Application of Derivatives Test

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Application of Derivatives Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

If the circle $$x^2+y^2+2gx+2fy+c=0$$ is touched by $$y=x$$ at P such that OP = $$6\sqrt{2}$$
then the value of c is

A
36

B
144

C
72

D
None of these

Solution

Let the point of contact be $$x_{1},y_{1}$$
However it lies on the line $$y=x$$
Hence
$$x_{1}=y_{1}$$
Applying distance formula, we get
$$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$$
$$2x_{1}^2=72$$
$$x_{1}=6$$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.
$$2x+2yy'+2g+2fy'=0$$
Now $$y'=1$$ since $$y'$$ is the slope of the line $$y=x$$.
By substituting, we get
$$x+y=-(g+f)$$
Now $$x_{1}=y_{1}=6$$
Hence
$$12=-(g+f)$$ ...(i)
Substituting $$x=y=6$$ in the equation of the circle, we get
$$36+36+2(6)(g+f)+c=0$$
$$72+12(-12)+c=0$$
$$c=144-72$$
$$c=72$$
Question 2
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The abscissas of points $$P$$ and $$Q$$ on the curve $$y=e^x+e^{-x}$$ such that tangents at $$P$$ and $$Q$$ make $$60^{\circ}$$ with the $$x$$-axis are

A
$$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {7}\right )$$ and $$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 5} {2}\right )$$

B
$$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$$

C
$$\ln \left (\displaystyle \frac {\sqrt 7-\sqrt 3} {7}\right )$$

D
$$\pm\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$$

Solution

$$y=e^{x}+e^{-x}$$
$$\displaystyle\frac{dy}{dx}=e^{x}-e^{-x}=\frac{e^{2x}-1}{e^{x}}$$
Let P$$(x_{1},y_{1})$$ and Q$$(x_2,y_2)$$ be the points on the given curve.
Slope of tangent at P is $$\displaystyle m_1=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$$
$$\Rightarrow \displaystyle \sqrt{3}=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$$
$$\displaystyle\Rightarrow e^{2x_1}-\sqrt{3}e^{x}-1=0$$
$$\displaystyle\Rightarrow (e^{x_1}-\frac{\sqrt{3}}{2})^{2}=\frac{7}{4}$$
$$\displaystyle\Rightarrow e^{x_1}=\pm\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}$$
$$\displaystyle\Rightarrow {x_1}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$$ as logarithm of negative numbers is not defined.
Similarly, $$\displaystyle\Rightarrow {x_2}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$$ as it makes same angle,$$60^{0}$$ with x-axis.
Question 3
1 / -0

The graphs $$y=2x^3-4x+2$$ and $$y=x^3+2x-1$$ intersect at exactly 3 distinct points. The slope of the line passing through two of these points

A
is equal to 4

B
is equal to 6

C
is equal to 8

D
is not unique

Solution

Given equation of curves
$$y=2x^3-4x+2$$
$$y=x^3+2x-1$$
Let $$(x_{1},y_{1})$$ be a point of intersection of the curves.
$$\Rightarrow y_{1}=2x_{1}^3-4x_{1}+2$$
And $$ y_{1}=x_{1}^3+2x_{1}-1$$
$$\Rightarrow y_1=8x_{1}-4$$
Let $$(x_{2},y_{2})$$ be another point of intersection of the curves.
$$\Rightarrow y_{2}=2x_{2}^3-4x_{2}+2$$
And $$y_{2}=x_{2}^3+2x_{2}-1$$
$$\Rightarrow y_2=8x_{2}-4$$
Now, slope $$= \displaystyle \dfrac{y_2-y_1}{x_2-x_1}$$
$$\Rightarrow m =\displaystyle \dfrac{8(x_2-x_1)}{x_2-x_1}$$
$$\Rightarrow m=8$$
Question 4
1 / -0

Let $$S$$ be a square with sides of length $$x$$. If we approximate the change in size of the area of $$S$$ by $$\displaystyle h.\frac{dA}{dx}|_{x=x_0}$$, when the sides are changed from $$x_0$$ to $$x_o+h$$, then the absolute value of the error in our approximation, is

A
$$h^2$$

B
$$2hx_0$$

C
$$x_0^2$$

D
$$h$$

Solution

$$ A=x^{2}$$
$$\displaystyle \dfrac{dA}{dx}=2x$$
Given , approximate change in size of area $$=\Delta A = \displaystyle h.\dfrac{dA}{dx}|_{x=x_0}$$
$$\Rightarrow \Delta A=2x_{0}h$$
Also, $$(A+\Delta A)-A=(x_{0}+h)^{2}-{x_0}^{2}$$
$$ =2x_{0}h+h^{2}$$
Hence, absolute value of error in our approximation is $$h^{2}$$
Question 5
1 / -0

The number of points on the curve $$x^{3/2}+y^{3/2}=a^{3/2}$$, where the tangents are equally inclined to the axes, is

A
$$2$$

B
$$1$$

C
$$0$$

D
$$4$$

Solution

If the tangents are equally inclined to the axes we get $$y'_{x,y}=1$$.
Therefore
$$\dfrac{3}{2}\sqrt{x}+\dfrac{3}{2}\sqrt{y}.y'=0$$

$$y'=1$$ implies
$$\sqrt{x}+\sqrt{y}=0$$

$$\sqrt{x}=-\sqrt{y}$$

$$x=y$$.
$$x_{1}=y_{1}$$.
Substituting in the equation we get
$$2x^{\frac{3}{2}}=a^{\frac{3}{2}}$$

$$x=\dfrac{a}{2^{\frac{2}{3}}}$$
Therefore
$$(x_{1},y_{1})=\left(\dfrac{a}{2^{\frac{2}{3}}},\dfrac{a}{2^{\frac{2}{3}}}\right)$$
Therefore there exists only one point.
Question 6
1 / -0

The point (s) on the curve $$\displaystyle y^{3}+3x^{2}= 12y,$$ where the tangent is vertical (i.e., parallel to the y-axis), is / true

A
$$\displaystyle \left ( \pm \frac{4}{\sqrt{3}},-2 \right )$$

B
$$\displaystyle \left ( \pm \frac{\sqrt{11}}{3},1 \right )$$

C
$$\displaystyle \left ( 0, 0 \right )$$

D
$$\displaystyle \left ( \pm \frac{4}{\sqrt{3}},2 \right )$$

Solution

We require the point where $$\dfrac{dx}{dy}=0$$
Or
$$3y^{2}.dy+6x.dx=12dy$$
Or
$$dy(3y^{2}-12)=-6xdx$$
Or
$$\dfrac{-dx}{dy}=\dfrac{3y^{2}-12}{6}$$
$$=0$$
Hence
$$3y^{2}-12=0$$
$$y^{2}-4=0$$
$$y=\pm 2$$
Now $$y=-2$$ does not satisfy the equation the curve.
Substituting in the equation of the curve, y=2,
$$8+3x^{2}=24$$
$$3x^{2}=16$$
$$x=\pm\dfrac{4}{\sqrt{3}}$$.
Question 7
1 / -0

Let the equation of a curve be $$x=a\left ( \theta +\sin \theta \right )$$, $$y=a\left ( 1-\cos \theta \right )$$. If $$\theta $$ changes at a constant rate $$k$$ then the rate of change of slope of the tangent to the curve at $$\displaystyle \theta =\frac{\pi }{2}$$ is

A
$$\displaystyle \frac{2k}{\sqrt{3}}$$

B
$$\displaystyle \frac{k}{\sqrt{3}}$$

C
$$k$$

D
none of these

Solution

As $$\theta $$ change with constant rate $$\Rightarrow \dfrac{d\theta}{dt}=k$$
$$x=a\left( \theta +sin\theta \right) $$
$$\cfrac { dx }{ d\theta } =a\left( 1+cos\theta \right) $$
$$y=a\left( 1-cos\theta \right) $$
$$\cfrac { dy }{ d\theta } =asin\theta $$

Then, $$\cfrac { dy }{ dx } =\cfrac { asin\theta }{ a\left( 1+cos\theta \right) } $$
Rate of change of slope $$=\dfrac{d}{dt}. \dfrac{dy}{dx} = \dfrac{1}{1+\cos\theta}.\dfrac{d\theta}{dt} = k $$ , at $$\theta =\dfrac{\pi}{2}$$
Question 8
1 / -0

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

A
$$\displaystyle \frac {1}{4} (\pi + 2t)$$

B
$$\displaystyle \frac {1 - \sin t}{ \cos t}$$

C
$$\displaystyle \frac {1}{4} (2t - \pi)$$

D
$$\displaystyle \frac {1 + \sin t}{\cos 2 t}$$

Solution

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

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

$$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\\$$
$$=\dfrac{2a\cos{t}\left(1+\sin{t}\right)}{a\left(1+\cos{2t}\right)}\\$$
$$=\dfrac{2\cos{t}\left(1+\sin{t}\right)}{2{\cos}^{2}{t}}\\$$
$$=\dfrac{\left(1+\sin{t}\right)}{\cos{t}}\\$$
$$=\dfrac{\left({\cos}^{2}{\dfrac{t}{2}}+{\sin}^{2}{\dfrac{t}{2}}+2\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}\right)}{{\cos}^{2}{\dfrac{t}{2}}-{\sin}^{2}{\dfrac{t}{2}}}\\$$
$$=\dfrac{{\left(\cos{\dfrac{t}{2}}+\sin{\dfrac{t}{2}}\right)}^{2}}{{\cos}^{2}{\dfrac{t}{2}}-{\sin}^{2}{\dfrac{t}{2}}}\\$$
$$=\dfrac{\left(\cos{\dfrac{t}{2}}+\sin{\dfrac{t}{2}}\right)}{\left(\cos{\dfrac{t}{2}}-\sin{\dfrac{t}{2}}\right)}\\$$
$$=\dfrac{1+\tan{\dfrac{t}{2}}}{1-\tan{\dfrac{t}{2}}}\\$$
$$=\dfrac{\tan{\dfrac{\pi}{4}}+\tan{\dfrac{t}{2}}}{1-\tan{\dfrac{\pi}{4}}\tan{\dfrac{t}{2}}}\\$$
$$=\tan{\left(\dfrac{\pi}{4}+\dfrac{t}{2}\right)}\\$$

If $$\theta$$ is the angle which the tangent at any point $$t$$ makes with the $$X$$ axis
then $$\tan{\theta}=\tan{\left(\dfrac{\pi}{4}+\dfrac{t}{2}\right)}$$

$$\therefore\,\theta=\dfrac{\pi}{4}+\dfrac{t}{2}=\dfrac{1}{4}\left(\pi+2t\right)$$
Question 9
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Directions For Questions

The number of points of intersection of the graphs of the functions $$y=f\left ( x \right )$$ and $$y=\phi \left ( x \right )$$ give the number of solutions of the equation $$f\left ( x \right )-\phi \left ( x \right )=0$$. Let $$f\left ( x \right )=ke^x$$ and $$\phi \left ( x \right )=x$$, where k is a real constant.

...view full instructions

The positive value of $$k$$ for which $$ke^x-x=0$$ has only one real solution is

A
$$\dfrac{1}{e}$$

B
1

C
e

D
$$\log_{e}2$$

Solution

$$ke^{ x }-x=0$$ will have only one solution when $$g(x)=x$$ is tangent to $$f(x)=ke^{x}$$
let $$P(a,b)$$ be a point on $$f(x)$$ and it also satisfy $$g(x)$$
Therefore, $$ke^{ a }=a$$ ...(1)
slope of $$g(x)=$$slope of $$f(x)$$ at point P
$$\Rightarrow 1=\dfrac { dy }{ dx } $$at point P$$=k{ e }^{ a }$$
From (1), we get $$a=1$$
Therefore, $$k=1/e$$

Ans: A
Question 10
1 / -0

Let $$\displaystyle \:f : R\rightarrow R$$ be a function such that $$\displaystyle \:f \left ( x \right )= ax+3\sin x+4\cos x.$$ Then $$\displaystyle \:f \left ( x \right )$$ is invertible if

A
$$\displaystyle \:a \epsilon \left ( -5, 5 \right )$$

B
$$\displaystyle \:a \epsilon \left ( -\infty , 5 \right )$$

C
$$\displaystyle \:a \epsilon \left ( -5, +\infty \right )$$

D
none of these

Solution

$$f(x)=ax+3sinx+4cosx$$
$$f'(x)=a+3cosx-4sinx$$
$$-5\leq3cosx-4sinx \geq5 $$
hence $$a\varepsilon (-5,5)$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 57

Result

Application of Derivatives Test - 57

Score

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Rank

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SHARING IS CARING

Question 1

36

144

72

None of these

Solution

Question 2

$$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {7}\right )$$ and $$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 5} {2}\right )$$

$$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$$

$$\ln \left (\displaystyle \frac {\sqrt 7-\sqrt 3} {7}\right )$$

$$\pm\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$$

Solution

Question 3

is equal to 4

is equal to 6

is equal to 8

is not unique

Solution

Question 4

$$h^2$$

$$2hx_0$$

$$x_0^2$$

$$h$$

Solution

Question 5

$$2$$

$$1$$

$$0$$

$$4$$

Solution

Question 6

$$\displaystyle \left ( \pm \frac{4}{\sqrt{3}},-2 \right )$$

$$\displaystyle \left ( \pm \frac{\sqrt{11}}{3},1 \right )$$

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

$$\displaystyle \left ( \pm \frac{4}{\sqrt{3}},2 \right )$$

Solution

Question 7

$$\displaystyle \frac{2k}{\sqrt{3}}$$

$$\displaystyle \frac{k}{\sqrt{3}}$$

$$k$$

none of these

Solution

Question 8

$$\displaystyle \frac {1}{4} (\pi + 2t)$$

$$\displaystyle \frac {1 - \sin t}{ \cos t}$$

$$\displaystyle \frac {1}{4} (2t - \pi)$$

$$\displaystyle \frac {1 + \sin t}{\cos 2 t}$$

Solution

Question 9

$$\dfrac{1}{e}$$

1

e

$$\log_{e}2$$

Solution

Question 10

$$\displaystyle \:a \epsilon \left ( -5, 5 \right )$$

$$\displaystyle \:a \epsilon \left ( -\infty , 5 \right )$$

$$\displaystyle \:a \epsilon \left ( -5, +\infty \right )$$

none of these

Solution

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