...(1)$$

To differentiate $$f(-x)$$, we use the chain rule formula as follows:

Let $$u = - x$$, hence $$\dfrac{d(f(-x))}{dx}=\dfrac{df(u)}{du}\cdot \dfrac{du}{dx}=f'(u)\cdot (-1)=-f'(-x)$$

Substituting in $$(1)$$ we get

$$f '(x) = - f '(- x)$$

That is $$f'(x)$$ is an odd function.
Question 7
1 / -0

For the curve $$x=t^2-1$$, $$y=t^2-t$$, the tangent is perpendicular to $$x$$-axis then

A
$$t=0$$

B
$$t=\dfrac{1}{2}$$

C
$$t=1$$

D
$$t=\dfrac{1}{\sqrt{3}}$$

Solution

Given curve is $$x=t^2-1, y=t^2-t$$

Derivating w.r.to $$t$$ we get

$$\dfrac{dx}{dt}=2t$$ -------(1)

and $$\dfrac{dy}{dt}=2t-1$$ -------(2)

dividing (2) by (1) we get

$$\dfrac{dy}{dx}=\dfrac{2t-1}{2t}$$

Therefore, the slope of the tangent is $$\dfrac{2t-1}{2t}$$

Given that the tangent is perpendicular to x-axis. Therefore, tangent is parallel to y-axis.

We know that slope of y-axis is infinity and the slopes of the two parallel lines are equal.

Therefore, slope of the tangent is infinity.

Hence, $$\dfrac{2t-1}{2t}=\dfrac{1}{0}$$

$$\implies 2t=0 \implies t=0$$
Question 8
1 / -0

Let $$f\left( x \right) = {\tan ^{ - 1}}x - \frac{{In\left| x \right|}}{2},x \ne 0.$$. Then $$f\left( x \right)$$ is increasing in

A
$$\left( {0,\infty } \right)$$

B
$$\left( { - \infty ,0} \right)$$

C
$$\left( {1,\infty } \right)$$

D
none of these

Solution

$$f(x)=\tan ^{-1} x-\frac{\ln |x|}{2}, x \neq 0$$

$$f^{\prime}(x)=\frac{1}{1+x^{2}}-\frac{1}{2 x}$$

$$\Rightarrow f^{\prime}(x)=0$$

$$\Rightarrow \frac{1}{1+x^{2}}-\frac{1}{2 x}=0$$

$$\Rightarrow \frac{1}{1+x^{2}}=\frac{1}{2 x}$$

$$\Rightarrow x^{2}+1=2 x \\$$
$$\Rightarrow x^{2}-2 x+1=0 \\$$
$$\Rightarrow(x-1)^{2}=0 \\$$
$$\therefore x=1$$
Hence
$$f(x)$$ is increasing in $$(-\infty, 0)$$
Question 9
1 / -0

If the tangent at $$({x_1},{y_1})$$ to the curve $${x^3} + {y^3} = {a^3}$$ meets the curve again at $$({x_2},{y_2})$$, then

A
$${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = - 1$$

B
$${{{x_2}} \over {{y_1}}} + {{{x_1}} \over {{y_2}}} = - 1$$

C
$${{{x_1}} \over {{x_2}}} + {{{y_1}} \over {{y_2}}} = - 1$$

D
$${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = 1$$

Solution

Given curve equation is $$x^3+y^3=a^3$$.

The curve passes through the points $$(x_1,y_1)$$ and $$(x_2,y_2)$$.

Therefore, $$x_1^2+y_1^2=a^3$$ -------(1)

and $$x_2^3+y_2^3=a^3$$ -------(2)

subtracting (1) from (2) we get

$$y_2^3-y_1^3=-(x_2^3-x_1^3)$$ -------(3)

Therefore, $$3x^2+3y^2(\dfrac{dy}{dx})=0$$

$$\implies \dfrac{dy}{dx}=-\dfrac{x^2}{y^2}$$

Therefore, the slope of the tangent at $$(x_1,y_1)$$ is $$-\dfrac{x_1^2}{y_1^2}$$

The tangent passes through the points $$(x_1,y_1)$$ and $$(x_2,y_2)$$.

Therefore, the slope of the tangent is $$\dfrac{y_2-y_1}{x_2-x_1}$$

On comparing two slopes we get,

$$-\dfrac{x_1^2}{y_1^2}=\dfrac{y_2-y_1}{x_2-x_1}$$

we know that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

$$\implies -\dfrac{x_1^2}{y_1^2}=\dfrac{y_2^3-y_1^3}{x_2^3-x_1^3}\times \dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}$$

$$\implies -\dfrac{x_1^2}{y_1^2}=-\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}$$

$$\implies x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_1^2$$

$$\implies x_1^2y_2^2-x_2^2y_1^2=x_1x_2y_1^2-x_1^2y_1y_2$$

$$\implies (x_1y_2-x_2y_1)(x_1y_2+x_2y_1)=x_1y_1(x_2y_1-x_1y_2)$$

$$\implies x-1y_2+x_2y_1=-x_1y_1$$

$$\implies \dfrac{x_2}{x_1}+\dfrac{y_2}{y_1}=-1$$
Question 10
1 / -0

If the error committed in measuring the radius of the circle is $$0.05\%$$, then the corresponding error in calculating the area is:

A
$$0.05\%$$

B
$$0.025\%$$

C
$$0.25\%$$

D
$$0.1\%$$

Solution

$$\dfrac { dr }{ r } =0.05\Rightarrow dr=(0.05)r$$
Area of circle $$=\pi r^2$$
$$\ A=\pi r^{ 2 }\Rightarrow \dfrac { dA }{ dr } =2\pi r\\ dA=2\pi rdr\Rightarrow \dfrac { dA }{ A } =\dfrac { 2\pi rdr }{ \pi r^{ 2 } } =\dfrac { 2dr }{ r } \\ \therefore \dfrac { dA }{ A } =2(0.05)^{ 2 }=0.1$$
$$\therefore$$ Corresponding error in area $$= 0.1\%$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 60

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Application of Derivatives Test - 60

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SHARING IS CARING

Question 1

$$\left( \cfrac { \pi }{ 4 } ,\cfrac { \pi }{ 2 } \right) $$

$$\left( 0,\cfrac { \pi }{ 4 } \right) $$

$$\left( 0,\cfrac { \pi }{ 2 } \right) $$

$$\left( \cfrac { \pi }{ 6 } ,\cfrac { \pi }{ 2 } \right) $$

Question 2

$$3e^{3x}-cosx+2x$$

$$e^{3x}+cosx+2x$$

$$3e^{3x}+cosx+x$$

$$3e^{3x}+sinx+2x$$

Question 3

2

24

15

19

Question 4

$${ \left( \cfrac { x }{ 2 } \right) }^{ 2 }+{ \left( \cfrac { y }{ 3 } \right) }^{ 2 }=2\quad $$

$$2y-3x=0$$

$$y=\cfrac { 6 }{ x } $$

$${ x }^{ 2 }+{ y }^{ 2 }=13$$

Solution

Question 5

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 6

$$f'(x)$$ is an even function

$$f'(x)$$ is an odd function

$$f'(x)$$ may be an even or odd function depending on the type of function

$$f'(x)$$ is a constant function

Solution

Question 7

$$t=0$$

$$t=\dfrac{1}{2}$$

$$t=1$$

$$t=\dfrac{1}{\sqrt{3}}$$

Solution

Question 8

$$\left( {0,\infty } \right)$$

$$\left( { - \infty ,0} \right)$$

$$\left( {1,\infty } \right)$$

none of these

Solution

Question 9

$${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = - 1$$

$${{{x_2}} \over {{y_1}}} + {{{x_1}} \over {{y_2}}} = - 1$$

$${{{x_1}} \over {{x_2}}} + {{{y_1}} \over {{y_2}}} = - 1$$

$${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = 1$$

Solution

Question 10

$$0.05\%$$

$$0.025\%$$

$$0.25\%$$

$$0.1\%$$

Solution

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