Application of Derivatives Test

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Application of Derivatives Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

The points of the curve $$y={ x }^{ 3 }+x-2$$ at which its tangent are parallel to the straight line $$y=4x-1$$ are

A
$$\left( 2,7 \right) ,\left( -2,-11 \right) $$

B
$$\left( 0,-2 \right) ,\left( { 2 }^{ 1/3 },{ 2 }^{ 1/3 } \right) $$

C
$$\left( { -2 }^{ 1/3 },{ -2 }^{ 1/3 } \right) \left( 0,-4 \right) $$

D
$$\left( 1,0 \right) ,\left( -1,-4 \right) $$

Solution

Given
$$y={ x }^{ 3 }+x-2...(i)$$

$$y=4x-1....(ii)$$

Slope of tangent to the curve (i)

$$\cfrac { dy }{ dx } =3{ x }^{ 2 }+1$$

slope of tangent at point $$\left( \alpha ,\beta \right) $$ is

$$\quad { \left| \cfrac { dy }{ dx } \right| }_{ \left( \alpha ,\beta \right) }^{ }=3{ \alpha }^{ 2 }+1...(iii)$$

Given, tangent of curve (i) is parallel to line (ii)

$$\therefore$$ Slope of line (ii) is $$4$$

$$\therefore$$ From Eq(iii) we get

$$3{ \alpha }^{ 2 }+1=4\Rightarrow \alpha =\pm 1$$

$$\therefore \left( \alpha ,\beta \right) $$ lie on curve (i)

$$\beta ={ \left( \pm 1 \right) }^{ 2 }+\left( \pm 1 \right) -2\Rightarrow \beta =0,-4\quad $$

$$\therefore$$ Points are $$(1,0)$$ and $$(-1,-4)$$
Question 2
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Consider the following statements:
Statement I:
$$x > \sin x$$ for all $$x > 0$$
Statement II:
$$f(x) = x - \sin x$$ is an increasing function for all $$x > 0$$
Which one of the following is correct in respect of the above statements?

A
Both Statements I and II are true and Statement II is the correct explanation of Statement I

B
Both Statements I and II are true and Statement II is the not correct explanation of Statement I

C
Statement I is true but Statement II is false

D
Statement I is true but Statement II is true

Solution

Assume $$f(x)=x-\sin x$$
$$\therefore f'(x)=1-\cos x\geq 0$$ and hence, $$x-\sin x$$ is an increasing function and hence, $$x>\sin x$$
So, both I and II are true and II is correct explanation for I.
Question 3
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The critical points of the function $$f(x)={ x }^{ 3/5 }\left( 4-x \right) ,x\in { R }^{ + }\cup \left\{ 0 \right\} $$ is _______

A
$$0,-\cfrac { 3 }{ 2 } $$

B
$$0,\cfrac { 3 }{ 2 } $$

C
$$0,\cfrac { 2 }{ 3 } $$

D
$$0,-\cfrac { 2 }{ 3 } $$

Solution

$$f(x)={ x }^{ \cfrac { 3 }{ 5 } }\left( 4-x \right) x\in { R }^{ + }$$
$$f(x)=4{ x }^{ \cfrac { 3 }{ 5 } }-{ x }^{ \cfrac { 8 }{ 5 } }$$
$$\therefore f'(x)=4\left( \cfrac { 3 }{ 5 } \right) { x }^{ -\cfrac { 2 }{ 3 } }=\cfrac { 8 }{ 5 } { x }^{ \cfrac { 3 }{ 5 } }$$
$$\therefore f'(x)=\cfrac { 4 }{ 5 } \left( \cfrac { 3 }{ { x }^{ \cfrac { 2 }{ 3 } } } -2{ x }^{ \cfrac { 3 }{ 5 } } \right) =\cfrac { 4 }{ 5 } \left( \cfrac { 3-2x }{ { x }^{ \cfrac { 2 }{ 3 } } } \right) \quad $$
For $$x=0;f'(x)$$ does not exists and for $$x=\cfrac { 2 }{ 3 } $$ we have $$f'(x)=0$$
$$\therefore$$ Critical points are $$0$$ and $$\cfrac{2}{3}$$
Question 4
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A point P moves such that sum of the slopes of the normal drawn from it to the hyperbola $$xy=16$$ is equal to the sum of the ordinates of the feet of the normal. Let 'P' lies on the curve C, then.
The equation of 'C' is?

A
$$x^2=4y$$

B
$$x^2=16y$$

C
$$x^2=12y$$

D
$$y^2=8x$$

Solution

Let P(h,k) be the point in the plane of hyperbola $$xy=16=4^2$$

The equation of the normal at a point $$(4t, \dfrac {4} {t})$$ to the hyperbola $$xy=4^2$$ is

$$xt^3-yt-4t^4+4=0$$

if if passes through $$P(h,k)$$ then,

$$ht^3-yt-4t^4+4=0$$

$$ \Rightarrow 4t^4-ht^3+kt-4=0$$

this is a fourth degree equation in t. So, it gives 4 values of t, $$t_1,t_2,t_3,t_4$$ corresponding to each value of t there is a point on hyperbola such that the normal passes through $$P(h,k)$$.

let the 4 points be $$ A(ct_1, \dfrac{c}{t_1}) $$,$$ B(ct_2, \dfrac{c}{t_2}) $$,$$ C(ct_3, \dfrac{c}{t_3}) $$ and $$ D(ct_4, \dfrac{c}{t_4}) $$

such that normal at these points pass through $$P(h,k)$$.

Since $$t_1,t_2,t_3,t_4$$ are roots of equation (1).

Therefore,
$$t_1+t_2+t_3+t_4 = \dfrac{h}{c}=\dfrac{h}{4}$$.......(2)

$$ \sum t_1t_2 =0 $$.......(3)

$$ \sum t_1t_2t_3 = - \dfrac{k}{4} $$.........(4)

$$ t_1t_2t_3t_4 =-1 $$.......(5)

It s given that the sum of the slopes of the normals at A,B,C and D is equal to the sum of the coordinates of these points.

Therefore
$$t_1^2+t_2^2+t_3^2+t_4^2 = \dfrac{c}{t_1}+\dfrac{c}{t_3}+\dfrac{c}{t_3}+\dfrac{c}{t_4}=\dfrac{4}{t_1}+\dfrac{4}{t_2}+\dfrac{4}{t_3}+\dfrac{4}{t_4}$$

$$(t_1+t_2+t_3+t_4 )^2-2\sum t_1t_2=4\left ( \dfrac {\sum t_1t_2t_3}{t_1t_2t_3t_4} \right )$$

$$\Rightarrow \dfrac{h^2}{4}-2\times 0 = 4 \times \left ( \dfrac{\dfrac {-k}{4}} {-1} \right ) $$

$$\Rightarrow \dfrac{h^2}{4} = k $$

$$\Rightarrow h^2 = 16k$$

so the locus of $$(h,k)$$ is $$x^2=16y$$, which is a parabola
Question 5
1 / -0

$$f(x) = |x\log_{e}x|$$ monotonically decreases in

A
$$(0, 1/e)$$

B
$$(1/e, 1)$$

C
$$(1, \infty)$$

D
$$(1/e, \infty)$$

Solution

Monotomic functions are those functions who either increase or decrease over a given domain.
Example: the function $$x^{3}$$ in an always increasing function.
The given equation is:
$$f(x) = |x\log_{e}x|$$
Differentiating once w.r.t to $$x$$ we get,
$$\Rightarrow f'(x) = \dfrac {|x\log_{e}x|}{x\log_{e}x}\times (\log_{e}x + 1)$$
Now for $$0 < x < 1$$ we have
$$\Rightarrow f'(x) = \dfrac {-x\log_{e}x}{x\log_{e}x} \times (\log_{e} x + 1)$$
$$\Rightarrow f'(x) = -(\log_{e}x + 1)$$
Now for monotonically decreasing function we have $$f'(x) < 0$$
$$\Rightarrow -(\log_{e}x + 1) < 0$$
$$\Rightarrow (\log_{e} x + 1) > 0$$
$$\Rightarrow x > \dfrac {1}{e}$$
Now for $$x > 1$$ we have,
$$\Rightarrow f'(x) = \dfrac {x\log_{e}x}{x\log_{e}x}\times (\log_{e}x + 1)$$
$$\Rightarrow f'(x) = (\log_{e}x + 1)$$
$$\Rightarrow (\log_{e} x + 1) < 0$$
This cannot be true as $$\log_{e} > 0$$ in this domain hence the inequality is never satisfied.
Hence the function decreases in the range $$\left (\dfrac {1}{e}, 1\right )$$.
Question 6
1 / -0

The percentage error in the surface area of a cube with edge x cm, when the edge is increased by $$11\%$$ is _________.

A
$$11$$

B
$$22$$

C
$$10$$

D
$$44$$

Solution

Surface area of cube $$=$$ S $$=6x^2$$
$$\therefore \dfrac{dS}{dt}=12x\dfrac{dx}{dt}$$
Side of cube increase $$11\%$$.
$$\therefore \dfrac{dx}{dt}=11\%$$ increment in side
$$=\dfrac{11x}{100}$$
$$\therefore \dfrac{dx}{dt}=12\left(\dfrac{11x}{100}\right)$$
$$=6\left(\dfrac{22}{100}\right)x^2$$
$$=6x^2\left(\dfrac{22}{100}\right)$$
$$=22\%$$ in surface area
$$\therefore$$ Surface area increase $$22\%$$.
Question 7
1 / -0

Let, $$f:A\rightarrow B$$ be an invertible function. If $$f(x)=2x^3+3x^2+x-1$$, then $$f'^{-1}(5)$$=

A
$$\dfrac{1}{13}$$

B
$$1$$

C
$$6$$

D
can not be determined

Solution

Given $$f$$ be an invertible function and $$f(x)=2x^3+3x^2+x-1$$.
Then $$f'^{-1}(5)=\dfrac{1}{f'(x)}\forall x\in f(A)$$ such that $$f(x)=5$$.
Now, $$f(x)=5$$ given only $$x=1$$ a real root, remaining are imaginary.
Now, $$f'(x)=6(x^2+x)+1$$.
$$\therefore $$$$f'^{-1}(5)=\dfrac{1}{f'(1)}=\dfrac{1}{13}$$
Question 8
1 / -0

The point on the curve $$y^{2}=x$$ where tangent makes $$45^{o}$$ angle with $$x-$$axis ?

A
$$(\dfrac{1}{2},\dfrac{1}{4})$$

B
$$(\dfrac{1}{4},\dfrac{1}{2})$$

C
$$(4,2)$$

D
$$(1,1)$$

Solution

$$ \dfrac{dy}{dx} = $$ Slope of the tangent at that point of the curve.

Given curve is $$ x = y^2 $$
Keep $$ y = a, \, then \, x = y^2 $$
point $$ (a^2, a) $$

$$ \displaystyle \dfrac{dy}{dx} (y^2 - x) = 0 \dfrac{dy}{dx} = \dfrac{1}{2y} = \dfrac{1}{2a} = \tan(45^0) $$

$$ \Rightarrow a= \dfrac{1}{2} \quad a^2= \dfrac{1}{4} $$

point $$ \left(\dfrac{1}{4}, \dfrac{1}{2}\right) $$
Question 9
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The tangent to $$\left( a{ t }^{ 2 },2at \right) $$ is perpendicular to X-axis at _____ point $$t\in R$$.

A
$$(4a,4a)$$

B
$$(a,2a)$$

C
$$(0,0)$$

D
$$(a,-2a)$$

Solution

Here $$\left( x,y \right) =\left( a{ t }^{ 2 },2at \right) $$
$$\therefore x=a{ t }^{ 2 };y=2at$$
$$\therefore \cfrac { dx }{ dt } =2at;\cfrac { dy }{ dt } =2a$$
$$\therefore \cfrac { dy }{ dx } \cfrac { dy/dt }{ dx/dt } =\cfrac { 2a }{ 2at } =\cfrac { 1 }{ t } $$
Since, tangent is perpendicular to X-axis slope is undefined
$$\therefore$$ $$t=0$$
$$\therefore t=0$$
$$\therefore \left( a{ t }^{ 2 },2at \right) =\left( 0,0 \right) $$
Thus, tangent is perpendicular to X-axis at $$(0,0)$$
Question 10
1 / -0

Line $$y=x$$ and curve $$y=x^2+bx+c$$ touches at $$(1, 1)$$ then __________.

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

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

C
$$b=1, c=1$$

D
$$b=0, c=1$$

Solution

Given line $$l : y =x$$
$$l : x-y=0$$
$$\therefore$$ slope of line $$l=1$$
Now $$y=x^2+bx+c$$
$$\therefore \dfrac{dy}{dx}=2x+b$$
$$\therefore \left(\dfrac{dy}{dx}\right)_{P(1, 1,)}=2(1)+b$$
$$=2+b$$
But slope of tangent of the curve $$=1$$.
$$\therefore 2+b=1$$
$$\therefore b=-1$$
Now $$(1, 1)\in y=x^2+bx+c$$
$$\therefore 1=1+b(1)+c$$
$$\therefore b+c=0$$
$$\therefore c=-b$$
$$c=-(-1)$$
$$\therefore c=1$$
$$\therefore$$ We have $$b=-1$$ and $$c=1$$.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 40

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Application of Derivatives Test - 40

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SHARING IS CARING

Question 1

$$\left( 2,7 \right) ,\left( -2,-11 \right) $$

$$\left( 0,-2 \right) ,\left( { 2 }^{ 1/3 },{ 2 }^{ 1/3 } \right) $$

$$\left( { -2 }^{ 1/3 },{ -2 }^{ 1/3 } \right) \left( 0,-4 \right) $$

$$\left( 1,0 \right) ,\left( -1,-4 \right) $$

Solution

Question 2

Both Statements I and II are true and Statement II is the correct explanation of Statement I

Both Statements I and II are true and Statement II is the not correct explanation of Statement I

Statement I is true but Statement II is false

Statement I is true but Statement II is true

Solution

Question 3

$$0,-\cfrac { 3 }{ 2 } $$

$$0,\cfrac { 3 }{ 2 } $$

$$0,\cfrac { 2 }{ 3 } $$

$$0,-\cfrac { 2 }{ 3 } $$

Solution

Question 4

$$x^2=4y$$

$$x^2=16y$$

$$x^2=12y$$

$$y^2=8x$$

Solution

Question 5

$$(0, 1/e)$$

$$(1/e, 1)$$

$$(1, \infty)$$

$$(1/e, \infty)$$

Solution

Question 6

$$11$$

$$22$$

$$10$$

$$44$$

Solution

Question 7

$$\dfrac{1}{13}$$

$$1$$

$$6$$

can not be determined

Solution

Question 8

$$(\dfrac{1}{2},\dfrac{1}{4})$$

$$(\dfrac{1}{4},\dfrac{1}{2})$$

$$(4,2)$$

$$(1,1)$$

Solution

Question 9

$$(4a,4a)$$

$$(a,2a)$$

$$(0,0)$$

$$(a,-2a)$$

Solution

Question 10

$$b=-1, c=1$$

$$b=1, c=2$$

$$b=1, c=1$$

$$b=0, c=1$$

Solution

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