Application of Derivatives Test

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Application of Derivatives Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

Find the equation of a line passing through $$(-2,3)$$ and parallel to tangent at origin for the circle $$\displaystyle x^{2}+y^{2}+x-y=0$$

A
$$x -2 y + 5 = 0$$

B
$$x -4 y + 3 = 0$$

C
$$x - y + 5 = 0$$

D
$$2x - y + 6 = 0$$

Solution

Given,$$x^{2}+y^{2}+x-y=0$$
Differentiating w.r.t $$x$$, we get
$$2x+2y\cfrac{dy}{dx}+1-\cfrac{dy}{dx}=0$$
$$\Rightarrow \cfrac{dy}{dx}=\cfrac{1+2x}{1-2y}$$
Thus slope of the tangent at origin is, $$m=\left(\cfrac{dy}{dx}\right)_{(0,0)}=1$$
Hence required line through $$(-2,3)$$ is, $$(y-3)=1(x+2)\Rightarrow x-y+5=0$$
Question 2
1 / -0

Stationary point of $$\displaystyle \mathrm{y}=\frac{\log \mathrm{x}}{\mathrm{x}}(\mathrm{x}>0)$$ is

A
$$(1, 0)$$

B
$$(\displaystyle \mathrm{e},\frac{1}{\mathrm{e}})$$

C
$$(\displaystyle \frac{1}{\mathrm{e}},-\mathrm{e})$$

D
$$(\displaystyle \frac{1}{\mathrm{e}'}\frac{1}{\mathrm{e}})$$

Solution

$$y'=\frac {1}{x^2}-\frac {log x}{x^2}$$ $$=(1-log x)\frac {1}{x^2}$$
and $$y'=0$$ for stationary point
$$\therefore x=e$$
Question 3
1 / -0

I: lf $$\mathrm{f}'(\mathrm{a})<0$$ then the function $$\mathrm{f}$$ is decreasing at $$\mathrm{x}=\mathrm{a}$$
II: lf $$\mathrm{f}$$ is decreasing at $$\mathrm{x}=\mathrm{a}$$ then $$\mathrm{f}'(\mathrm{a})<0$$
Which of the above statements are true ?

A
onlyI

B
only II

C
both I and II

D
neither I nor II

Solution

(I) $$f'(a)<0$$ Which implies f is decreasing at $$x=0$$
(II) If f is decresing at $$x=a$$
which implies $$f'(a)<0$$
has slope is negative for decreasing function
Question 4
1 / -0

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

A
$$\displaystyle \frac{1}{2}$$

B
$$1$$

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

D
none of these

Solution

$$y=\int _{ 0 }^{ x }{ \cfrac { dx }{ 1+{ x }^{ 3 } } } \\ \cfrac { dy }{ dx } =\cfrac { 1 }{ 1+{ x }^{ 3 } } =\cfrac { 1 }{ 2 } $$
Question 5
1 / -0

The value of $$\mathrm{x}$$ at which $$\mathrm{f}(\mathrm{x})=$$ cosx is stationary are given by

A
$$\mathrm{n}\pi,\ \forall n\in Z$$

B
$$(2\displaystyle \mathrm{n}+1)\frac{\pi}{2},\ \forall n\in Z$$

C
$$\displaystyle \frac{\mathrm{n}\pi}{4},\ \forall n\in Z$$

D
$$\displaystyle \frac{\mathrm{n}\pi}{2},\ \forall n\in Z$$

Solution

$$f'(x)=\sin x$$ ($$f'(x)=0$$ means stationary)
$$f'(x)=0$$
when $$x=n\pi$$
Question 6
1 / -0

The number of stationary points of $$\mathrm{f}(\mathrm{x})=\cos \mathrm{x}$$ in $$[0, 2{\pi}]$$ are

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$f'(x)=\sin x$$
$$f'(x)=0$$ when $$x=0, \pi, 2\pi$$
Question 7
1 / -0

The curve $$\displaystyle y-e^{xy}+x=0$$ has a vertical tangent at

A
$$(1, 1)$$

B
$$(0, 1)$$

C
$$(1, 0)$$

D
$$(0,0)$$

Solution

Given, $$\displaystyle y-e^{xy}+x=0$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}-e^{xy}(y+x\cfrac{dy}{dx})+1=0$$
$$\Rightarrow \cfrac{dy}{dx}=\cfrac{1-ye^{xy}}{xe^{xy}-1}$$
Thus for vertical tangent, $$\cfrac{dy}{dx}=\infty$$
$$\Rightarrow xe^{xy}-1=0$$
Hence required point is $$(1,0)$$
Question 8
1 / -0

The stationary point of $$\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-10\mathrm{x}+43$$ is

A
(5, 18)

B
(18, 5)

C
(5, 5)

D
(5, 15)

Solution

$$f'(x)=2x-10$$
$$f'(x)=0$$ at $$x=5$$
$$f(5)=25-50+43$$
$$=18$$
So stationary point $$=(5, 18)$$
Question 9
1 / -0

The point on the curve $$\displaystyle y=x^{2}-3x+2$$ at which the tangent is perpendicular to the line $$y = x$$ is -

A
$$(0, 2)$$

B
$$(1, 0)$$

C
$$(-1, 6)$$

D
$$(2, -2)$$

Solution

Let the point be $$P(a,b)$$
Now the given curve is $$y=x^2-3x+2$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}=2x-3$$
Thus slope of tangent at P is $$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=2a-3$$
But given the tangent is perpendicular to line $$y=x$$
$$\Rightarrow 2a-3=-1\Rightarrow a=1$$
Also the point P lies on the given curve,
$$b=a^2-3a+2=0$$
Therefore, the point P is $$(1,0)$$
Hence, option 'B' is correct.
Question 10
1 / -0

If tangent to curve at a point is perpendicular to $$x$$ - axis then at that point -

A
$$\displaystyle \frac{dy}{dx}=0 $$

B
$$\displaystyle \frac{dx}{dy}=0 $$

C
$$\displaystyle \frac{dy}{dx}=1 $$

D
$$\displaystyle \frac{dy}{dx}=-1 $$

Solution

We know slope of tangent to the any curve $$'y'$$ is $$m=\cfrac{dy}{dx}$$
Now if line is perpendicular to x-axis this means its slope is $$\infty$$
$$\Rightarrow \cfrac{dy}{dx} \to \infty \Rightarrow \cfrac{dx}{dy}=0$$
Hence, option 'C' is correct.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 13

Result

Application of Derivatives Test - 13

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SHARING IS CARING

Question 1

$$x -2 y + 5 = 0$$

$$x -4 y + 3 = 0$$

$$x - y + 5 = 0$$

$$2x - y + 6 = 0$$

Solution

Question 2

$$(1, 0)$$

$$(\displaystyle \mathrm{e},\frac{1}{\mathrm{e}})$$

$$(\displaystyle \frac{1}{\mathrm{e}},-\mathrm{e})$$

$$(\displaystyle \frac{1}{\mathrm{e}'}\frac{1}{\mathrm{e}})$$

Solution

Question 3

onlyI

only II

both I and II

neither I nor II

Solution

Question 4

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

$$1$$

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

none of these

Solution

Question 5

$$\mathrm{n}\pi,\ \forall n\in Z$$

$$(2\displaystyle \mathrm{n}+1)\frac{\pi}{2},\ \forall n\in Z$$

$$\displaystyle \frac{\mathrm{n}\pi}{4},\ \forall n\in Z$$

$$\displaystyle \frac{\mathrm{n}\pi}{2},\ \forall n\in Z$$

Solution

Question 6

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 7

$$(1, 1)$$

$$(0, 1)$$

$$(1, 0)$$

$$(0,0)$$

Solution

Question 8

(5, 18)

(18, 5)

(5, 5)

(5, 15)

Solution

Question 9

$$(0, 2)$$

$$(1, 0)$$

$$(-1, 6)$$

$$(2, -2)$$

Solution

Question 10

$$\displaystyle \frac{dy}{dx}=0 $$

$$\displaystyle \frac{dx}{dy}=0 $$

$$\displaystyle \frac{dy}{dx}=1 $$

$$\displaystyle \frac{dy}{dx}=-1 $$

Solution

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