Application of Derivatives Test

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Application of Derivatives Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

At what values of $$a$$, the curve $$x^4+3ax^3+6x^2+5$$ is not situated below any of its tangent lines

A
$$|a|\,>\,\displaystyle\frac{4}{3}$$

B
$$|a|\,<\,\displaystyle\frac{4}{3}$$

C
$$|a|\,>\,1$$

D
$$|a|\,<\,\displaystyle\frac{1}{3}$$

Solution

$$y=4x^3+3ax^3+6x^2+5$$
For given situation curve must be concave so, $$\displaystyle\frac{d^2y}{dx^2}\,>\,0$$
$$y'=4x^3+9ax^2+12x$$
$$y''=12x^2+18ax+12\,>\,0=6(2x^2+3ax+2)\,>\,0$$
$$D\,<\,0\;\;\Rightarrow\;\;9a^2-16\,<\,0$$
$$\Rightarrow |a|\,<\,\displaystyle\frac{4}{3}$$
Question 2
1 / -0

The minimum value of the polynomial.
$$p(x)=3{ x }^{ 2 }-5x+2$$

A
$$-\frac { 1 }{ 6 }$$

B
$$\frac { 1 }{ 6 }$$

C
$$\frac { 1 }{ 12 }$$

D
$$-\frac { 1 }{ 12 }$$

Solution

$$p(x)=3{ x }^{ 2 }-5x+2$$
$$=3\left( x2-\frac { 5 }{ 3 } x+\frac { 2 }{ 3 } \right) $$
$$=3\left[ { x }^{ 2 }-\frac { 5 }{ 3 } x+\left( { \frac { 5 }{ 6 } } \right) ^{ 2 }-\left( { \frac { 5 }{ 6 } } \right) ^{ 2 }+\frac { 2 }{ 3 } \right] $$
$$=3\left[ \left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }-\frac { 25 }{ 36 } +\frac { 2 }{ 3 } \right] $$
$$=3\left[ \left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }+\frac { -25+24 }{ 36 } \right] $$
$$=3\left[ \left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }-\frac { 1 }{ 36 } \right] =3\left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }-\frac { 1 }{ 12 } $$
Question 3
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If the function $$\displaystyle f\left( x \right)=\left( { a }^{ 2 }-3a+2 \right) \cos { \frac { x }{ 2 } } +\left( a-1 \right) x$$ possesses critical points, then $$a$$ belongs to the interval

A
$$\left( -\infty ,0 \right) \cup \left( 4,\infty \right) $$

B
$$(-\infty ,0]\cup [4,\infty )$$

C
$$(-\infty ,0]\cup \left\{ 1 \right\} \cup [4,\infty )$$

D
None of these

Solution

Given
$$\displaystyle f\left( x \right) =\left( { a }^{ 2 }-3a+3 \right) \cos { \frac { x }{ 2 } } +\left( a-1 \right) x$$
$$\displaystyle \Rightarrow f'\left( x \right) =\frac { -1 }{ 2 } \left( a-1 \right) \left( a-2 \right) \sin { \left( \frac { x }{ 2 } \right) } +\left( a-1 \right) $$
$$\displaystyle \Rightarrow f'\left( x \right) =\left( a-1 \right) \left[ 1-\frac { 1 }{ 2 } \left( a-2 \right) \sin { \left( \frac { x }{ 2 } \right) } \right] $$
If $$f(x)$$ possesses critical points, then
$$f'(x)=0$$ $$\displaystyle \Rightarrow \left( a-1 \right) \left[ 1-\left( \frac { a-2 }{ 2 } \right) \sin { \frac { x }{ 2 } } \right] =0$$
$$\Rightarrow a=1$$ and $$\displaystyle 1-\left( \frac { a-2 }{ 2 } \right) \sin { \frac { x }{ 2 } } =0$$
$$\Rightarrow a=1$$ and $$\displaystyle \sin { \left( \frac { x }{ 2 } \right) } =\frac { 2 }{ a-2 } $$
$$\Rightarrow a=1$$ and $$\displaystyle \left| \frac { 2 }{ a-2 } \right| \le 1$$
$$\Rightarrow a=1$$ and $$\left| a-2 \right| \ge 2$$
$$\Rightarrow a=1$$ and $$a-2\ge 2$$ or $$a-2\le -2$$
$$\Rightarrow a=1$$ and $$a\ge 4$$ or $$a\le 0$$
$$\Rightarrow a=1$$ and $$a\in (-\infty ,0]\cup [4,\infty )$$
Therefore, $$a\in (-\infty ,0]\cup \left\{ 1 \right\} \cup [4,\infty )$$
Question 4
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The curve which passes through $$(1, 2)$$ and whose tangent at any point has a slope that is half of slope of the line joining origin to the point of contact, is -

A
A rectangle hyperbola

B
A circle

C
A parabola

D
A straight line through origin

E
Answer required

Solution

Let the arbitrary point be $$x,y.$$
Now the slope of the line joining the origin and the point is
$$=\dfrac{y-0}{x-0}$$
$$=\dfrac{y}{x}$$.
The equation of the curve
$$y=f(x)$$.
Now slope of the tangent
$$=\dfrac{dy}{dx}$$
$$=\dfrac{y}{2x}$$ ... as per the given condition.
Hence
$$2\dfrac{dy}{y}=\dfrac{dx}{x}$$
Integrating both sides we get
$$2\ln y=\ln x+\ln c$$
Now $$y_{x=1}=2$$
Hence
$$2\ln (2)=\ln (1)+\ln (c)$$
Or
$$c=2^{2}=4$$
Hence
$$2\ln y=\ln x+\ln 4$$
Or
$$\ln (y^{2})=\ln 4x$$
Or
$$y^{2}=4x$$ is the required equation.
This is an equation of parabola.
Question 5
1 / -0

The lines tangent to the curve $$x^3-y^3+x^2y-yx^2+3x-2y=0$$ and $$x^5-y^4+2x+3y=0$$ at the origin intersect at an angle $$\theta$$ equal to

A
$$\displaystyle\frac{\pi}{6}$$

B
$$\displaystyle\frac{\pi}{4}$$

C
$$\displaystyle\frac{\pi}{3}$$

D
$$\displaystyle\frac{\pi}{2}$$

Solution

Given equation of curve $$x^3-y^3+x^2y-yx^2+3x-2y=0$$
Differentiating we get
$$\displaystyle 3{ x }^{ 2 }-3{ y }^{ 2 }\frac { dy }{ dx } +x^{ 2 }\frac { dy }{ dx } +2xy-2xy-x^{ 2 }\frac { dy }{ dx } +3-2\frac { dy }{ dx } =0$$

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

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

Slope of tangent to curve at (0,0) is $$m_1=\dfrac{3}{2}$$

Given equation of other curve $$x^5-y^4+2x+3y=0$$
On differentiation ,
$$\displaystyle 5{ x }^{ 4 }-4{ y }^{ 3 }\frac { dy }{ dx } +2+3\frac { dy }{ dx } =0$$

$$\Rightarrow \dfrac { dy }{ dx } =\dfrac { 5{ x }^{ 4 }+2 }{ (4{ y }^{ 3 }-3) } $$
Slope of tangent to the curve at (0,0) is $$m_2=-\dfrac{2}{3}$$

Here, $$m_1m_2=-1$$
Hence, the lines are perpendicular.
Question 6
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A curve $$\displaystyle y=f\left( x \right) ;\left( y>0 \right) $$ passes thorugh $$(1,1)$$ and at point $$\displaystyle P(x,y)$$ tangents cuts x-axis and y-axis at A and B respectively. If P divides AB internally in the ratio $$3 : 2$$, then the value of $$\displaystyle f\left( \frac { 1 }{ 8 } \right) $$ is

A
$$4$$

B
$$\displaystyle \frac { 1 }{ 4 } $$

C
$$\displaystyle 16\sqrt { 2 } $$

D
$$\displaystyle \frac { 1 }{ 16\sqrt { 2 } } $$

Solution

$$\displaystyle \frac { dy }{ dx } =-\frac { k }{ h } =-\frac { 2y }{ 3x } $$
$$\displaystyle \Rightarrow \quad 3\frac { dy }{ y } +2\frac { dx }{ x } =0$$
$$\displaystyle \Rightarrow \quad 3\ln { y }^{ 3 }{ x }^{ 3 }=C$$
$$\displaystyle \because \quad $$passing through $$(1,1)$$
$$\displaystyle \Rightarrow \quad c=0$$
$$\displaystyle \Rightarrow \quad { y }^{ 3 }{ x }^{ 2 }=1$$
at $$\displaystyle x=\frac { 1 }{ 8 }$$
$$ \Rightarrow \quad y=4$$
Question 7
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Directions For Questions

Consider the function $$f(x) = b \ln x - x$$ on the interval $$(0,\infty) $$ where $$b$$ is positive real constant.
On the basis of above information, answer the following questions

...view full instructions

If the line $$x -y = 0$$ is tangent to $$f(x) = b \ln x - x$$, then $$b$$ lies in the interval

A
$$(1, 3)$$

B
$$(0, 1)$$

C
$$(4, 6)$$

D
$$(6, 8)$$

Solution

Given equation of curve is
$$y=b\ln x-x$$
Also, given $$x-y=0$$ is tangent to the curve.
So, let $$P(x_1,x_1)$$ be the point of tangency.
$$x_1=b\ln {x_1}-{x_1}$$
$$2x_1= b\ln {x_1}$$ ....(1)

Slope of tangent to the curve at P$$=\dfrac{b}{x_1}-1$$
Slope of given tangent $$=1$$
$$\Rightarrow \dfrac{b}{x_1}-1=1$$
$$\Rightarrow x_1=\dfrac{b}{2}$$
Substitute this value in (1), we get
$$b=b\ln {\dfrac{b}{2}}$$
$$\Rightarrow \ln {\dfrac{b}{2}}=1$$
$$\Rightarrow \dfrac{b}{2}=e$$
$$\Rightarrow b=2e$$
Since, $$2< e<3$$
$$\Rightarrow 4<b<6$$
Question 8
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If f(x) = $$\dfrac{x}{ sin x}$$ and g(x) = $$\dfrac{x}{tanx}$$ where 0<x $$\leq$$ 1, then in this interval $$f(x)$$ is

A
both f(x) and g(x) are increasing functions

B
both f(x) and g(x) are decreasing functions

C
f(x) is an increasing function

D
g(x) is an increasing function

Solution

We have f(x) $$= \displaystyle \dfrac{x}{ sin x }$$ and g(x) $$=\displaystyle \dfrac{x}{tan x}$$
$$\therefore f'(x) =\displaystyle \dfrac{sin x - x cos x }{sin^2 x }$$ and
$$g'(x)=\displaystyle \dfrac{tan x - x sec^2 x}{tan^2 x}$$
Let $$\Phi (x0) = sin x - x cos x $$ and $$\Psi (x) = tan x - x sec^2 x$$
Then, f'(x) = $$\Phi (x) / sin^2 x $$ and $$g'(x) = \Psi (x)/ tan^2$$
Now, $$\Phi(x) = cos x - cos x + x sin x = x sin x$$
and $$\Psi (x) = sec^2 x - sec^2 x - 2x sec^2 x tan x = -2x sec^2 x tan x$$
For $$0 < x \leq 1$$, we have x > 0 , sin x >0, tan x >0, sec x >0
$$\therefore \Phi (x) = x sin x >x $$ and $$\Psi (x( <0$$ for $$0<x \leq 1$$
$$\Rightarrow \Phi (x) $$ is increasing on (0,1) and $$\Psi (x)$$ is decreasing on (0,1)
$$\Rightarrow \Phi (x) > Phi(0) $$ and $$\Psi (x) < \Psi (0) \Rightarrow \Phi (x) >0$$ and $$\Psi (x) <0$$
$$\therefore f(x) =\Phi (x) /sin^2 x >0$$ & g(x) = $$\Phi (x)/ tan^2 x <0$$
$$\Rightarrow $$ f(x) is increasing on (0,1) and g(x) is decreasing on (0,1)
Question 9
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If the tangent to the curve $$x = a(8 + sin \theta), y = a(1 + cos \theta )$$ at $$\theta = \displaystyle \frac{\pi}{3}$$ makes an angle $$\alpha$$ with x-axis, then $$\alpha$$ is equal to

A
$$\displaystyle \frac{\pi}{3}$$

B
$$\displaystyle \frac{2\pi}{3}$$

C
$$\displaystyle \frac{\pi}{6}$$

D
$$\displaystyle \frac{5\pi}{6}$$

Solution

$$x=a(8+\sin \theta)$$

$$\frac{d x}{d \theta}=a \cos \theta$$

$$y=a(1+\cos \theta)$$
$$\frac{d y}{d \theta}=-a \sin \theta$$

Now,
$$\frac{d y}{d \theta}=-a \sin \theta$$
$$\frac{d x}{d \theta}=a \cos \theta$$
$$\frac{d y}{d x}=-\tan \theta$$
$$\frac{d y}{d x}\left(\operatorname{at} \theta=\frac{\pi}{3}\right)=-\sqrt{3}=m$$

Now,
$$m=\tan \alpha$$
$$\Rightarrow \tan \alpha=-\sqrt{3}$$

Hence,
$$\alpha=\frac{2 \pi}{3}$$
Question 10
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Determine the intervals over which the function is decreasing, increasing, and constant.

A
Increasing $$[3, \infty )$$; Decreasing $$(-\infty , 3]$$

B
Increasing $$(-\infty , 3]$$; Decreasing $$[3, \infty )$$

C
Increasing $$(-\infty , 3]$$; Decreasing $$(-\infty , 3]$$

D
Increasing $$[3, \infty)$$; Decreasing $$[3, \infty )$$

Solution

As we can see from the graph, when $$x$$ is increasing till $$3$$, $$y$$ is also increasing
Further increment in $$x$$ reduces the $$y$$'s values.
It means that upto $$x=3$$, function is increasing and after $$x=3$$ function is decreasing.
Hence, increasing in the interval $$(-\infty,3]$$ ; decreasing in the interval $$[3,\infty)$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 32

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Application of Derivatives Test - 32

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Question 1

$$|a|\,>\,\displaystyle\frac{4}{3}$$

$$|a|\,<\,\displaystyle\frac{4}{3}$$

$$|a|\,>\,1$$

$$|a|\,<\,\displaystyle\frac{1}{3}$$

Solution

Question 2

$$-\frac { 1 }{ 6 }$$

$$\frac { 1 }{ 6 }$$

$$\frac { 1 }{ 12 }$$

$$-\frac { 1 }{ 12 }$$

Solution

Question 3

$$\left( -\infty ,0 \right) \cup \left( 4,\infty \right) $$

$$(-\infty ,0]\cup [4,\infty )$$

$$(-\infty ,0]\cup \left\{ 1 \right\} \cup [4,\infty )$$

None of these

Solution

Question 4

A rectangle hyperbola

A circle

A parabola

A straight line through origin

Answer required

Solution

Question 5

$$\displaystyle\frac{\pi}{6}$$

$$\displaystyle\frac{\pi}{4}$$

$$\displaystyle\frac{\pi}{3}$$

$$\displaystyle\frac{\pi}{2}$$

Solution

Question 6

$$4$$

$$\displaystyle \frac { 1 }{ 4 } $$

$$\displaystyle 16\sqrt { 2 } $$

$$\displaystyle \frac { 1 }{ 16\sqrt { 2 } } $$

Solution

Question 7

$$(1, 3)$$

$$(0, 1)$$

$$(4, 6)$$

$$(6, 8)$$

Solution

Question 8

both f(x) and g(x) are increasing functions

both f(x) and g(x) are decreasing functions

f(x) is an increasing function

g(x) is an increasing function

Solution

Question 9

$$\displaystyle \frac{\pi}{3}$$

$$\displaystyle \frac{2\pi}{3}$$

$$\displaystyle \frac{\pi}{6}$$

$$\displaystyle \frac{5\pi}{6}$$

Solution

Question 10

Increasing $$[3, \infty )$$; Decreasing $$(-\infty , 3]$$

Increasing $$(-\infty , 3]$$; Decreasing $$[3, \infty )$$

Increasing $$(-\infty , 3]$$; Decreasing $$(-\infty , 3]$$

Increasing $$[3, \infty)$$; Decreasing $$[3, \infty )$$

Solution

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