Application of Derivatives Test

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Application of Derivatives Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The slope of the tangent to the curve $$x={t}^{2}+3t-8$$, $$y=2{t}^{2}-2t-5$$ at the point $$(2,-1)$$ is

A
$$\cfrac{22}{7}$$

B
$$\cfrac{6}{7}$$

C
$$\cfrac{7}{6}$$

D
$$\cfrac{-6}{7}$$

E
answer required

Solution

Slope to any curve is given by $$\dfrac{dy}{dx}$$

Here, $$\cfrac{dx}{dt}=2t+3$$ ...(1)
And $$\cfrac{dy}{dt}=4t-2$$ ...(2)
Dividing (2) by (1), we get $$\cfrac {dy}{dx} = \cfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{4t-2}{2t+3}$$
At point $$(2,-1)$$, $$t=2$$
So slope is given by,
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{4\times 2-2}{2\times 2+3}=\dfrac{6}{7}$$
Question 2
1 / -0

The line $$y=mx+1$$ is a tangent to the curve $${y}{^2}=4x$$, if the value of $$m$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$\cfrac{1}{2}$$

E
answer required

Solution

Lets put $$y=mx+1$$ in $${y}^{2}=4x$$ we get
$${(mx+1)}^{2}=4x$$
$$\Rightarrow {m}^{2}{x}^{2}+1+2mx-4x=0$$
Tangent touches a curve at one point so descriminant of the equation should be zero.
$$\Rightarrow {(2m-4)}^{2}-4\times 1\times {m}^{2}=0$$
$$\Rightarrow 4{m}^{2}+16-16m-4{m}^{2}=0$$
$$\Rightarrow m=1$$
Question 3
1 / -0

The abscissa of the points, where the tangent to curve $$y={x}^{3} - 3{x}^{2} - 9x+5$$ is parallel to x-axis, are

A
$$x=0$$ and $$0$$

B
$$x=1$$ and $$-1$$

C
$$x=1$$ and $$-3$$

D
$$x=-1$$ and $$3$$

Solution

Given, $$y={ x }^{ 3 }-3{ x }^{ 2 }-9x+5$$
$$\Rightarrow \dfrac { dy }{ dx } =3{ x }^{ 2 }-6x-9$$
We know that, this equation gives the slope of the tangent to the curve. The tangent is parallel to x-axis,
$$\therefore \dfrac { dy }{ dx } =0$$
$$\Rightarrow 3{ x }^{ 2 }-6x-9=0$$
$$\Rightarrow x=-1,3$$
Question 4
1 / -0

The points on the curve $$9{y}^{2}={x}^{3}$$, where the normal to the curve makes equal intercepts with the axes are

A
$$\left( 4,\pm \cfrac { 8 }{ 3 } \right) $$

B
$$\left( 4,\cfrac { -8 }{ 3 } \right) $$

C
$$\left( 4,\pm \cfrac { 3 }{ 8 } \right) $$

D
$$\left( \pm 4,\cfrac { 3 }{ 8 } \right) $$

E
answer required

Solution

Let the point on curve be $$(h,k)$$
$$y^2= \dfrac{x^3}{9}\\
2y.\dfrac{dy}{dx}= \dfrac{1}{9} .3 x^2\\
\therefore \dfrac{dy}{dx}= \dfrac{x^2}{6y}$$
$$\dfrac{dy}{dx}$$ gives slope of curve at given point
$$\dfrac{dy}{dx}\;at\;(h,k) = \dfrac{h^2}{6k}$$
but this is the slope of tangent and we need slope of normal
As we know that slope of normal$$\times$$ slope of tangent is $$-1$$
$$\therefore$$ slope of normal is $$\dfrac{-6k}{h^2}$$
Given in question that normal makes equal intercepts on axes which means slope of normal is either $$1$$ or $$-1$$
$$-6k=h^2$$ and $$6k=h^2$$
Now go through the option you will that option A satisfy our equation
Question 5
1 / -0

Angle between $${ y }^{ 2 }=x$$ and $${ x }^{ 2 }=y$$ at the origin is

A
$$2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $$

B
$$\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $$

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

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

Solution

It is clear from the graph that both the curves have a tangent at the coordinate axes, so the angle between the curves is $$\dfrac { \pi }{ 2 } $$.

Alternative
Given curves are $${ y }^{ 2 }=x$$ and $${ x }^{ 2 }=y$$

On differentiating with respect to $$x$$, we get

$$2y\dfrac { dy }{ dx } =1$$ and $$2x=\dfrac { dy }{ dx } $$

$$\Rightarrow \dfrac { dy }{ dx } =\dfrac { 1 }{ 2y } $$ and $$\dfrac { dy }{ dx } =2x$$

At $$\left( 0,0 \right) $$

$${ m }_{ 1 }=\dfrac { dy }{ dx } =\infty $$ and $${ m }_{ 2 }=\dfrac { dy }{ dx } =0$$

$$\therefore \tan { \theta } =\dfrac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } =\infty $$

$$\Rightarrow \theta =\dfrac { \pi }{ 2 } $$
Question 6
1 / -0

If the line $$\alpha\,x+by+c=0$$ is a tangent to the curve $$xy=4$$, then

A
$$a < 0,\,b > 0$$

B
$$a \le o,\,b > 0$$

C
$$a < 0,\,b < 0$$

D
$$a \le 0,\,b < 0$$

Solution

Slope of line $$ax+by+c=0$$ is
$$\Rightarrow\;-\dfrac{a}{b}$$
$$y=\dfrac{4}{x}=1,\,\dfrac{dy}{dx}=-\dfrac{4}{x^2}$$,
$$-\dfrac{a}{b}=-\dfrac{4}{x^2}\Rightarrow\,\dfrac{a}{b}=\dfrac{4}{x^2} > 0$$
$$a < 0,\,b < 0$$
Question 7
1 / -0

Let $$y=e^{x^2}$$ and $$y=e^{x^2}\sin\, x$$ be two given curves. Then, angle between the tangents to the curves at any point their intersection is

A
$$0$$

B
$$\pi$$

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

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

Solution

For intersecting points, $$e^{x^2}=e^{x^2}\sin\,x$$
$$\Rightarrow e^{x^2}(\sin x-1)=0$$
$$\Rightarrow e^{x^2}=0$$ or $$\sin\,x=1$$

But $$e^{x^2}\neq 0\Rightarrow \sin x=1$$
$$\Rightarrow x=\dfrac{\pi}{2}$$

Now,
$$y=e^{x^2}$$

$$\therefore \dfrac{dy}{dx}=e^{x^2}.2x=2xe^{x^2}$$ $$[\because \sin\,x=1, \cos \, x = 0]$$

Since, both the curves has equal slope.
Hence, angle between the tangents at intersecting point is $$0$$.
Question 8
1 / -0

The slope at any point of a curve $$y=f\left( x \right) $$ is given by $$\dfrac { dy }{ dx } =3{ x }^{ 2 }$$ and it passes through $$\left( -1,1 \right) $$. The equation of the curve is

A
$$y={ x }^{ 3 }+2$$

B
$$y=-{ x }^{ 3 }-2$$

C
$$y=3{ x }^{ 3 }+4$$

D
$$y=-{ x }^{ 3 }+2$$

Solution

Given, $$\dfrac { dy }{ dx } =3{ x }^{ 2 }$$
$$\Rightarrow dy=3{ x }^{ 2 }dx$$
On integrating, we get
$$y=\dfrac { 3{ x }^{ 3 } }{ 3 } +c$$
$$\Rightarrow y={ x }^{ 3 }+c$$
It passes through $$\left( -1,1 \right) $$.
$$\therefore 1={ \left( -1 \right) }^{ 3 }+c$$
$$\Rightarrow c=2$$
$$\therefore y={ x }^{ 3 }+2$$
Question 9
1 / -0

Suppose that the equation $$f\left( x \right) ={ x }^{ 2 }+bx+c=0$$ has two distinct real roots $$\alpha $$ and $$\beta $$. The angle between the tangent to the curve $$y=f\left( x \right) $$ at the point $$\left( \dfrac { \alpha +\beta }{ 2 } ,f\left( \dfrac { \alpha +\beta }{ 2 } \right) \right) $$ and the positive direction of the $$x$$-axis is

A
$${ 0 }^{ }$$

B
$${ 30 }^{ }$$

C
$${ 60 }^{ }$$

D
$${ 90 }^{ }$$

Solution

Since, $$\alpha $$ and $$\beta $$ are the roots of
$$f\left( x \right) ={ x }^{ 2 }+bx+c$$
$$\therefore \alpha +\beta =-b$$ and $$\alpha \beta =c$$
$$\therefore f\left( \dfrac { \alpha +\beta }{ 2 } \right) ={ \left( \dfrac { \alpha +\beta }{ 2 } \right) }^{ 2 }+b\left( \dfrac { \alpha +\beta }{ 2 } \right) +c$$
$$={ \left( -\dfrac { b }{ 2 } \right) }^{ 2 }+b\left( -\dfrac { b }{ 2 } \right) +c$$
$$=\dfrac { { b }^{ 2 } }{ 4 } -\dfrac { { b }^{ 2 } }{ 2 } +c=-\dfrac { { b }^{ 2 } }{ 4 } +c$$

Now, $$\dfrac { dy }{ dx } =f^{ \prime }\left( x \right) =2x+b$$

At point, $$\left( \dfrac { \alpha +\beta }{ 2 } ,f\left( \dfrac { \alpha +\beta }{ 2 } \right) \right) $$, i.e., $$\left( -\dfrac { b }{ 2 } ,-\dfrac { { b }^{ 2 } }{ 4 } +c \right) $$,

$$\dfrac { dy }{ dx } =2\left( -\dfrac { b }{ 2 } \right) +b=0$$

Hence, the slope of the tangent to the curve and the positive direction of $$x$$-axis is $${ 0 }^{ }$$.
Question 10
1 / -0

The equation of one of the curves whose slope at any point is equal to $$y+2x$$ is

A
$$y=2(e^x+x-1)$$

B
$$y=2(e^x-x-1)$$

C
$$y=2(e^x-x+1)$$

D
$$y=2(e^x+x+1)$$

Solution

Given, $$\dfrac{dy}{dx}=y+2x$$
Put $$y+2x=z$$
$$\Rightarrow\;\dfrac{dy}{dx}+2=\dfrac{dz}{dx}$$
$$\Rightarrow\;\dfrac{dy}{dx}=\dfrac{dz}{dx}-2\;\dots(ii)$$
From Eqs. (i) and (ii)
$$\dfrac{dz}{dx}-2=z$$
$$\Rightarrow\;\int\dfrac{dz}{.z+2}=\int\,dx$$
$$\Rightarrow\;log(z+2)=x+c$$
$$\Rightarrow\;log(y+2x+2)=x+c$$
$$\Rightarrow\;y+2x+2=e^{x+c}$$
$$\Rightarrow\;y+2x+2=e^x.e^c$$
$$\Rightarrow\;y=2[e^x-x-1]$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 33

Result

Application of Derivatives Test - 33

Score

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Rank

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SHARING IS CARING

Question 1

$$\cfrac{22}{7}$$

$$\cfrac{6}{7}$$

$$\cfrac{7}{6}$$

$$\cfrac{-6}{7}$$

answer required

Solution

Question 2

$$1$$

$$2$$

$$3$$

$$\cfrac{1}{2}$$

answer required

Solution

Question 3

$$x=0$$ and $$0$$

$$x=1$$ and $$-1$$

$$x=1$$ and $$-3$$

$$x=-1$$ and $$3$$

Solution

Question 4

$$\left( 4,\pm \cfrac { 8 }{ 3 } \right) $$

$$\left( 4,\cfrac { -8 }{ 3 } \right) $$

$$\left( 4,\pm \cfrac { 3 }{ 8 } \right) $$

$$\left( \pm 4,\cfrac { 3 }{ 8 } \right) $$

answer required

Solution

Question 5

$$2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $$

$$\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $$

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

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

Solution

Question 6

$$a < 0,\,b > 0$$

$$a \le o,\,b > 0$$

$$a < 0,\,b < 0$$

$$a \le 0,\,b < 0$$

Solution

Question 7

$$0$$

$$\pi$$

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

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

Solution

Question 8

$$y={ x }^{ 3 }+2$$

$$y=-{ x }^{ 3 }-2$$

$$y=3{ x }^{ 3 }+4$$

$$y=-{ x }^{ 3 }+2$$

Solution

Question 9

$${ 0 }^{ }$$

$${ 30 }^{ }$$

$${ 60 }^{ }$$

$${ 90 }^{ }$$

Solution

Question 10

$$y=2(e^x+x-1)$$

$$y=2(e^x-x-1)$$

$$y=2(e^x-x+1)$$

$$y=2(e^x+x+1)$$

Solution

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