Application of Derivatives Test

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Application of Derivatives Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

Directions For Questions

We are given the curves $$y=\displaystyle \int_{-\infty}^{\infty}f(t)dt $$ through the point $$\left( 0,\dfrac{1}{2}\right) $$ any y=f(x), where f(x)>0 and f(x) is differentiable, $$\forall x \in R$$ through (0,1). Tangents drawn to both the curves at the points with equal abscissae intersect on the same point on the X-axis.

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The function f(x) is

A
increasing for all x

B
non-monotonic

C
decreasing for all x

D
None of these

Solution

We have the equations of the tangents of the curve
$$y=\displaystyle \int_{-∞}^{x} f(t)dt $$ and $$y=f(x)$$ at arbitrary points on them are
$$Y-\displaystyle \int_{-∞}^{x} f(t)dt=f(x)(X-x)$$
and $$Y-f(x)=f'(x)(X-x)$$
As Eqs.(i) and (ii) intersect at the same point on the X-axis
Putting Y=0 and equating x-coordinates, we have
$$x-\dfrac{f(x)}{f'(x)}=x-\dfrac{\displaystyle \int_{-∞}^{x}f(t)dt}{f(x)}$$
$$\Rightarrow \dfrac{f(x)}{\displaystyle \int_{-∞}^{x}f(t)dt}=\dfrac{f'(x)}{f(x)}$$
$$\Rightarrow \displaystyle \int_{-∞}^{x}f(t)dt=cf(x)$$
As, f(0)=1 $$\Rightarrow\displaystyle \int_{-∞}^{x}f(t)dt=c\times1\Rightarrow c=\dfrac{1}{2}$$
$$\Rightarrow \displaystyle \int_{-∞}^{x}f(t)dt =\dfrac{1}{2}f(x); $$ differentiating both the sides and integrating and using boundary conditions, we get $$f(x)=e^{2x}; y=2ex$$ is tangent to $$y=e^{2x}$$
$$\therefore $$ Number of solutions=1
Clearly, f(x) is increasing for all x.
$$\therefore lim_{x→∞}(e^{2x})^{e^{-2x}}=1$$
Question 2
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Directions For Questions

Let $$y=\displaystyle \int_{u(x)}^{v(x)} f(t)dt, $$ let us define $$\dfrac{dy}{dx}$$ as $$\dfrac{dy}{dx}=v'(x)f^2(v(x))-u'(x)f^2(u(x))$$ and the equation of tangent at $$(a,b)$$ and$$ y-b=\left( \dfrac{dy}{dx}\right)_{(a,b)}(x-a).$$

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If $$ f(x) = \displaystyle \int_{1}^{x} e^{t^2/2}(1-t^2)dt, $$ then $$\dfrac{d}{dx} f(x) $$ at x=1 is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$-1$$

Solution

We have, $$f(x)=\displaystyle \int_{1}^{x}e^{t^2/2}(1-t^2)dt$$
$$f'(x)=[e^{x^2/2}(1-x^2)]^2$$
$$f'(1)=e^{1/2}.0=0$$
Question 3
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The tangent to the curve $$y = e^{x}$$ drawn at the point $$(c, e^{c})$$ intersects the line joining the points $$(c-1, e^{c-1})$$ and $$(c+1, e^{c+1})$$

A
on the left of x =c

B
on the right of x = c

C
at no point

D
at all point

Solution
Question 4
1 / -0

Directions For Questions

For certain curves $$y=f(x)$$ satisfying $$\dfrac{d^2y}{dx^2}=6x-4,f(x)$$ has
local minimum values $$5$$ when $$x=1$$

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Number of critical point for $$y=f(x)$$ for $$x \in [0,2]$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Integrating, $$\dfrac{d^2y}{dx^2}=6x-4,$$
we get $$\dfrac{dy}{dx}=3x^2-4x+A$$
When $$x=1, \dfrac{dy}{dx}=0$$ so that $$A=1$$.
Hence, $$\dfrac{dy}{dx}=3x^2-4x+1$$
Integrating we get $$y=x^3-2x^2+x+B$$.
When $$x=1, y=5$$ so that $$B=5$$
Thus, we have $$y=x^3-2x^2+x+5$$
From Eq. (i) we get the critical points $$x=\dfrac{1}{3}, x=1$$
At the critical point $$x=\dfrac{1}{3}, \dfrac{d^2y}{dx^2} $$ is negative.
Therefore, at $$x=\dfrac{1}{3},y$$ has a local maximum.
At $$x=1, \dfrac{d^2y}{dx^2}$$ is positive. THerefore, at $$x=1,y$$ has a local
minimum.
Also, $$f(1)=5,f\left(\dfrac{1}{3}\right)=\dfrac{157}{27}, f(0)=5, f(2)=7$$
Hence, the global value$$=7$$
And the global minimum value $$=5$$
Question 5
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If the line ax +by + c = 0 is a normal to the curve xy = 1, then

A
$$a > 0, b> 0$$

B
$$a > 0, b < 0$$

C
$$a < 0, b > 0$$

D
$$a < 0, b < 0$$

Solution
Question 6
1 / -0

The equation of the curves through the point $$(1,0)$$ and whose slope is $$ \dfrac{y -1}{x^{2} + x} $$ is

A
$$ (y - 1)(x + 1) + 2x = 0 $$

B
$$ 2x(y - 1) + x + 1 = 0 $$

C
$$ x(y - 1)(x + 1) + 2 = 0 $$

D
None of these

Solution

Slope $$ = \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{y - 1}{x^{2} + x} $$
$$ \implies \dfrac{dy}{y-1} = \dfrac{dx}{x^{2} + x} $$
$$ \implies \int\dfrac{1}{y-1}dy = \int(\dfrac{1}{x} - \dfrac{1}{x+1})dx + C $$
$$\implies \log(y-1)=\log(\dfrac{x}{x+1})+\log{c}$$
$$ \implies \dfrac{(y-1)(x+1)}{x} = k $$
Putting $$ x=1, y=0 $$, we get $$ k=-2 $$
The equation is $$ (y-1)(x+1) + 2x = 0 $$
Question 7
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The point of the curve $$ y^2 = x$$ where the tangent makes an angle of $$ \frac { \pi}{4} $$ with x-axis is

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

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

C
$$ (4,2 ) $$

D
$$ (1,1) $$

Solution

Given curve is $$ y^2 = x$$
Therefore as per given condition
$$ \dfrac{dy}{dx}=\dfrac{1}{2y} = tan \dfrac{\pi}{4}=1 \Rightarrow y=\dfrac{1}{2} \\ \Rightarrow x= \dfrac{1}{4} $$
Therefore, correct answer is $$ B $$.
Question 8
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The abscissa of the point on the curve $$ 3y=6x- 5x^3 $$ the normal at which passes through origin is :

A
$$ 1 $$

B
$$ \frac {1}{3} $$

C
$$ 2 $$

D
$$ \frac {1}{2} $$

Solution

Let $$ (x_1,y_1) $$ be the point on the given curve $$ 3y=6x-5x^3 $$ at which the normal passes through the origin. Then we have $$ \left( \frac { dy }{ dx } \right) _{ (x_{ 1 },y_{ 1 }) } =2-5x_1^2 .$$
Again the equation of the normal at $$ (x_1,y_1) $$ passing through the origin gives
$$ 2-5x_1^2= \dfrac{-x_1}{y_1}=\dfrac{-3}{6-5x_1^2} $$
Since. $$ x_1=1 $$ satisfies the equation, therefore, correct answer is $$ (A) .$$
Question 9
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The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is :

A
an ellipse

B
parabola

C
circle

D
rectangular hyperbola

Solution

$${\textbf{Step -1: Differentiate slope}}{\textbf{.}}$$

$${\text{Given: slope of tangent is equal to ratio of abscissa to the ordinate of the point}}{\text{.}}$$

$${\text{Let slope of the tangent be }}\dfrac{{dy}}{{dx}}.$$

$$\text{According to the question}$$

$$\dfrac{{dy}}{{dx}} = \dfrac{x}{y}$$

$${\textbf{Step -2: Solve further to find the equation}}{\textbf{.}}$$

$${\text{Separating the variables}}{\text{.}}$$

$$ydy = xdx$$

$${\text{Upon integration we get,}}$$

$$\int {ydy = \int {xdx} } $$

$$\Rightarrow\dfrac{{{y^2}}}{2} = \dfrac{{{x^2}}}{2} + C$$

$$\Rightarrow{y^2} = {x^2} + 2C$$

$$ \Rightarrow {x^2} - {y^2} + 2C = 0$$

$$ \Rightarrow {x^2} - {y^2} = k$$ $$\textbf{[ k = - 2 C]} $$

$${\textbf{Hence, the given equation is of a rectangular hyperbola}}{\textbf{.}}$$
Question 10
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The curve $$ y=x^{\frac{1}{5}} $$ has at $$ (0,0) $$

A
a vertical tangent (parallel to y-axis)

B
a horizontal tangent (parallel to x-axis)

C
an oblique tangent

D
no tangent

Solution

We have, $$ y=x^{1/5} $$
$$ \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{5}x^{\tfrac{1}{5}-1}=\frac{1}{5}x^{-4/5} $$
$$ \therefore \left( \dfrac { dy }{ dx } \right) _{ (0,0) }= \tfrac{1}{5} \times (0)^{-4/5}= \infty $$
So, the curve $$ y=x^{1/5} $$ has vertical tangent at $$ (0,0) $$, which is parallel to y-axis.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 53

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Application of Derivatives Test - 53

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SHARING IS CARING

Question 1

increasing for all x

non-monotonic

decreasing for all x

None of these

Solution

Question 2

$$0$$

$$1$$

$$2$$

$$-1$$

Solution

Question 3

on the left of x =c

on the right of x = c

at no point

at all point

Solution

Question 4

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 5

$$a > 0, b> 0$$

$$a > 0, b < 0$$

$$a < 0, b > 0$$

$$a < 0, b < 0$$

Solution

Question 6

$$ (y - 1)(x + 1) + 2x = 0 $$

$$ 2x(y - 1) + x + 1 = 0 $$

$$ x(y - 1)(x + 1) + 2 = 0 $$

None of these

Solution

Question 7

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

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

$$ (4,2 ) $$

$$ (1,1) $$

Solution

Question 8

$$ 1 $$

$$ \frac {1}{3} $$

$$ 2 $$

$$ \frac {1}{2} $$

Solution

Question 9

an ellipse

parabola

circle

rectangular hyperbola

Solution

Question 10

a vertical tangent (parallel to y-axis)

a horizontal tangent (parallel to x-axis)

an oblique tangent

no tangent

Solution

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