Application of Derivatives Test

Result

Application of Derivatives Test - 15

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The slope of the tangent to the curve $$y=\displaystyle\int_{0}^{x}\dfrac{dt}{1+t^3}$$ at the point where x=1 is

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{3}$$

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

D
1

Solution

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

$$m=\left ( \dfrac{dy}{dx} \right )x=1=\dfrac{1}{1+1}=\dfrac{1}{2}$$
Question 2
1 / -0

Consider the curve $$y = e^{2x}$$.What is the slope of the tangent to the curve at (0, 1) ?

A
0

B
1

C
2

D
4

Solution

We know that slope of tangent to a curve $$y=f(x)$$ at a point $$(x_1,y_1)$$ equals $$f^{'}(x_1)$$
$$\Rightarrow $$ Slope of tangent to the curve $$y=e^{2x}$$ equals $$\cfrac{d(e^{2x})}{dx} =2e^{2x}$$
$$\Rightarrow $$Slope of tangent at $$(0,1)=f^{'}(0)=2$$
Question 3
1 / -0

If a tangent to the curve $$\displaystyle y=6x-{ x }^{ 2 }$$ is parallel to the line $$\displaystyle 4x-2y-1=0$$, then the point of tangency on the curve is:

A
(2, 8)

B
(8, 2)

C
(6, 1)

D
(4, 2)

Solution

$$y=6x-x^2$$
$$\Rightarrow y' =6-2x\ as\ tangent \ is\ parallel \ to \ line 4x-2y-1=0 \ $$ slope $$=2$$
$$\Rightarrow 6-2x=2$$
$$\Rightarrow x=2$$
So $$y=6\cdot 2-2^2=8$$
Hence the point of tangency is $$(2,8)$$
Question 4
1 / -0

The function $$x^{x}$$ is increasing, when

A
$$x > \dfrac {1}{e}$$

B
$$x < \dfrac {1}{e}$$

C
$$x < 0$$

D
For all $$x$$

Solution

Let $$y = x^{x}$$
$$\Rightarrow \dfrac {dy}{dx} = x^{x}(1 + \log x)$$
For increasing function, $$\dfrac {dy}{dx} > 0$$
$$\Rightarrow x^{x} (1 + \log x) > 0$$
$$\Rightarrow 1 + \log x > 0$$
$$\Rightarrow \log_{e} x > \log_{e} \dfrac {1}{e}$$
$$\Rightarrow x > \dfrac {1}{e}$$
Therefore, the function is increasing, when $$x > \dfrac {1}{e}$$.
Question 5
1 / -0

The slope of the tangent to the curve $$y = \int_{0}^{x} \dfrac {dt}{1 + t^{3}}$$ at the point where $$x = 1$$ is

A
$$\dfrac {1}{4}$$

B
$$\dfrac {1}{3}$$

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

D
$$1$$

Solution

$$m=\dfrac {dy}{dx} = \dfrac {1}{1 + x^{3}}$$
$$m = \left (\dfrac {dy}{dx}\right )_ {x = 1} = \dfrac {1}{1 + 1} = \dfrac {1}{2}$$
Question 6
1 / -0

If tangent to the curve $$\displaystyle x={ at }^{ 2 },y=2at$$ is perpendicular to $$x$$-axis, then its point of contact is:

A
$$(a, a)$$

B
$$(0, a)$$

C
$$(0, 0)$$

D
$$(a, 0)$$

Solution

We have, $$x=at^2\Rightarrow \dfrac{dx}{dt}=2at$$
And $$y=2at\Rightarrow \dfrac{dy}{dt}=2a$$
$$\therefore \dfrac{dy}{dx}=\dfrac{1}{t}=\infty$$, for tangent to the curve perpendicular to $$x-$$axis
$$\Rightarrow t = 0$$
so the point of contact is $$(at^2,2at)=(0,0)$$
Question 7
1 / -0

Equation of normal drawn to the graph of the function defined as $$f\left( x \right) =\dfrac { \sin { { x }^{ 2 } } }{ x } ,x\neq 0$$ and $$f\left( 0 \right) =0$$ at the origin is

A
$$x+y=0$$

B
$$x-y=0$$

C
$$y=0$$

D
$$x=0$$

Solution

Differentiating $$f(x)$$ w.r.t $$x$$, we get
$$ f'(x)=\dfrac{x\cos(x^2).2x -\sin x^2.1}{x^2}$$
$$ \Rightarrow f'(x)= \dfrac{2x^2\cos x^2- \sin x^2}{x^2} $$

Find the value of $$f'(x)$$ at $$x=0$$, we get
$$f'(x)$$ is in $$\dfrac{0}{0}$$ indeterminate form

Apply L'Hospital rule
$$ \lim_{x \to0}f'(x)= \dfrac{2x\cos x^2-4x^2\sin x^2}{2x}$$
$$ \Rightarrow \lim_{x\to0} f'(x)=\dfrac{2x\times(\cos x^2-2x^2\sin x^2)}{2x}$$
$$\Rightarrow \lim_{x\to0} \cos x^2- 2x^2\sin x^2$$
$$ \Rightarrow 1 - 0$$
$$ \Rightarrow 1$$

The slope of the tangent at $$x=0$$ on the curve $$ f(x)=\dfrac{sin x^2}{x}$$ is $$1$$.

We know that, product of slope of tangent and normal is $$-1$$.
Therefore, $$ m_1\times m_2 =-1$$
where, $$ m_1 =1$$
Hence, $$ m_2 = -1$$

Slope of normal $$ = -1$$
$$ y =mx+c$$
Here, $$m=-1$$

To find c: at $$x=0$$ we have $$y=f(x)=0$$
Substituting these values in above equation of a line, we get
$$ 0 = -1 \times 0 +c$$
$$ \Rightarrow c=0$$

Therefore, equation of normal is $$y=-x$$ or $$y+x=0$$

Hence, correct answer id option (A)
Question 8
1 / -0

The gradient of the tangent line at the point $$(a cos \alpha, a sin \alpha)$$ to the circle $$x^2 + y^2 = a^2$$, is

A
$$tan (\pi - \alpha)$$

B
$$ tan \alpha$$

C
$$ cot \alpha$$

D
- $$ cot \alpha$$

Solution

Equation of the circle, $$x^2+y^2=a^2$$

At a point $$(a \cos \alpha, a \ sin \alpha)$$

Gradient of radius $$= \dfrac{y_2-y_1}{x_2-x_1}$$

$$= \dfrac{a \ sin \alpha-0}{a \ cos \alpha-0}$$

$$=\ tan \alpha $$

gradient of radius $$\times$$ gradient of tangent = -1

$$\ tan \alpha \times$$ gradient of tangent = -1

$$\Rightarrow $$gradient of tangent$$ = -\ tan \alpha $$

$$\Rightarrow $$gradient of tangent$$ = \ tan (\pi - \alpha)$$
Question 9
1 / -0

Consider the following statements:
1. $$\dfrac {dy}{dx}$$ at a point on the curve gives slope of the tangent at that point.
2. If $$a(t)$$ denotes acceleration of a particle, then $$\displaystyle \int a(t) dt + c$$ give velocity of the particle.
3. If $$s(t)$$ gives displacement of a particle at time $$t$$, then $$\dfrac {ds}{dt}$$ gives its acceleration at that instant.
Which of the above statements is/ are correct?

A
$$1$$ and $$2$$ only

B
$$2$$ only

C
$$1$$ only

D
$$1, 2$$ and $$3$$

Solution

Let's check each statement one by one.

1. $$\dfrac { dy }{ dx }$$ at a particular point will give slope of tangent at that point so this statement is true.

2. If $$ a(t)$$ denote acceleration of a particle, then $$\int a(t)dt+c$$ give velocity of the particle so this statement is true.

3. If $$s(t)$$ gives displacement of a particle at time ‘t’, then $$\dfrac { ds }{ dt }$$ gives velocity not acceleration so this statement is false.

So Statement 1 and 2 are correct.
Question 10
1 / -0

If the product of the slope of tangent to curve at $$(x,y)$$ and its y-co-ordinate is equal to the x-co-ordinate of the point, then it represent.

A
circle

B
parabola

C
ellipse

D
rectangular hyperbola

Solution

Let slope of a tangent at point $$(x,y)$$ to the curve is $$=\cfrac { dy }{ dx } $$
$$\therefore y\cfrac { dy }{ dx } =x$$ (given)
$$\therefore ydy=xdx\quad $$
Integrating both side $$\int { y } dy=\int { x } dx$$
$$\therefore \cfrac { { y }^{ 2 } }{ 2 } =\cfrac { { x }^{ 2 } }{ 2 } +c'$$
$$\therefore { y }^{ 2 }-{ x }^{ 2 }+2c'$$
$$\therefore { y }^{ 2 }-{ x }^{ 2 }=c$$
where $$2c'=c$$
Which is a rectangular hyperbola

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 15

Result

Application of Derivatives Test - 15

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\dfrac{1}{4}$$

$$\dfrac{1}{3}$$

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

1

Solution

Question 2

0

1

2

4

Solution

Question 3

(2, 8)

(8, 2)

(6, 1)

(4, 2)

Solution

Question 4

$$x > \dfrac {1}{e}$$

$$x < \dfrac {1}{e}$$

$$x < 0$$

For all $$x$$

Solution

Question 5

$$\dfrac {1}{4}$$

$$\dfrac {1}{3}$$

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

$$1$$

Solution

Question 6

$$(a, a)$$

$$(0, a)$$

$$(0, 0)$$

$$(a, 0)$$

Solution

Question 7

$$x+y=0$$

$$x-y=0$$

$$y=0$$

$$x=0$$

Solution

Question 8

$$tan (\pi - \alpha)$$

$$ tan \alpha$$

$$ cot \alpha$$

- $$ cot \alpha$$

Solution

Question 9

$$1$$ and $$2$$ only

$$2$$ only

$$1$$ only

$$1, 2$$ and $$3$$

Solution

Question 10

circle

parabola

ellipse

rectangular hyperbola

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.