Application of Derivatives Test

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Application of Derivatives Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

Equation of the tangent at (1, -1) to the curve
$${ x }^{ 3 }-x{ y }^{ 2 }-4{ x }^{ 2 }-xy+5x+3y+1=0$$ is

A
$$x-4y-5=0$$

B
$$x+1=0$$

C
$$y-1=0$$

D
$$y+1=0$$

Solution
Question 2
1 / -0

The angle between the curves $$y = \sin x$$ and $$y = \cos x$$ is

A
$$\tan^{-1}(2\sqrt{2})$$

B
$$\tan^{-1}(3\sqrt{2})$$

C
$$\tan^{-1}(3\sqrt{3})$$

D
$$\tan^{-1}(5\sqrt{2})$$

Solution

C₁ : $$y = \sin x$$, C₂ : $$y = \cos x$$.
Equating the $$y$$' s,
$$\sin x = \cos x$$ ∴ $$x = \dfrac{\pi}{4} $$
∴ curves intersect each other at the point P : $$x = \dfrac{\pi}{4}$$.
Now, differentiating w.r.t. $$x$$,
C₁ gives : $$\dfrac{dy}{dx} = \cos x$$
C₂ gives : $$\dfrac{dy}{dx} = - \sin x$$.
Hence slopes $m₁ and m₂ of C₁ and C₂ at P : $$x = \dfrac{π}{4}$$ are
m₁ $$= \cos \dfrac{π}{4} = \dfrac{1}{√2}$$
m₂ =$$ - \sin \dfrac{π}{4} = -\dfrac{1}{√2}$$.
If $$θ$$ is the acute angle between them at $$P$$, then
$$tan θ =\dfrac{ | ( m₁ - m₂ )|}{ |( 1 + m₁m₂ ) |}$$
$$=\dfrac{ | ((\dfrac{1}{√2}) - (-\dfrac{1}{√2}))| }{ |( 1 + (\dfrac{1}{√2})(-\dfrac{1}{√2})) | }$$
$$=\dfrac{ | 2( \dfrac{1}{√2 })|}{ |( \dfrac{(2-1)} { (2)} | }$$
$$= 2√2 $$
∴ $$θ = \tanֿ¹ ( 2√2 ) $$
Question 3
1 / -0

If $$-4\le x\le 4$$, then critical points of $$f\left( x \right) ={ x }^{ 2 }-6\left| x \right| +4$$ are

A
$$3, -2$$

B
$$6, -6$$

C
$$3, -3, 0$$

D
$$0, 1, 3$$

Solution

For $$x\geq 0$$:
$$f(x)=x^2-6x+4$$
$$f'(X)=2x-6$$
For critical point $$f'(x)=0$$
Hence $$x=3$$

For $$x\leq 0$$:
$$f(x)=x^2+6x+4$$
$$f'(X)=2x+6$$
For critical point $$f'(x)=0$$
Hence $$x=-3$$

Also the curve is similar about line $$x=0$$ i.e. $$y-axis$$
Hence function have a critical point at $$x=0$$

So the critical points are $$0,3,-3$$
Question 4
1 / -0

The tangent to the curve, $$y = xe^{x^2}$$ passing through the point $$(1, e)$$ also passes through the point:

A
$$\left(\dfrac{4}{3}, 2e\right)$$

B
$$(2, 3e)$$

C
$$\left(\dfrac{5}{3}, 2e\right)$$

D
$$(3, 6e)$$

Solution

$$y = xe^{x^2}$$

$$\dfrac{dy}{dx}\left|_{(1,e)} = \left(x\cdot e^{x^2} \cdot 2x+e^{x^2}\right)\right|_{1,e} = 2\cdot e + e = 3e$$

$$T : y - e = 3e (x - 1)$$

$$y = 3ex - 3e + e$$

$$y = (3e) x - 2e$$

Out of the options only option:A

$$\left(\dfrac{4}{3}, 2e\right)$$ lies on it as $$2e=3e\times \dfrac 43-2e\Rightarrow 2e=2e$$
Question 5
1 / -0

The angle made by the tangent at any point on the curve $$x=a(t+\sin { t } \cos { t } ),y=a{ (1+\sin { t } ) }^{ 2 }$$ with x-axis is

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

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

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

D
$$\dfrac { \pi }{ 4 } +\dfrac { t }{ 2 } $$

Solution

Given,

$$x=a(t+\sin t \cos t)$$

$$\dfrac{dx}{dt}=\dfrac{d}{dt}a(t+\sin t \cos t)$$

$$\therefore \dfrac{dx}{dt}=a[1+\cos ^2t-\sin ^2t]=a\left(1+\cos \left(2t\right)\right)=2a\cos ^2t$$

$$y=a(1+\sin t)^2$$

$$\dfrac{dy}{dt}=\dfrac{d}{dt}a(1+\sin t)^2$$

$$\therefore \dfrac{dy}{dt}=2a\left(1+\sin \left(t\right)\right)\cos \left(t\right)$$

$$\Rightarrow \dfrac{dy}{dx}$$

$$=\dfrac{2a\left(1+\sin \left(t\right)\right)\cos \left(t\right)}{2a\cos ^2t}$$

$$=\dfrac {1+\sin t}{\cos t}$$

$$=\dfrac{\cos \frac{t}{2}+\sin \frac{t}{2}}{\cos \frac{t}{2}-\sin \frac{t}{2}}$$

$$=\dfrac{1+\tan \frac{t}{2}}{1 -\tan \frac{t}{2}}$$

$$=\tan \left ( \dfrac{\pi }{4}+\dfrac{t}{2} \right )$$

angle made by tangent,

$$\tan \theta =\tan \left ( \dfrac{\pi }{4}+\dfrac{t}{2} \right )$$

$$\Rightarrow \theta =\dfrac{\pi }{4}+\dfrac{t}{2}$$
Question 6
1 / -0

The equation of the normal to the curve$$y=\left( 1+x \right) ^{ y }+{ sin }^{ -1 }\left( { sin }^{ 2 }x \right) at\quad x=0$$ is

A
x+y=1

B
x-y+1=0

C
2x+y=2

D
2x-y+1

Solution

Given,

$$y=(1+x)^y+\sin ^{-1}(\sin ^2x)$$

when $$x=0$$

$$y=1^y+0=1$$

$$\therefore (x_1,y_1)=(0,1)$$

$$\dfrac{dy}{dx}=(1+x)^y\left [ \ln (1+x)\dfrac{dy}{dx}+\dfrac{y}{1+x} \right ]+\dfrac{2\sin x\cos x}{\sqrt{1-\sin ^4x}}$$

$$\dfrac{dy}{dx}_{(0,1)}=1[1]+0$$

$$\therefore \dfrac{dy}{dx}_{(0,1)}=1$$

Therefore, slope of normal is $$-1$$

Equation of normal,

$$y-y_1=\dfrac{-1}{\frac{dy}{dx}}(x-x_1)$$

$$y-1=(-1)(x-0)$$

$$y-1=-x$$

$$x+y=1$$ is the equation of normal
Question 7
1 / -0

If tangent at any point on the curve $${ y }^{ 2 }=1+{ x }^{ 2 }\ makes\ an\ angle\ \theta $$ with positive direction of the x-axis then

A
$$\left| \tan { \theta } \right| >1$$

B
$$\left| \tan { \theta } \right| <1$$

C
$$\left| \tan { \theta } \right| \ge1$$

D
$$\left| \tan { \theta } \right| \le 1$$

Solution
Question 8
1 / -0

A particle moves along a line by $$s = \dfrac {1}{3} t^{3} - 3t^{2} + 8t + 5$$, it changes its direction when

A
$$t = 1, t = 2$$

B
$$t =2, t = 4$$

C
$$t =0, t = 4$$

D
$$t = 2, t = 3$$

Solution

A particle changes its direction, when, velocity is equal to zero

Given,

$$s=\dfrac{1}{3}t^3-3t^2+8t+5$$

Velocity,

$$v=\dfrac{ds}{dt}=\dfrac{1}{3}(3t^2)-6t+8$$

$$\therefore v=t^2-6t+8$$

when $$v=0$$ particle changes its direction,

$$\Rightarrow t^2-6t+8=0$$

$$(t-2)(t-4)=0$$

$$\therefore t=2,4$$
Question 9
1 / -0

Length of the normal to the curve at any point on the curve $$y=\dfrac { a\left( { e }^{ x/a }+{ e }^{ -x/a } \right) }{ 2 } $$ varies as

A
$$x$$

B
$${ x }^{ 2 }$$

C
$$y$$

D
$${ y }^{ 2 }$$

Solution

Given,

$$y=\dfrac{a(e^{\frac{x}{a}}+e^{-\frac{x}{a}})}{2}$$

$$\Rightarrow y=a\cos h\left ( \dfrac{x}{a} \right )$$

$$\dfrac{dy}{dx}=a \sin h\left ( \dfrac{x}{a} \right )\dfrac{1}{a}$$

$$=\sin h\left ( \dfrac{x}{a} \right )$$

Length of normal $$=y\left [ 1+\left (\dfrac{dy}{dx} \right )^2 \right ]^{\frac{1}{2}}$$

$$=a\cos h\left ( \dfrac{x}{a} \right )\left [ 1+ \sin ^2h\left ( \dfrac{x}{a} \right )\right ]^{\frac{1}{2}}$$

$$=a\cos h\left ( \dfrac{x}{a} \right )\left [ \cos ^2h\left ( \dfrac{x}{a} \right ) \right ]^{\frac{1}{2}}$$

$$=a\cos ^2h\left ( \dfrac{x}{a} \right )$$

$$=a \times \dfrac{y^2}{a^2}$$

$$=\dfrac{y^2}{a}$$
Question 10
1 / -0

Function $$f(x)=\dfrac { \lambda sinx+6cosx }{ 2sinx+3cosx } $$ is monotonic increasing If

A
$$\lambda >4$$

B
$$\lambda >1$$

C
$$\lambda <4$$

D
$$\lambda <2$$

Solution

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 47

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Application of Derivatives Test - 47

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SHARING IS CARING

Question 1

$$x-4y-5=0$$

$$x+1=0$$

$$y-1=0$$

$$y+1=0$$

Solution

Question 2

$$\tan^{-1}(2\sqrt{2})$$

$$\tan^{-1}(3\sqrt{2})$$

$$\tan^{-1}(3\sqrt{3})$$

$$\tan^{-1}(5\sqrt{2})$$

Solution

Question 3

$$3, -2$$

$$6, -6$$

$$3, -3, 0$$

$$0, 1, 3$$

Solution

Question 4

$$\left(\dfrac{4}{3}, 2e\right)$$

$$(2, 3e)$$

$$\left(\dfrac{5}{3}, 2e\right)$$

$$(3, 6e)$$

Solution

Question 5

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

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

$$\pi +\dfrac { t }{ 2 } $$

$$\dfrac { \pi }{ 4 } +\dfrac { t }{ 2 } $$

Solution

Question 6

x+y=1

x-y+1=0

2x+y=2

2x-y+1

Solution

Question 7

$$\left| \tan { \theta } \right| >1$$

$$\left| \tan { \theta } \right| <1$$

$$\left| \tan { \theta } \right| \ge1$$

$$\left| \tan { \theta } \right| \le 1$$

Solution

Question 8

$$t = 1, t = 2$$

$$t =2, t = 4$$

$$t =0, t = 4$$

$$t = 2, t = 3$$

Solution

Question 9

$$x$$

$${ x }^{ 2 }$$

$$y$$

$${ y }^{ 2 }$$

Solution

Question 10

$$\lambda >4$$

$$\lambda >1$$

$$\lambda <4$$

$$\lambda <2$$

Solution

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