Application of Derivatives Test

Result

Application of Derivatives Test - 19

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

For the parabola $$y^{2}=8x$$, tangent and normal are drawn at $$P(2, 4)$$ which meet the axis of the parabola in $$A$$ and $$B$$, then the length of the diameter of the circle through $$A, P, B$$ is

A
$$2$$

B
$$4$$

C
$$8$$

D
$$6$$

Solution

Considering the equation of the parabola
$$y^{2}=8x$$
The axis of symmetry is the positive $$x-$$axis.
By, differentiating with respect to $$x$$, we get
$$2y.y'=8$$
$$y'=\dfrac{4}{y}$$
Hence, $$y'_{(2,4)}=1$$
Thus slope of the tangent at the point of contact $$P(2,4)$$ is $$1$$ while that of normal is $$-1$$.
Hence, equation of the tangent will be
$$\dfrac{y-4}{x-2}=1$$
$$-x+y=2$$ ...(i)
Hence, the tangent meets the x axis at $$(-2,0)$$
Therefore, $$A=(-2,0)$$.

Equation of normal will be
$$\dfrac{y-4}{x-2}=-1$$
$$x+y=6$$ ...(ii)
Hence the normal meets the x axis at $$(6,0)$$.
Thus, $$B=(6,0)$$

Therefore the three points through which the circle passes are
$$(-2,0),(6,0),(2,4)$$
Now let the equation of the circle be
$$(x-h)^{2}+(y-k)^{2}=r^{2}$$
Therefore
$$(-2-h)^{2}+k^{2}=r^{2}$$
$$\rightarrow (2+h)^{2}+k^{2}=r^{2}$$
$$(6-h)^{2}+k^{2}=r^{2}$$
Subtracting {ii} from {i}, we get
$$2h-4=0$$
$$h=2$$
Therefore the equation of the circle reduces to
$$(x-2)^{2}+(y-k)^{2}=r^{2}$$
Now
$$(2-2)^{2}+(4-k)^{2}=r^{2}$$
$$\rightarrow (4-k)^{2}=r^{2}$$
$$(6-2)^{2}+k^{2}=r^{2}$$
$$\rightarrow 16+k^{2}=r^{2}$$
Subtracting i from ii, we get
$$16+k^{2}-(k-4)^{2}=0$$
$$16+(2k-4)(4)=0$$
$$4+2k-4=0$$
$$k=0$$
Thus the centre of the circle lies at $$C=(2,0)$$.
Hence the radius of the circle is
$$CA=CP=CB$$
Now
$$CA=\sqrt{(2-(-2))^{2}+0^{2}}$$
$$=2-(-2)$$
$$=4$$
$$=r$$

Hence, the diameter of the circle is $$8$$ units.
Question 2
1 / -0

I. lf $$f'(\mathrm{a})>0$$ then $$\mathrm{f}$$ is increasing at $$\mathrm{x}=\mathrm{a}$$
II: If f is increasing at $$\mathrm{x}=\mathrm{a}$$ then $$f'(\mathrm{a})$$ need not to be positive

A
only I

B
only II

C
both I and II

D
neither I nor II

Solution

$$f(x)$$ is increasing when $$f'(x)>0$$
Question 3
1 / -0

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

A
(1, 2)

B
(2, 3)

C
(1, 3)

D
(0, 2)

Solution

$$f'(x)=e^x-e^{-x}$$
$$f'(x)=0$$ at $$x=0$$
$$f(0)=2$$
Question 4
1 / -0

$$\mathrm{f}(\mathrm{x})=(\sin^{-1}\mathrm{x})^{2}+(\cos^{-1}\mathrm{x})^{2}$$ is stationary at

A
$$\displaystyle \mathrm{x}=\frac{1}{\sqrt{2}}$$

B
$$\displaystyle \mathrm{x}=\frac{\pi}{4}$$

C
$$\mathrm{x} = 1$$

D
$$\mathrm{x} = 0$$

Solution

$$f'(x)=2 sin^{-1}(x)\dfrac {1}{\sqrt {1-x^2}}+2 cos^{-1}(x)(\dfrac {-1}{\sqrt {1-x^2}})$$
$$=2\dfrac {1}{\sqrt{1-x^2}}(sin^{-1}(x)-cos^{-1}(x))$$
$$f'(x)=0$$ when $$x=\dfrac {1}{\sqrt 2}$$
Question 5
1 / -0

The arrangment of the slopes of the normals to the curve $$y=e^{\log(cosx)}$$ in the ascending order at the points given below.
$$A) \displaystyle x=\frac{\pi}{6}, B) \displaystyle x=\frac{7\pi}{4}, C)x=\frac{11\pi}{6}, D)x=\frac{\pi}{3}$$

A
$$C, B, D, A$$

B
$$B, C, A, D$$

C
$$A, D, C, B$$

D
$$D, A, C, B$$

Solution

We have, $$y = e^{\log \cos x} =(\cos x)^{\log e}[\because a^{\log_bc} =c^{\log_ba}] =\cos x$$
$$\Rightarrow \dfrac{dy}{dx} =-\sin x=m$$(say)
Thus slope of normal to the given curve at any point is,
$$m'=- \dfrac{1}{m}=\dfrac{1}{\sin x}$$
Now, $$m'_{\frac{\pi}{6}}=2, m'_{\frac{7\pi}{4}}=-\sqrt{2}, m'_{\frac{\pi}{6}}=-2,m'_{\frac{\pi}{6}}=\dfrac{\sqrt{3}}{2}$$
Clearly ascending order is, $$C,B,D,A$$
Note: Given function is not defined where $$\cos x<0$$
Question 6
1 / -0

Assertion(A): The tangent to the curve $$y=x^{3}-x^{2}-x+2$$ at (1, 1) is parallel to the x axis.
Reason(R): The slope of the tangent to the above curve at (1, 1) is zero.

A
Both A and R are true R is the correct explanation of A

B
Both A and R are true but R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

$$y^{1}=3x^{2}-2x-1$$
$$y^{1}|_{x=1}=0=m$$
$$m=0$$
$$\therefore$$ the curve is parallel to x-axis.
Question 7
1 / -0

Assertion A: The curves $$x^{2}=y,\ x^{2}=-y$$ touch each other at (0, 0).
Reason R: The slopes of the tangents at (0, 0) for both the curves are equal.

A
Both A and R are true R is the correct explanation of A

B
Both A and R are true but R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

$$y'=2x$$
$$y'|_{x=0}=0=m_{1}$$
$$y'=-2x$$
$$y'|_{x=0}=0=m_{2}$$
$$m_{1}=m_{2}$$
$$\therefore$$ both the curve touches each other
Question 8
1 / -0

Match the points on the curve $$2y^{2}=x+1$$ with the slope of normals at those points and choose
the correct answer.
Point
Slope of normal
I : $$(7, 2)$$

$$a){-4\sqrt{2}}$$

II: $$(0, \displaystyle \frac{1}{\sqrt{2}})$$

$$b) -8$$
III : $$(1, 1)$$
$$c) -4$$
IV: $$(3, \sqrt{2})$$

$$d){-2\sqrt{2}}$$

A
$$i-b,ii- d,iii- c,iv- a$$

B
$$i-b,ii- a,iii- d,iv- c$$

C
$$i-b,ii- c,iii- d,iv- a$$

D
$$i-b,ii- d,iii- a,iv- c$$

Solution

$$\displaystyle y'=\frac{1}{4y}=m$$

Slope of normal $$m'=-4y$$

For given points slope is
(1) $$m'_{1}=-4\times 2 ==-8$$
(2) $$m'_{2}=-4\times \dfrac{1}{\sqrt 2}=-2\sqrt{2}$$
(3) $$m'_{3}=-4\times 1=-4$$
(4) $$m'_{4}=-4\times {\sqrt 2}=-4\sqrt{2}$$

Option A is correct answer
Question 9
1 / -0

P(1, 1) is a point on the parabola $$y=x^{2}$$ whose vertex is A. The point on the curve at which the tangent drawn is parallel to the chord $$\overline{AP}$$ is

A
$$(\displaystyle \frac{1}{2}, \frac{1}{4})$$

B
$$(\displaystyle \frac{-1}{2}, \frac{1}{4})$$

C
$$(2, 4)$$

D
$$(4, 2)$$

Solution

$$tan \theta =\dfrac{(1-0)}{(1-0)}=1$$
$${y}'=2x$$
$$2x=1$$
$$x=\dfrac{1}{2}, y=\dfrac{1}{4}$$
Question 10
1 / -0

lf the parametric equation of a curve given by $$x=e^{t}\cos t,\ y=e^{t}\sin t$$, then the tangent to the curve at the point $$t=\dfrac{\pi}{4}$$ makes with axis of $$x$$ the angle.

A
$$0$$

B
$$\dfrac{\pi}{4}$$

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

D
$$\dfrac{\pi}{2}$$

Solution

$$at\ t={\pi}/{4}$$
$$y=e^{t}sint$$
$$\dfrac{dy}{dl}=e^{t}(sin\ t+cos\ t)$$
$$\dfrac{dx}{dl}=d^{t}(cos\ t+sin\ t)$$
$$\dfrac{dy}{dx}=\dfrac{sin \ t+cos\ t}{cos\ t-sin\ t}$$
$$= \alpha$$
$$\therefore tan^{-1}(\infty)={\pi}/{2}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Point	Slope of normal
I : $$(7, 2)$$	$$a){-4\sqrt{2}}$$
II: $$(0, \displaystyle \frac{1}{\sqrt{2}})$$	$$b) -8$$
III : $$(1, 1)$$	$$c) -4$$
IV: $$(3, \sqrt{2})$$	$$d){-2\sqrt{2}}$$

Application of Derivatives Test - 19

Result

Application of Derivatives Test - 19

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2$$

$$4$$

$$8$$

$$6$$

Solution

Question 2

only I

only II

both I and II

neither I nor II

Solution

Question 3

(1, 2)

(2, 3)

(1, 3)

(0, 2)

Solution

Question 4

$$\displaystyle \mathrm{x}=\frac{1}{\sqrt{2}}$$

$$\displaystyle \mathrm{x}=\frac{\pi}{4}$$

$$\mathrm{x} = 1$$

$$\mathrm{x} = 0$$

Solution

Question 5

$$C, B, D, A$$

$$B, C, A, D$$

$$A, D, C, B$$

$$D, A, C, B$$

Solution

Question 6

Both A and R are true R is the correct explanation of A

Both A and R are true but R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 7

Both A and R are true R is the correct explanation of A

Both A and R are true but R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 8

$$i-b,ii- d,iii- c,iv- a$$

$$i-b,ii- a,iii- d,iv- c$$

$$i-b,ii- c,iii- d,iv- a$$

$$i-b,ii- d,iii- a,iv- c$$

Solution

Question 9

$$(\displaystyle \frac{1}{2}, \frac{1}{4})$$

$$(\displaystyle \frac{-1}{2}, \frac{1}{4})$$

$$(2, 4)$$

$$(4, 2)$$

Solution

Question 10

$$0$$

$$\dfrac{\pi}{4}$$

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

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

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.