Application of Derivatives Test

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Application of Derivatives Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

Identify the false statement:

A
All the stationary numbers are critical numbers

B
At the stationary point the first derivative is zero

C
At critical numbers the first derivative need not exist

D
All the critical numbers are stationary numbers

Solution

The numbers where $$f'\left( x \right) =0$$ are called stationary numbers
The numbers where $$f'\left( x \right) =0$$ or $$f'\left( x \right) $$ doesnt exist are called critical numbers
Therefore all critical numbers are not stationary numbers
So option $$D$$ is false statement
Question 2
1 / -0

The percentage error in the $$11^{th}$$ root of the number $$28$$ is approximately ____________ times the percentage error in $$28$$

A
$$\dfrac { 1 }{ 28 } $$

B
$$\dfrac { 1 }{ 11 } $$

C
$$11$$

D
$$28$$

Solution

Suppose $$y = x^{11}$$
$$dy = 11x^{10}dx$$
Dividing by y on both sides, we have $$ \cfrac{\Delta y}{y} = \cfrac{11x^{10}\Delta x}{y} = \cfrac{11x^{10}\Delta x}{x^{11}}$$
$$\therefore \cfrac{\Delta y}{y} = \cfrac{11 \Delta x}{x}$$

Here, when $$y = 28$$, $$x$$ will be the $$11^{th}$$ root of $$28$$.
$$\therefore \cfrac{\Delta (28)}{28} = \cfrac{11 \Delta (\sqrt[11]{28})}{\sqrt[11]{28}}$$
Hence, the percentage error in the $$11$$th root of $$28$$ would approximately be $$\cfrac{1}{11}$$ times the error in $$28$$
Question 3
1 / -0

The slope of the normal to the curve $$y = 3x^2$$ at the point whose $$x$$-coordinate $$2$$ is

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

B
$$\dfrac{1}{14}$$

C
$$\dfrac{-1}{12}$$

D
$$\dfrac{1}{12}$$

Solution

$$y=3x^{2}$$. Differentiating the equation of the curve with respect to $$x$$.
$$\dfrac{dy}{dx}=6x$$
Now
Slope$$_{x=2}=\dfrac{dy}{dx}|_{x=2}$$
$$=6(2)$$
$$=12$$.
Hence the slope of the tangent at $$x=2$$ is $$12$$. Since the normal is perpendicular to the tangent, therefore, the
Slope$$_\text{normal}\times$$Slope$$_\text{tangent}=-1$$ or Slope$$_\text{normal}=\dfrac{-1}{\text{Slope}_\text{tangent}}=\dfrac{-1}{12}$$
Question 4
1 / -0

The slope of the tangent to the curve $$y=3{ x }^{ 2 }+3\sin { x } $$ at $$x=0$$ is

A
$$3$$

B
$$2$$

C
$$1$$

D
$$-1$$

Solution

$$y=3x^2+3\sin x$$
$$\Rightarrow \dfrac{dy}{dx}=6x+3\cos x$$
Thus slope of tangent to the given curve at $$x=0$$ is
$$=\dfrac{dy}{dx}\bigg|_{x=0}=(6x+3\cos x)\bigg|_{x=0}=6(0)+3\cos 0=0+3\cdot 1=3$$
Question 5
1 / -0

Consider the following statements:
$$1. y = \dfrac {e^{x} + e^{-x}}{2}$$ is an increasing function on $$[0, \infty)$$.
$$2. y = \dfrac {e^{x} - e^{-x}}{2}$$ is an increasing function on $$(-\infty, \infty)$$.
Which of the above statements is/are correct?

A
$$1$$ only

B
$$2$$ only

C
Both $$1$$ and $$2$$

D
Neither $$1$$ nor $$2$$

Solution

$$y=\dfrac { { e }^{ x }+{ e }^{ -x } }{ 2 } =\cosh x$$
$$y=\dfrac { { e }^{ x }-{ e }^{ -x } }{ 2 } =\sinh x$$
Both are increasing functions
Hence, both the statements are correct.
Question 6
1 / -0

What is the slope of the tangent to the curve $$ x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$$ at t = 2 ?

A
$$\dfrac{7}{6}$$

B
$$\dfrac{6}{7}$$

C
1

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

Solution

Differentiating $$x,y$$ w.r.t $$t,$$ we get
$$\dfrac { dy }{ dt } =\dfrac { d }{ dt } (2t^2-2t-5)=4t-2\\ \dfrac { dx }{ dt } =\dfrac { d }{ dt } (t^2+3t-8)=2t+3\\$$
$$\therefore \dfrac { dy }{ dx } =\dfrac { \left( \dfrac { dy }{ dt } \right) }{ \left( \dfrac { dx }{ dt } \right) } =\dfrac { 4t-2 }{ 2t+3 } $$
At $$t=2$$ derivative will be $$\dfrac {dy}{dx}=\dfrac {4\times 2-2}{2\times 2+3}=\dfrac {6}{7}$$
Thus, the slope of tangent is $$\dfrac{6}{7}$$.
Question 7
1 / -0

If the tangent to the function $$y = f(x)$$ at $$(3, 4)$$ makes an angle of $$\dfrac {3\pi}{4}$$ with the positive direction of x-axis in anticlockwise direction then $$f'(3)$$ is

A
$$-1$$

B
$$1$$

C
$$\dfrac {1}{\sqrt {3}}$$

D
$$\sqrt {3}$$

Solution

Given $$y=f(x)$$
Differentiating w.r.t x, we get
$$y' = f'(x)$$ which is the slope of the tangent

$$f'(3) = \tan \dfrac {3\pi}{4} = -\tan \dfrac {\pi}{4} = -1$$
Hence, $$f'(3)=-1$$.
Question 8
1 / -0

How many tangents are parallel to x-axis for the curve $$ y = x^2 - 4x + 3$$ ?

A
1

B
2

C
3

D
No tangent is parallel to x-axis.

Solution

Slope of the tangent is $$\dfrac {dy} {dx}$$

So, $$\dfrac{dy}{dx}=2x-4$$

Tangent parallel to x-axis. So, slope of tangent should be $$0.$$

$$\Rightarrow 2x-4=0$$

$$\Rightarrow x=2$$ is the only point where slope is parallel to x-axis.

So, only 1 tangent exists.
Question 9
1 / -0

The slope of the tangent to the curve given by $$x = 1 - \cos { \theta }$$, $$y = \theta -\sin { \theta } $$ at $$\theta = \dfrac { \pi }{ 2 } $$ is

A
$$0$$

B
$$-1$$

C
$$1$$

D
Not defined

Solution

$$y=\theta -\sin\theta$$
$$\Longrightarrow \dfrac { dy }{ d\theta } =1-\cos\theta$$
$$x=1-\cos\theta$$
$$\Longrightarrow \dfrac { dx }{ d\theta } =sin\theta $$
Slope of the tangent is $$\dfrac{dy}{dx} $$
$$\Longrightarrow \dfrac{dy}{dx} =\dfrac { 1-\cos\theta}{\sin\theta}$$

At $$\theta =\dfrac { \pi }{ 2 } $$
$$\Longrightarrow \dfrac { dy }{ dx } =\dfrac { 1-\cos\frac { \pi }{ 2 } }{ \sin\frac { \pi }{ 2 } } =1$$
Question 10
1 / -0

Consider the following in respect of the function $$f(x) = \left\{\begin{matrix}2+ x, & x \geq 0\\ 2 - x, & x < 0\end{matrix}\right.$$
1. $$\displaystyle \lim_{x \rightarrow 1} f(x)$$ does not exist.
2. f(x) is differentiable at x = 0.
3. f(x) is continuous at x = 0.
Which of the above statements is/are correct?

A
$$1$$ only

B
$$3$$ only

C
$$2$$ and $$3$$ only

D
$$1$$ and $$3$$ only

Solution

Left hand limit$$ = \lim_{x\to 1^-}{f(x)}$$
$$= \displaystyle \lim_{h\to 0}{f(1-h)}$$
$$=\displaystyle \lim_{h\to 0}2+(1-h)=2+1=3$$
Right hand limit$$ = \displaystyle \lim_{x\to 1^+}{f(x)}$$
$$= \displaystyle \lim_{h\to 0}{f(1+h)}$$
$$=\displaystyle \lim_{h\to 0}2+(1+h)=2+1=3$$
So, the limit exists at x=1

Left hand limit$$ =\displaystyle \lim_{x\to 0^-}{f(x)}$$
$$= \displaystyle \lim_{h\to 0}{f(0-h)}$$
$$=\displaystyle \lim_{h\to 0} 2-(0-h)=2-0=2$$
Right hand limit$$ = \displaystyle \lim_{x\to 0^+}{f(x)}$$
$$= \displaystyle \lim_{h\to 0}{f(0+h)}$$
$$=\displaystyle \lim_{h\to 0}2+(0+h)=2+0=2$$
So, LHL=RHL=f(0)
So, f(x) is continuous at x=0

Right hand limit$$ = \displaystyle \lim_{h\to 0}\dfrac{f(0+h)-f(0)}{h}$$
$$= \displaystyle \lim_{h\to 0}\dfrac{(2+(0+h))-2}{h}$$
$$=\displaystyle \lim_{h\to 0}\dfrac{h}{h}=1$$
Left hand limit$$ = \displaystyle \lim_{h\to 0}\dfrac{f(0-h)-f(0)}{-h}$$
$$= \displaystyle \lim_{h\to 0}\dfrac{(2-(0-h))-(2)}{-h}$$
$$=\displaystyle \lim_{h\to 0}\dfrac{h}{-h}=-1$$
LHL$$\neq$$RHL
So, f(x) is not differentiable at x=0

The answer is option (B)

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 35

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Application of Derivatives Test - 35

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SHARING IS CARING

Question 1

All the stationary numbers are critical numbers

At the stationary point the first derivative is zero

At critical numbers the first derivative need not exist

All the critical numbers are stationary numbers

Solution

Question 2

$$\dfrac { 1 }{ 28 } $$

$$\dfrac { 1 }{ 11 } $$

$$11$$

$$28$$

Solution

Question 3

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

$$\dfrac{1}{14}$$

$$\dfrac{-1}{12}$$

$$\dfrac{1}{12}$$

Solution

Question 4

$$3$$

$$2$$

$$1$$

$$-1$$

Solution

Question 5

$$1$$ only

$$2$$ only

Both $$1$$ and $$2$$

Neither $$1$$ nor $$2$$

Solution

Question 6

$$\dfrac{7}{6}$$

$$\dfrac{6}{7}$$

1

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

Solution

Question 7

$$-1$$

$$1$$

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

$$\sqrt {3}$$

Solution

Question 8

1

2

3

No tangent is parallel to x-axis.

Solution

Question 9

$$0$$

$$-1$$

$$1$$

Not defined

Solution

Question 10

$$1$$ only

$$3$$ only

$$2$$ and $$3$$ only

$$1$$ and $$3$$ only

Solution

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