Application of Derivatives Test

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Application of Derivatives Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=x^{3/2}(3x-10), x\ge-0)$$, then $$f(x)$$ is decreasing in

A
$$(-\infty,0)\cup(0,\infty)$$

B
$$(2,\infty)$$

C
$$(-\infty,-1]\cup[1,\infty)$$

D
$$(-\infty,0]\cup[2,\infty)$$

Solution

R.E.F image

$$ f'(x) = 3x^{3/2}+(3x-10)\dfrac{3}{2}x^{1/2} $$

Putting $$ f'(x) = 0 \Rightarrow 3x^{3/2}+(3x-10)\dfrac{3}{2}x^{1/2} = 0 $$

$$ \Rightarrow 2x^{3/2} = (10-3x)x^{1/2} $$
$$ \Rightarrow 2x = 10-3x $$
$$ \Rightarrow 5x = 10 $$
$$ \Rightarrow 5x = 10 $$

$$ \Rightarrow x = 2 $$

value of $$x$$ Intervals sign of $$f'(x) $$

$$x > 2$$ $$x\epsilon (2,\infty )$$ $$ < 0 $$ decreasing
$$x< 2 $$ $$ x\epsilon (-\infty ,2) $$ $$ > 0 $$ Increasing
Question 2
1 / -0

The numbers of tangent to the curve $$y - 2 = {x^5}$$ which are drawn
from point $$\left( {2,2} \right)$$ is/are

A
$$30$$

B
$$80$$

C
$$20$$

D
$$50$$

Solution

$$y-2=x^5 +2$$
$$y=x^5+2$$
$$y^1=5x^4$$
$$y^1=5(2)^4$$
$$y^1=5\times 16=80$$
Question 3
1 / -0

If the curves $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{4}=1$$ and $$y^{2}=16x$$ intersect at right angles then value of $$a^{2}$$ is

A
$$2$$

B
$$4$$

C
$$\dfrac{8}{3}$$

D
$$\dfrac{16}{3}$$

Solution
Question 4
1 / -0

Tangents are drawn from a point on the circle $$x^2+y^2=25$$ to the ellipse $$9x^2+16y^2-144=0$$ then the angle between the tangents is

A
$$\frac{\pi}{4}$$

B
$$\frac{3\pi}{4}$$

C
$$\frac{\pi}{2}$$

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

Solution
Question 5
1 / -0

Slope of the line $$ \sqrt { { x }^{ 2 }+{ 4y }^{ 2 }-4xy+4 } +x-2y=1$$ equals to

A
$$ \dfrac { 1 }{ 2 }$$

B
$$2$$

C
$$ -\dfrac { 1 }{ 2 }$$

D
$$None\ of\ these$$

Solution
Question 6
1 / -0

The radius of the sphere is measured as $$ \left( {10 \pm 0.02} \right)cm$$. The error in the measurement of its volume is

A
$$25.1 cc$$

B
$$25.21 cc$$

C
$$2.51 cc$$

D
$$251.2 cc$$

Solution

Let $$r$$ be the radius of the sphere.

$$\Rightarrow$$ $$r=10$$

Error in the measurement of radius $$=\Delta r$$

$$\therefore$$ $$\Delta r=0.02\,m$$

$$\Rightarrow$$ Volume of the sphere $$(V)=\dfrac{4}{3}\pi r^3$$

We need to find error in calculating the volume that is $$\Delta V$$

$$\Delta V=\dfrac{dv}{dr}\times \Delta r$$

$$=\dfrac{d\left(\dfrac{4}{3}\pi r^3\right)}{dr}\times \Delta r$$

$$=\dfrac{4}{3}\pi\dfrac{d(r^3)}{dr}\times \Delta r$$

$$=\dfrac{4}{3}\pi(3r^2)\times (0.0.2)$$

$$=4\pi r^2\times 0.02$$

$$=4\times 3.14\times (10)^3\times 0.02$$

$$=251.2\,cm^3$$ i.e. $$251.2\,cc$$
Question 7
1 / -0

If the normal to the curve $$y=f\left( x \right)$$ at $${(3,4)}$$ makes angle $$\dfrac {3\pi}{4}$$ with $$\bar {OX}$$ then $$f^{ 1 }\left( 3 \right)=$$

A
$$-1$$

B
$$1$$

C
$${(-3/4)}$$

D
$${(4/3)}$$

Solution

$$\because y=f\left(x\right)$$
$${y}^{‘}={f}^{‘}\left(x\right)$$
Slope of tangent $$={f}^{‘}\left(x\right)$$
Slope of normal $$=-\dfrac{1}{{f}^{‘}\left(x\right)}$$
$$\Rightarrow \tan{\dfrac{3\pi}{4}}=-\dfrac{1}{{f}^{‘}\left(3\right)}$$
$$\Rightarrow -1=-\dfrac{1}{{f}^{‘}\left(3\right)}$$
$$\Rightarrow \boxed{{f}^{‘}\left(3\right)}=1$$
Question 8
1 / -0

Area of the triangle formed by the tangent, normal to the curve $$x^{2}/a^{2}+y^{2}/b^{2}=1$$ at the point $$(a/\sqrt{2} , b/\sqrt{2})$$ and the $$x-$$axis is

A
$$\dfrac{ab}{4}\sqrt{a^{2}+b^{2}}$$

B
$$4ab$$

C
$$\dfrac{b}{4a}({a^{2}+b^{2}})$$

D
$$none$$

Solution
Question 9
1 / -0

The inclination of the tangent at $$\theta = \dfrac {\pi}{3}$$ on the curve $$x = a(\theta + \sin \theta), y = a(1 + \cos \theta)$$ is

A
$$\dfrac {\pi}{3}$$

B
$$\dfrac {\pi}{6}$$

C
$$\dfrac {2\pi}{3}$$

D
$$\dfrac {5\pi}{6}$$

Solution

For the curve ,

$$\displaystyle \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{\dfrac{d(a(1+cos\theta))}{d\theta}}{\dfrac{d(a(\theta+sin\theta))}{d\theta}} = \dfrac{-asin\theta}{a+acos\theta} = \dfrac{-sin\theta}{1+cos\theta}$$

So, If the inclination of tangent at $$\theta = \dfrac{\pi}{3}$$ be $$y$$ ,

Then, $$\tan(y) =$$ $$\dfrac{-sin(\dfrac{\pi}{3})}{1+cos\dfrac{\pi}{3}} = \dfrac{-1}{\sqrt{3}}$$

Thus,$$ y = \pi-\dfrac{\pi}{6}= \dfrac{5\pi}{6} $$
Question 10
1 / -0

The greatest slope among the lines represented by the equation $$4x^2 - y^2 + 2y - 1 = 0 $$ is -

A
$$-3$$

B
$$-2$$

C
$$2$$

D
$$3$$

Solution

$$4{x}^{2} - {y}^{2} + 2y - 1 = 0$$
$${y}^{2} - 2y + 1 = 4{x}^{2}$$
$${\left( y - 1 \right)}^{2} = {\left( 2x \right)}^{2}$$
$$y - 1 = \pm 2x$$
$$y = 2x + 1 \text{ or } y = -2x + 1$$
Comparing the above equations with $$y = mx + c$$, we have
Slope $$\left( m \right) = 2, -2$$
Hence the greatest slope will be $$2$$.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 43

Result

Application of Derivatives Test - 43

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SHARING IS CARING

Question 1

$$(-\infty,0)\cup(0,\infty)$$

$$(2,\infty)$$

$$(-\infty,-1]\cup[1,\infty)$$

$$(-\infty,0]\cup[2,\infty)$$

Solution

Question 2

$$30$$

$$80$$

$$20$$

$$50$$

Solution

Question 3

$$2$$

$$4$$

$$\dfrac{8}{3}$$

$$\dfrac{16}{3}$$

Solution

Question 4

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

$$\frac{3\pi}{4}$$

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

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

Solution

Question 5

$$ \dfrac { 1 }{ 2 }$$

$$2$$

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

$$None\ of\ these$$

Solution

Question 6

$$25.1 cc$$

$$25.21 cc$$

$$2.51 cc$$

$$251.2 cc$$

Solution

Question 7

$$-1$$

$$1$$

$${(-3/4)}$$

$${(4/3)}$$

Solution

Question 8

$$\dfrac{ab}{4}\sqrt{a^{2}+b^{2}}$$

$$4ab$$

$$\dfrac{b}{4a}({a^{2}+b^{2}})$$

$$none$$

Solution

Question 9

$$\dfrac {\pi}{3}$$

$$\dfrac {\pi}{6}$$

$$\dfrac {2\pi}{3}$$

$$\dfrac {5\pi}{6}$$

Solution

Question 10

$$-3$$

$$-2$$

$$2$$

$$3$$

Solution

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