Application of Derivatives Test

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Application of Derivatives Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

Given function $$f(x)=\left(\displaystyle\frac{e^{2x}-1}{e^{2x}+1}\right)$$ is.

A
Increasing

B
Decreasing

C
Even

D
None of these

Solution

$$f\left( x \right) =\cfrac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 } $$

$$f\left( x \right) =\cfrac { \left[ { e }^{ 2x }+1 \right] \left[ 2{ e }^{ 2x } \right] -\left[ { e }^{ 2x }-1 \right] \left[ 2{ e }^{ 2x } \right] }{ { \left[ { e }^{ 2x }+1 \right] }^{ 2 } } $$

$$f^{ 1 }\left( x \right) =\cfrac { 2{ e }^{ 2x }\left[ { e }^{ 2x }+1-{ e }^{ 2x }+1 \right] }{ { \left[ { e }^{ 2x }+1 \right] }^{ 2 } } $$

$$f^{ 1 }\left( x \right) =\cfrac { 4{ e }^{ 2x } }{ { \left[ { e }^{ 2x }+1 \right] }^{ 2 } } $$

Now, $${ \left[ { e }^{ 2x }+1 \right] }^{ 2 }>0$$ and $${ e }^{ 2x }>0$$
$$\therefore f^{ 1 }\left( x \right) >0$$
$$\therefore f\left( x \right)$$ is increasing function.
Question 2
1 / -0

The coordinates of the points(s) at which the tangents to the curve $$\displaystyle y=x^{3}-3x^{2}-7x+6$$ cut the positive semi axis OX a line segment half that on the negative semi axis OY is/are given by

A
$$(-1, 9)$$

B
$$(3, -15)$$

C
$$(1, -3)$$

D
none

Solution

Let the point of tangency be
$$(x_{1}+y_{1})$$
Then the equation of tangent is
$$\dfrac{y-y_{1}}{x-x_{1}}=3x_{1}^{2}-6x_{1}-7$$
when
$$y=0,x=x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7}$$
when
$$x=0,y=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$$
$$-2(x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$$
$$-2(\dfrac{x_{1}(3x_{1}-6x_{1}-7)-y_{1}}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$$
$$2=3x_{1}-6x_{1}-7$$
$$3x_{1}-6x_{1}-9=0$$
$$x_{1}=-1 or 3$$
when
$$x_{1}=-1,y_{1}=(-1)^{3}-3(-1)^{2}-7(-1)+6=9$$
and the y-intercept is
$$9-(-1)[3(-1)^{2}-6(-1)-7]=11> 0$$
when
$$x_{1}=3,y_{1}=(3)^{3}-3(3)^{2}-7(3)+6=-15$$
and the y-intercept is
$$-15-(3)[3(3)^{2}-6(3)-7]=-21< 0$$
So,the point of tangency is $$(-1,9)$$, this is the correct answer.
Question 3
1 / -0

If $$y = 4x - 5$$ is a tangent to the curve $$y^{2} = px^{3} + q$$ at $$(2, 3)$$, then

A
$$p = 2, q = -7$$

B
$$p = -2, q = 7$$

C
$$p = -2, q = -7$$

D
$$p = 2, q = 7$$

Solution

Since, $$(2, 3)$$ lies on the curve $$y^{2} = px^{3} + q$$
Therefore, $$9 = 8p + q ..... (i)$$
$$y^{2} = px^{3} + q$$
$$\Rightarrow 2y \dfrac {dy}{dx} = 3px^{2}$$
$$\dfrac {dy}{dx} = \dfrac {3px^{2}}{2y}$$
$$\Rightarrow \left (\dfrac {dy}{dx}\right )_{(2, 3)} = \dfrac {12p}{6} = 2p$$
Since, $$y = 4x - 5$$ is tangent to $$y^{2} = px^{3} + q$$ at $$(2, 3)$$. Therefore,
$$\therefore \left (\dfrac {dy}{dx}\right )_{(2, 3)} =$$ Slope of the line $$y = 4x - 5$$
$$\Rightarrow 2p = 4 \Rightarrow p = 2$$
Putting $$p = 2$$ in Eq. (i), we get $$q = -7$$.
Question 4
1 / -0

If $$f(x)=e^x(x-2)^2$$ then

A
f is increasing in $$(-\infty, 0)$$ and $$(2, \infty)$$ deceasing in $$(0, 2)$$

B
f is increasing in $$(-\infty, 0)$$ and deceasing in $$(0, \infty)$$

C
f is increasing in $$(2, \infty)$$ and deceasing in $$(-\infty, 0)$$

D
f is increasing in $$(0, 2)$$ and deceasing in $$(-\infty, 0)$$ and $$(2, \infty)$$

Solution

$$f\left( x \right) ={ e }^{ x }{ (x-2) }^{ x }$$
For increasing or decreasing, we need to find $$f'\left( x \right) $$
$$f'\left( x \right) ={ e }^{ x }\times 2(x-2)+{ e }^{ x }{ (x-2) }^{ x }$$
$$={ e }^{ x }(x-2)\left[ 2+x-2 \right] $$
$$f'\left( x \right) ={ e }^{ x }x(x-2)$$
It will have two solutions, $$x=0,x=2$$
plotting it on number line
(Image)
Therefore $$f'\left( x \right) >0$$ for $$\left( -\infty ,0 \right) \bigcup { (2,\infty ) } $$ and
$$f'\left( x \right) <0$$ for $$(0,2)$$
Thus function increases for $$\left( -\infty ,0 \right) \bigcup { (2,\infty ) } $$ and function decreases for $$(0,2)$$
Question 5
1 / -0

All the critical points of $$f(x)=\dfrac {|2-x|}{x^2}$$ is/are:

A
$$x=0,2$$

B
$$x=2,4$$

C
$$x=2,-4$$

D
None of the above.

Solution

The critical point for $$f(x)$$ is a value $$x_0$$ where its derivative is zero. i.e., $$f^{'}(x_0)=0$$

Given that $$f(x)=\dfrac{|2-x|}{x^2}$$

Therefore, $$f^{'}(x)=\dfrac{x^2(\dfrac{|2-x|}{2-x})-|2-x|(2x)}{x^4}=0$$

$$\implies \dfrac{|2-x|(x^2-2x(2-x))}{x^4(2-x)}=0$$

$$\implies |2-x|(x^2-2x(2-x))=0$$ and $$x^4\neq 0$$ and $$2-x\neq 0$$

$$\implies |2-x|=0$$ and $$(3x-4)x=0$$ and $$x\neq 0$$ and $$x\neq 2$$

$$\implies x=2$$ and $$x=0$$ and $$x=\dfrac 43$$and $$x\neq 0$$ and $$x\neq 2$$

Therefore, $$x=\dfrac 43$$ is the critical point.
Question 6
1 / -0

If the slope of the curve $$y=\cfrac { ax }{ b-x } $$ at the point $$(1,1)$$ is $$2$$, then the values of $$a$$ and $$b$$ are respectively

A
$$1,-2$$

B
$$-1,2$$

C
$$1,2$$

D
None of these

Solution

$$y=\cfrac { ax }{ b-x } $$
Slope= $$\cfrac { dy }{ dx } $$
So,$$\cfrac { dy }{ dx } $$= $$\cfrac { a(b-x)-(-1)(ax) }{ { (b-1) }^{ 2 } } \quad \longrightarrow (1)$$
for $$(1,1)$$
put in equation of curve-
$$1=\cfrac { a }{ (b-1) } \quad \Rightarrow (b-1)=a\quad \longrightarrow (2)$$
From slop after putting (1,1)
We get,
$$\\ 2=\cfrac { a(b-1)+a }{ { (b-1) }^{ 2 } } $$
put $$(b-1)=a$$ from equation(2)
$$2=\cfrac { { a }^{ 2 }+a }{ { a }^{ 2 } } \quad \quad \\ \Rightarrow 2{ a }^{ 2 }={ a }^{ 2 }+a\\ \quad \Rightarrow a(a-1)=0\\ \quad \Rightarrow a=0,1$$
put in (2),
We get
$$b=1,2$$
In option only one value is given
$$\therefore a=1,b=2$$
Question 7
1 / -0

If $$f:[1, 10]\rightarrow[1,10]$$ is a non-decreasing function and $$g:[1,10] \rightarrow [1,10]$$ is a non-increasing function, Let $$h(x) = f(g(x))$$ with $$h(1)=1$$. then, $$h(2)$$

A
less than $$1$$

B
is more than two

C
is equal to $$2$$

D
is not defined

Solution
Question 8
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
$$(-1,-1)$$

Solution

Given curve is $$x+y={ e }^{ xy }\quad $$
On differentiating w.r.t $$x$$ we get
$$1+\cfrac { dy }{ dx } ={ e }^{ xy }\left\{ y+x\cfrac { dy }{ dx } \right\} $$
$$\Rightarrow \cfrac { dy }{ dx } =\cfrac { y{ e }^{ xy }-1 }{ 1-x{ e }^{ xy } } $$
Since, tangent is parallel to y-axis
Here, $$\cfrac { dy }{ dx } =\infty \Rightarrow 1-x{ e }^{ xy }=0$$
$$\Rightarrow 1-x(x+y)=0$$
This holds for $$x=1,y=0$$
Question 9
1 / -0

Directions For Questions

Tangent at any point $$p_{1}$$ (other than $$(0, 0)$$) on the curve $$y = x^{3}$$ meets the curve again at $$p_{2}$$.
Tangent at $$p_{2}$$ meets the curve again at $$p_{3}$$ and so on.

...view full instructions

Abscissa of $$p_{1}, p_{2}, p_{3} .... p_{n}$$ are in

A
A.P.

B
G.P.

C
H.P

D
None

Solution
Question 10
1 / -0

If $$g(x)$$ is continuous function at $$x = a$$, such that $$g(a) > 0$$ and $$f'(x) (g(x))(x^{2} - ax + a^{2}) \forall x\epsilon R$$, then $$f(x)$$ is

A
Increasing in the neighbourhood of $$x = a$$

B
Decreasing in the neighbourhood of $$x = a$$

C
Constant in the neighbourhood of $$x = a$$

D
Maximum at $$x = a$$

Solution

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Application of Derivatives Test - 59

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Application of Derivatives Test - 59

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SHARING IS CARING

Question 1

Increasing

Decreasing

Even

None of these

Solution

Question 2

$$(-1, 9)$$

$$(3, -15)$$

$$(1, -3)$$

none

Solution

Question 3

$$p = 2, q = -7$$

$$p = -2, q = 7$$

$$p = -2, q = -7$$

$$p = 2, q = 7$$

Solution

Question 4

f is increasing in $$(-\infty, 0)$$ and $$(2, \infty)$$ deceasing in $$(0, 2)$$

f is increasing in $$(-\infty, 0)$$ and deceasing in $$(0, \infty)$$

f is increasing in $$(2, \infty)$$ and deceasing in $$(-\infty, 0)$$

f is increasing in $$(0, 2)$$ and deceasing in $$(-\infty, 0)$$ and $$(2, \infty)$$

Solution

Question 5

$$x=0,2$$

$$x=2,4$$

$$x=2,-4$$

None of the above.

Solution

Question 6

$$1,-2$$

$$-1,2$$

$$1,2$$

None of these

Solution

Question 7

less than $$1$$

is more than two

is equal to $$2$$

is not defined

Solution

Question 8

$$(0,1)$$

$$(1,0)$$

$$(1,1)$$

$$(-1,-1)$$

Solution

Question 9

A.P.

G.P.

H.P

None

Solution

Question 10

Increasing in the neighbourhood of $$x = a$$

Decreasing in the neighbourhood of $$x = a$$

Constant in the neighbourhood of $$x = a$$

Maximum at $$x = a$$

Solution

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