Application of Derivatives Test

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Application of Derivatives Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The equation of normal of $$x^2+y^2-2x+4y-5=0$$ at $$(2,\,1)$$ is

A
$$y=3x-5$$

B
$$2y=3x-4$$

C
$$y=3x+4$$

D
$$y=x+1$$

Solution

Given equation is $$x^2+y^2-2x+4y-5=0$$
On differentiating, we get
$$2x+2y\dfrac{dy}{dx}-2+4\dfrac{dy}{dx}=0$$
$$\Rightarrow\;(y+2)\dfrac{dy}{dx}=1-x$$
$$\Rightarrow\,\begin{pmatrix}\dfrac{dy}{dx}\end{pmatrix}_{(2,\,1)}=\dfrac{1-2}{1+2}=\dfrac{-1}{3}$$
$$\Rightarrow\,-\begin{pmatrix}\dfrac{dx}{dy}\end{pmatrix}_{(2,\,1)}=3$$
Now, equation of normal is
$$(y-1)=3(x-2)$$
$$y-1=3x-6$$
$$\Rightarrow\;y=3x-5$$
Question 2
1 / -0

The coordinates of the point P on the curve $$x = a(\theta + \sin \theta), y = a(1 - \cos \theta)$$ where the tangent is inclined at an angle $$\dfrac {\pi}{4}$$ to the x-axis, are

A
$$\left (a\left (\dfrac {\pi}{2} - 1\right ), a\right )$$

B
$$\left (a\left (\dfrac {\pi}{2} + 1\right ), a\right )$$

C
$$\left (a \dfrac {\pi}{2}, a\right )$$

D
$$(a, a)$$

Solution

$$x = a(\theta + \sin \theta), y = a(1 - \cos \theta)$$
$$\dfrac {dy}{d\theta} = a\sin \theta, \dfrac {dx}{d\theta} = a(1 + \cos \theta)$$
$$\dfrac {dy}{dx} = \dfrac {\sin \theta}{1 + \cos \theta} = \tan \dfrac {\theta}{2}$$
$$\Rightarrow \tan \dfrac {\theta}{2} = 1$$
$$\Rightarrow \dfrac {\theta}{2} = \dfrac {\pi}{4} \Rightarrow \theta = \dfrac {\pi}{2}$$
$$x = a(\theta + \sin \theta)$$
$$y = a(1 - \cos \theta)$$
$$x = a (\theta + \sin \theta)$$
$$y = a(1 - \cos \theta)$$
$$(x, y) \equiv \left (a \left (\dfrac {\pi}{2} + 1\right ), a\right )$$.
Question 3
1 / -0

The slope of the tangent to the curve $$x=3t^2+1, y=t^3-1$$ at $$x=1$$ is

A
$$\dfrac{1}{2}$$

B
$$0$$

C
$$-2$$

D
$$\infty$$

Solution

Given, $$x=3t^2+1, y=t^3-1$$
Slope of the tangent to the given curve is $$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx}$$
$$= 3t^2 \times \dfrac{1}{6t}$$
$$=\dfrac{t}{2}$$
Since the slope has to be calculated at $$x = 1, $$ i.e. at $$3t^2 + 1 = 1$$, we get $$t = 0$$
Thus, the required slope is $$0.$$
Question 4
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The function $$f(x)=\frac{x}{3}+\frac{3}{x}$$ decreaes in the interval.

A
(-3, 3)

B
$$(-\infty , 3)$$

C
$$(3, \infty )$$

D
$$(-9, 9)$$

E
$$(-3,3) - \{0\}$$

Solution

We have $$f(x)=\dfrac{x}{3}+\dfrac{3}{x}$$, Domain is $$R-\{0\}$$
Differentiate it,
$$f'(x)=\dfrac{1}{3}-\dfrac{3}{x^2}$$
Now for f to be decreasing $$f'(x)<0$$
$$\Rightarrow \dfrac{1}{3}-\dfrac{3}{x^2}<0$$
$$\Rightarrow \dfrac{x^2-9}{3x^2}<0$$
$$\Rightarrow x^2-9<0$$, since $$3x^2>0\forall x\in $$ Domain
$$\Rightarrow (x+3)(x-3)<0$$
$$\Rightarrow x\in (-3,3)-\{0\}$$, since $$0$$ is not in the domain
Question 5
1 / -0

Find the equation of the quadratic function $$f$$ whose graph increases over the interval $$(-\infty, -2)$$ and decreases over the interval $$(-2,+\infty)$$, $$f(0)=23$$ and $$f(1)=8$$

A
$$f(x)=3({x+2)}^{2}+35$$

B
$$f(x)=-3({x+2)}^{2}-35$$

C
$$f(x)=-3({x-2)}^{2}+35$$

D
$$f(x)=-3({x+2)}^{2}+35$$

Solution

Let the quadratic equation be $$f(x)=ax^{2}+bx+c$$
Given $$f(0)=23$$
Put $$x=0$$ in $$f(x)$$ we get $$c=23$$
Given $$f(1)=8$$
Put $$x=1$$ in $$f(x)$$, we get
$$a+b+23=8$$
$$a+b=-15$$ ......(i)
Now we can see from the interval that maximum value of quadratic equation is at $$x=-2$$
So $$f'(x)=0$$ at $$x=-2$$
$$f'(x)=2ax+b$$
Put $$x=-2$$, we get
$$2a(-2)+b=0$$
$$b=4a$$
Put it in equation (i), we get
$$a+4a=-15$$
$$a=-3$$
So $$b=4a=4(-3)=-12$$
So quadratic equation is
$$=-3x^{2}-12x+23$$
$$=-3(x^2+4x+4-4)+23$$
$$=-3((x+2)^2-4)+23$$
$$=-3(x+2)^2+12+23$$
$$=-3(x+2)^2+35$$
Question 6
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The focal length of a mirror is given by $$\displaystyle \frac{2}{f}\, =\, \displaystyle \frac{1}{v}\, -\, \displaystyle \frac{1}{u}$$. In finding the values of u and v, the errors are equal and equal to 'p'. The the relative error in f is

A
$$\displaystyle \frac{p}{2}\, \left ( \displaystyle \frac{1}{u}\, +\, \displaystyle \frac{1}{v} \right )$$

B
$$p \left ( \displaystyle \frac{1}{u}\, +\, \displaystyle \frac{1}{v} \right )$$

C
$$\displaystyle \frac{p}{2}\, \left ( \displaystyle \frac{1}{u}\, -\, \displaystyle \frac{1}{v} \right )$$

D
$$p \left ( \displaystyle \frac{1}{u}\, -\, \displaystyle \frac{1}{v} \right )$$

Solution

$$\dfrac{2}{f}=\dfrac{1}{v}-\dfrac{1}{u}$$
$$-2\dfrac{\Delta f}{f^2}=\dfrac{-\Delta V}{v^2}+\dfrac{\Delta u}{u^2}$$
$$\Rightarrow \dfrac{2\Delta f}{f^2}=\dfrac{\Delta v}{v^2}-\dfrac{\Delta v}{u^2}$$
$$\Rightarrow \dfrac{\Delta f}{f}=\dfrac{f P}{2}[\dfrac{1}{v}-\dfrac{1}{u}][\dfrac{1}{v}+\dfrac{1}{u}]$$
$$\dfrac{\Delta f}{f}=\dfrac{f p}{2}\times \dfrac{2}{f}[\dfrac{1}{v}+\dfrac{1}{u}]$$
$$\dfrac{\Delta f}{f} = p[\dfrac{1}{v}+\dfrac{1}{u}]$$
Question 7
1 / -0

Suppose that $$f$$ is a polynomial of degree $$3$$ and that $$f''(x)\neq 0$$ at any of the stationary point. Then

A
$$f$$ has exactly one stationary point

B
$$f$$ must have no stationary point

C
$$f$$ must have exactly $$2$$ stationary point

D
$$f$$ has exactly $$0$$ or $$2$$ stationary point.

Solution

The given information is a polynomial of degree $$3$$.

Let the function be $$f(x)=ax^3 +bx^2 + cx +d$$.

To find the stationary points we differentiate and equate it to zero.

$$f'(x) = 3ax^2 + 2bx + x$$

The differentiated function is a quadratic equation which will give two real or two complex roots. If we get two real roots then we will have exactly $$2$$ stationary points. Now if the roots are complex as they always occur in pair then we have no stationary points. Thus f has exactly $$0$$ or $$2$$ stationary points. .....Answer
Question 8
1 / -0

Let $$f:R\rightarrow R$$ be a differentiable function for all values of $$x$$ and has the peoperty that $$f(x)$$ and $$f'(x)$$ have opposite signs for all values of $$x$$. Then,

A
$$f(x)$$ is an increasing function

B
$$f(x)$$ is a decreasing function

C
$$f ^2 (x)$$ is a decreasing function

D
$$|f(x)|$$ is an increasing function

Solution

Given $$f(x) $$ and $$f^{'} (x) $$ have opposite signs

$$\Rightarrow f(x) f^{'} (x) <0$$

We cannot say f(x) is increasing or decreasing because we don't know the sign of f(x)

Let us take option C;

If $$f^{2}(x)$$ is increasing $$\Rightarrow \dfrac {d} {dx} (f^{2}(x))>0$$

$$\Rightarrow 2f(x)f^{'}(x)>0$$

$$\Rightarrow f(x) f^{'} (x) >0$$ which is the given condition

$$\therefore $$option C is correct
Question 9
1 / -0

The function $$f(x) = x^2$$ is decreasing in

A
$$(- \infty, \infty)$$

B
$$(- \infty, 0)$$

C
$$(0, \infty)$$

D
$$(-2, \infty)$$

Solution

$$f(x)=x^{2}$$,
$$f'(x)=2x$$.
$$f'(x)>0$$ implies $$2x>0$$ or $$x>0$$. Hence $$f(x)$$ is increasing in the interval of $$(0,\infty)$$ and decreasing in the interval $$(-\infty,0)$$.
Question 10
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Identify the correct statements
(a) Every constant function is an increasing function.
(b) Every constant function is a decreasing function.
(c) Every identify function is an increasing function.
(d) Every identify function is a decreasing function.

A
$$(a), (b)$$ and $$(c)$$

B
$$(a)$$ and $$(c)$$

C
$$(c)$$ and $$(d)$$

D
$$(a), (c)$$ and $$(d)$$

Solution

Increasing function means $$f'\left( x \right) \ge 0$$ and decreasing function means $$f'\left( x \right) \le 0$$
For constant function , derivative will be zero.
Therefore constant function is both increasing and decreasing.
Identity function means $$f(x)=x$$ , its derivative is always $$1 \ge 0$$
Therefore identity function is increasing function
So $$a,b,c$$ are correct
Therefore the correct option is $$A$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 34

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Application of Derivatives Test - 34

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SHARING IS CARING

Question 1

$$y=3x-5$$

$$2y=3x-4$$

$$y=3x+4$$

$$y=x+1$$

Solution

Question 2

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

$$\left (a\left (\dfrac {\pi}{2} + 1\right ), a\right )$$

$$\left (a \dfrac {\pi}{2}, a\right )$$

$$(a, a)$$

Solution

Question 3

$$\dfrac{1}{2}$$

$$0$$

$$-2$$

$$\infty$$

Solution

Question 4

(-3, 3)

$$(-\infty , 3)$$

$$(3, \infty )$$

$$(-9, 9)$$

$$(-3,3) - \{0\}$$

Solution

Question 5

$$f(x)=3({x+2)}^{2}+35$$

$$f(x)=-3({x+2)}^{2}-35$$

$$f(x)=-3({x-2)}^{2}+35$$

$$f(x)=-3({x+2)}^{2}+35$$

Solution

Question 6

$$\displaystyle \frac{p}{2}\, \left ( \displaystyle \frac{1}{u}\, +\, \displaystyle \frac{1}{v} \right )$$

$$p \left ( \displaystyle \frac{1}{u}\, +\, \displaystyle \frac{1}{v} \right )$$

$$\displaystyle \frac{p}{2}\, \left ( \displaystyle \frac{1}{u}\, -\, \displaystyle \frac{1}{v} \right )$$

$$p \left ( \displaystyle \frac{1}{u}\, -\, \displaystyle \frac{1}{v} \right )$$

Solution

Question 7

$$f$$ has exactly one stationary point

$$f$$ must have no stationary point

$$f$$ must have exactly $$2$$ stationary point

$$f$$ has exactly $$0$$ or $$2$$ stationary point.

Solution

Question 8

$$f(x)$$ is an increasing function

$$f(x)$$ is a decreasing function

$$f ^2 (x)$$ is a decreasing function

$$|f(x)|$$ is an increasing function

Solution

Question 9

$$(- \infty, \infty)$$

$$(- \infty, 0)$$

$$(0, \infty)$$

$$(-2, \infty)$$

Solution

Question 10

$$(a), (b)$$ and $$(c)$$

$$(a)$$ and $$(c)$$

$$(c)$$ and $$(d)$$

$$(a), (c)$$ and $$(d)$$

Solution

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