Application of Derivatives Test

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Application of Derivatives Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

What is the slope of the tangent to the curve $$y=sin^{-1}(sin^2x)$$ at $$x=0$$ ?

A
0

B
1

C
2

D
None of the above

Solution

Slope of tangent to a curve f(x), at a point $$({x}_{o},{y}_{o})$$ is given by $$f'(x)$$,
where $$f'(x)$$ is derivative of f(x) at $$x={x}_{o}$$
$$f(x) =\sin^{-1}({\sin ^{ 2 }{ x } })$$
$${x}_{o}=0$$
$$f'\left( x \right) =\cfrac { df\left( x \right) }{ dx } =\cfrac { df(x) }{ d\sin ^{ 2 }{ x } } \times \cfrac { d\sin ^{ 2 }{ x } }{ d\sin x } \times \cfrac { d\sin x }{ dx } $$
$$f'\left( x \right) =\cfrac { 1 }{ \sqrt { 1-\sin ^{ 2 }{ x } } } \times 2\sin x\times \cos x$$
$$f'\left( 0 \right) =0$$
Slope of tangent to the curve is 0
Question 2
1 / -0

Find the slope of the normal to the curve $$4x^3+6x^2-5xy-8y^2+9x+14=0$$T the point $$-2, 3$$.

A
$$\infty$$

B
$$11$$

C
$$\displaystyle\frac{9}{19}$$

D
$$\displaystyle-\frac{19}{9}$$

Solution

The given curve is $$4x^3+6x^2-5xy-8y^2+9x+14=0$$

Differentiating above equation w.r.t $$x$$, we get

$$12{ x }^{ 2 }+12x-5x\cfrac { dy }{ dx } -5y-16y\cfrac { dy }{ dx } +9=0$$

$$\Rightarrow \cfrac { dy }{ dx } =\cfrac { 12x^{ 2 }+12x-5y+9 }{ \left( 5x+16y \right) } $$

At $$\left( x,y \right) \equiv \left( -2,3 \right) $$

$$m=\cfrac { dy }{ dx } =\dfrac{12(4)+12(-2)-5(3)+9}{5(-2)+16(3)}=\dfrac{18}{38}=\dfrac{9}{19}$$ is the slope of tangent to the curve
Slope of normal is $$-\dfrac{1}{m}=-\dfrac{19}{9}$$
Question 3
1 / -0

A mirror in the first quadrant is in the shape of a hyperbola whose equation is xy = 1. A light source in the second quadrant emits a beam of light that hits the mirror at the point (2,1/2). If the reflected ray is parallel to the y-axis the slope of the incident beam is

A
$$\dfrac{13}{8}$$

B
$$\dfrac{7}{4}$$

C
$$\dfrac{15}{8}$$

D
$$2$$

Solution

Slope of tangent at $$\left ( 2, \frac{1}{2} \right )$$
$$m=-\dfrac{1}{4}$$
$$tan \theta =-\dfrac{1}{y}$$

$$\theta =\phi -90^o$$
$$tan\theta =4$$
Slope of incident ray $$= m$$
$$\left | \dfrac{m-\left ( -\dfrac{1}{4} \right )}{1+m\left ( -\dfrac{1}{4} \right )} \right |=tan \theta =4$$

$$\left | \dfrac{4m+1}{4-m} \right |=4$$

$$4m + 1 = 16 - 4m$$
$$m=\dfrac{15}{8}$$
Question 4
1 / -0

The value of $$K$$ in order that $$f(x) = \sin x - \cos x - Kx + 5$$ decreases for all positive real values of $$x$$ is given by

A
$$K < 1$$

B
$$K\geq 1$$

C
$$K > \sqrt {2}$$

D
$$K < \sqrt {2}$$

Solution

Given : $$f(x)=\sin x - \cos x - Kx+5$$
$$f(x)$$ decrease for all $$x$$ if $$f'(x)<0$$
$$f'(x) = \cos x + \sin x - K<0$$
$$\therefore k > \cos x + \sin x$$
We know that, $$\cos x+\sin x=\sqrt{2}\left({\dfrac{1}{\sqrt2}\cos x+\dfrac{1}{\sqrt2}\sin x}\right)=\sqrt2\left({\sin\dfrac{\pi}{4}\cos x}+\cos\dfrac{\pi}{4}\sin x\right)=\sqrt2(\sin(\dfrac{\pi}{4}+x))$$
and,
$$-1\le\sin(\dfrac{\pi}{4}+x)\le1\Rightarrow -\sqrt2\le\sqrt2\cdot\sin(\dfrac{\pi}{4}+x)\le\sqrt2$$
$$\therefore -\sqrt2\le\cos x+\sin x\le\sqrt2$$
$$max(\cos x + \sin x) = \sqrt {2}$$
$$\therefore K > \sqrt {2}$$
Question 5
1 / -0

The function $$f(x)=\cfrac { \sin { x } }{ x } $$ is decreasing in the interval

A
$$\left( -\cfrac { \pi }{ 2 } ,0 \right) $$

B
$$\left(0, \cfrac { \pi }{ 2 } \right) $$

C
$$\left( -\cfrac { \pi }{ 4 } ,0 \right) $$

D
None of these

Solution

$$f(x)=\cfrac { \sin { x } }{ x } $$
Let $$f'(x)=\cfrac { x\cos { x } -\sin { x } }{ { x }^{ 2 } } $$
Let $$g(x)=x\cos { x } -\sin { x } $$
$$\quad \quad \quad =\cos { x(x-\tan { x) } } $$
where $$\cos x>0$$
$$\quad \quad x-\tan { x } <0$$
$$\Rightarrow g(x)<0$$
$$\Rightarrow f'(x)<0$$
$$\therefore f(n)$$ is a decreasing function in $$\left( 0,\cfrac { \pi }{ 2 } \right) $$
Question 6
1 / -0

Consider the curve $$y = e^{2x}$$.Where does the tangent to the curve at (0, 1) meet the x-axis ?

A
$$(1, 0)$$

B
$$(2, 0)$$

C
$$\left(-\dfrac{1}{2}, 0\right)$$

D
$$\left(\dfrac{1}{2}, 0\right)$$

Solution

Slope of tangent at any point to the curve is given by $$\left|{ \cfrac { df\left( x \right) }{ dx } }\right|_{ ({ x }_{ 0 }, { y }_{ 0 }) }$$
Here $$f(x)=y={e}^{2x}$$
$$\Rightarrow \cfrac{dy}{dx}=2{e}^{2x}$$
At point $$(0,1)$$
$$\cfrac{dy}{dx}=2{e}^{2(0)}$$
$$\cfrac{dy}{dx}=2$$
Equation of tangent is at point $$({x}_{0},{y}_{0})$$
$$y-{y}_{0}=m(x-{x}_{0})$$
So, tangent at (0,1) is
$$y-1=2(x-0)$$
$$\therefore y=2x+1$$
When this tangent meets x-axis, $$y$$ co-ordinate becomes zero
Putting $$y$$ in above equation as zero, we get
$$0=2x+1$$
$$\Rightarrow x=\cfrac{-1}{2}$$
Thus, the tangent to the curve meets at $$\left(-\dfrac{1}{2},0\right)$$
Question 7
1 / -0

The approximate value of $$f(x)={ x }^{ 3 }+5{ x }^{ 2 }-7x+9=0$$ at $$x=1.1$$ is

A
$$8.6$$

B
$$8.5$$

C
$$8.4$$

D
$$8.3$$

Solution

Given, $$f(x)={ x }^{ 3 }+5{ x }^{ 2 }-7x+9=0$$
Let $$x=1$$ and $$\Delta x=0.1$$
Approximate of a function $$f(x+\Delta x)\approx f(x)+f'(x)\Delta x$$
$$ =f(1)+f'(1)\Delta x$$
$$ =8+0.6=8.6 $$
Question 8
1 / -0

The equation to the normal to the hyperbola $$\dfrac {x^{2}}{16} - \dfrac {y^{2}}{9} = 1$$ at $$(-4, 0)$$ is.

A
$$2x - 3y = 1$$

B
$$x = 0$$

C
$$x = 1$$

D
$$y = 0$$

Solution

$$\dfrac { { x }^{ 2 } }{ 16 } -\dfrac { { y }^{ 2 } }{ 9 } =1$$
Differentiating above equation, we get
$$ \dfrac { { 2x } }{ 16 } -\dfrac { 2{ y } }{ 9 } .\dfrac { dy }{ dx } =0$$
$$\implies \dfrac{dy}{dx}=\dfrac{9x}{16y}$$
$$\implies m=\left|\dfrac { dy }{ dx }\right|_{(-4,0)} =\dfrac{9(-4)}{ 16(0)} =\infty$$ is the slope of tangent
Slope of the normal is $$-\dfrac{1}{m} =0$$
Slope of the normal is zero so it is parallel to $$y$$-axis.
Equation of normal is $$y-0=0(x-(-4))$$
$$\therefore y=0$$ is the equation of normal.
Question 9
1 / -0

Let $$f(x)=2{ x }^{ 3 }-5{ x }^{ 2 }-4x+3,\cfrac { 1 }{ 2 } \le x\le 3$$. The point at which the tangent to the curve is parallel to the X-axis is

A
$$(1,-4)$$

B
$$(2,-9)$$

C
$$(2,-4)$$

D
$$(2,-1)$$

E
$$(2,-5)$$

Solution

Given, $$f(x)=2{ x }^{ 3 }-5{ x }^{ 2 }-4x+3$$

$$f'(x)=6{x}^{2}-10x-4$$

If tangent is parallel to X-axis, then $$\cfrac { dy }{ dx } =0\quad $$

$$\Rightarrow 6{ x }^{ 2 }-10x-4=0\quad $$

$$\Rightarrow x=\cfrac { 5\pm \sqrt { 25+24 } }{ 6 } =\cfrac { 5\pm 7 }{ 6 } $$

$$\Rightarrow x=2,\cfrac { -1 }{ 3 } $$

$$\quad \therefore y=f(x)=2{ (2) }^{ 3 }-5{ (2) }^{ 2 }-4(2)+3=-9$$
Question 10
1 / -0

The equation of the tangent to the curve $$y={ x }^{ 3 }-6x+5$$ at $$(2,1)$$ is

A
$$6x-y-11=0$$

B
$$6x-y-13=0$$

C
$$6x+y+11=0$$

D
$$6x-y+11=0$$

Solution

The equation of the curve
$$y={ x }^{ 3 }-6x+5$$
$$\Rightarrow \cfrac { dy }{ dx } =3{ x }^{ 2 }-6$$
$$\Rightarrow { \left( \cfrac { dy }{ dx } \right) }_{ 2,1 }=6$$
Now, equation of the tangent at $$(2,1)$$ is
$$(y-1)=6(x-2)$$
$$\Rightarrow 6x-y-11=0$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 36

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Application of Derivatives Test - 36

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SHARING IS CARING

Question 1

0

1

2

None of the above

Solution

Question 2

$$\infty$$

$$11$$

$$\displaystyle\frac{9}{19}$$

$$\displaystyle-\frac{19}{9}$$

Solution

Question 3

$$\dfrac{13}{8}$$

$$\dfrac{7}{4}$$

$$\dfrac{15}{8}$$

$$2$$

Solution

Question 4

$$K < 1$$

$$K\geq 1$$

$$K > \sqrt {2}$$

$$K < \sqrt {2}$$

Solution

Question 5

$$\left( -\cfrac { \pi }{ 2 } ,0 \right) $$

$$\left(0, \cfrac { \pi }{ 2 } \right) $$

$$\left( -\cfrac { \pi }{ 4 } ,0 \right) $$

None of these

Solution

Question 6

$$(1, 0)$$

$$(2, 0)$$

$$\left(-\dfrac{1}{2}, 0\right)$$

$$\left(\dfrac{1}{2}, 0\right)$$

Solution

Question 7

$$8.6$$

$$8.5$$

$$8.4$$

$$8.3$$

Solution

Question 8

$$2x - 3y = 1$$

$$x = 0$$

$$x = 1$$

$$y = 0$$

Solution

Question 9

$$(1,-4)$$

$$(2,-9)$$

$$(2,-4)$$

$$(2,-1)$$

$$(2,-5)$$

Solution

Question 10

$$6x-y-11=0$$

$$6x-y-13=0$$

$$6x+y+11=0$$

$$6x-y+11=0$$

Solution

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