Application of Derivatives Test

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Application of Derivatives Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

A function $$y=f(x)$$ has a second-order derivative $$f''(x)=6(x-1)$$. If its graph passes through the point $$(2,1)$$ and at the point tangent to the graph is $$y=3x-5$$, then the value of $$f(0)$$ is

A
$$1$$

B
$$-1$$

C
$$2$$

D
$$0$$

Solution

$$f''(x)=6x-6$$
$$\int { f''(x)dx } =\int { (6x-6)dx } $$
$$\Rightarrow f'(x)=3x^{2}-6x+C$$
Since, it passes through (2,1) and tangent is $$y=3x-5$$
$$\Rightarrow C=3$$
So, $$f'(x)=3x^{2}-6x+3$$
$$\int { f'(x)dx } =\int { (3{ x }^{ 2 }-6x+3)dx } $$
$$\Rightarrow f(x)=x^{3}-3x^{2}+3x+C_1$$
Since, it passes through $$ (2,1)$$
$$\Rightarrow C_1=-1$$
Hence, $$f(x)=x^{3}-3x^{2}+3x-1$$
$$f(0)=-1$$
Question 2
1 / -0

The angle made by the tangent of the curve $$x=a (t+\sin t \cos t)$$, $$y=a(1+sint)^2$$ 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
$$\displaystyle \frac {1+\sin t}{\cos 2t}$$

Solution

$$x=a (t+sin t cos t)$$
$$\dfrac{dx}{dt}=2acos^{2}t$$
$$y=a(1+sint)^2$$
$$\dfrac{dy}{dt}=2a(cost+costsint)$$
$$\Rightarrow \displaystyle \dfrac{dy}{dx}=\dfrac{1+sint}{cost}$$
$$\displaystyle = \dfrac{(cos\dfrac{t}{2}+sin\dfrac{t}{2})^{2}}{cos^{2}\dfrac{t}{2}-sin^{2}\dfrac{t}{2}}$$
$$\displaystyle = \dfrac{1+tan{\dfrac{t}{2}}}{1-tan{\dfrac{t}{2}}}$$
$$\displaystyle =tan(\dfrac{\pi}{4}+\dfrac{t}{2})$$
Slope of tangent to the curve $$= tan(\dfrac{\pi}{4}+\dfrac{t}{2})$$
So, the angle made by the tangent to the curve with the x-axis $$=\dfrac{\pi}{4}+\dfrac{t}{2}$$ or $$\dfrac{1}{4}(\pi+2t)$$
Question 3
1 / -0

The value of $$x$$ at which tangent to the curve $$y=x^3-6x^2+9x+4, 0\leq x \leq 5$$ has maximum slope is

A
$$0$$

B
$$2$$

C
$$\dfrac{5}{2}$$

D
$$5$$

Solution

$$y=x^3-6x^2+9x+4$$
$$\dfrac{dy}{dx}=3x^2-12x+9$$

Slope of tangent,
$$m=f(x)=3x^2-12x+9$$

For maxima or minima,
$$f'(x)= 0$$
$$\Rightarrow 6x-12=0$$
$$\Rightarrow x=2$$.

Now,
$$m(0)=9, m(2)=-3$$ and $$m(5)=24$$
Hence,
The maximum value is attained at $$x = 5$$.

Hence, option D.
Question 4
1 / -0

The point on the curve $$y^{2} = x ,$$ the tangent at which makes an angle of $$45^{0}$$ with positive direction of $$x -$$ axis will be given by

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

B
$$\left ( \displaystyle \frac{1}{2}, \displaystyle \frac{1}{2} \right )$$

C
$$(2,4)$$

D
$$\left ( \displaystyle \frac{1}{4}, \displaystyle \frac{1}{2} \right )$$

Solution

Given equation of curve is $$y^{2} = x$$
$$\displaystyle \frac{dy}{dx}=\frac{1}{2y}$$
Slope of tangent $$=\dfrac{1}{2y}$$
Also, given slope of tangent $$=tan 45^{\circ}$$
$$\displaystyle\frac{1}{2y}=1\Rightarrow y=\frac{1}{2}$$
$$\displaystyle \Rightarrow x=\frac{1}{4}$$
Hence, the required point is $$\left ( \dfrac{1}{4},\dfrac{1}{2} \right )$$
Question 5
1 / -0

If the curve represented parametrically by the equations $$x=2 \ln\cot t+1$$ and $$y=\tan t+ \cot t$$

A
tangent and normal intersect at the point $$(2,1)$$

B
normal at $$t=\displaystyle \frac{\pi}{4}$$ is parallel to the $$y$$ axis

C
tangent at $$t=\displaystyle \frac{\pi}{4}$$ is parallel to the line $$y=x$$

D
tangent at $$t=\displaystyle \frac{\pi}{4}$$ is parallel to the $$x$$ axis

Solution

$$x=2 \ln\cot t+1$$
$$\displaystyle \frac{dx}{dt}=-\frac{2cosec^{2}t}{\cot t}=-\frac{2}{\sin t\cos t}$$
$$y=\tan t+ \cot t$$
$$\displaystyle \frac{dy}{dt}=sec^{2}t-cosec^{2}t=\frac{sin^{2}t-cos^{2}t}{sin^{2}t cos^{2}t}$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{sin^{2}t-cos^{2}t}{sin2t}$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=\cot 2t$$
Slope of tangent at $$t=\dfrac{\pi}{4}$$ is 0.
Hence, tangent is parallel to x-axis.
Question 6
1 / -0

The period of oscillation $$T$$ of a pendulum of length $$l$$ at a place of acceleration due to gravity $$g$$ is given by $$T=2\pi \sqrt {\dfrac {l}{g}}$$. If the calculated length is $$0.992$$ times the actual length and if the value assumed for $$g$$ is $$1.002$$ times its actual value, the relative error in the computed value of $$T$$ is

A
$$0.005$$

B
$$-0.005$$

C
$$0.003$$

D
$$-0.003$$

Solution

Relative error will calculate by
$$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{\Delta L}{L}-\dfrac{\Delta g}{g}\right]$$

$$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{0.992L}{L}-\dfrac{1.002 g}{g}\right]$$

$$\Delta T=-0.005T$$
Question 7
1 / -0

The focal length of a mirror is given by $$\dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$$. If equal errors ($$\alpha$$) are made in measuring $$u$$ and $$v$$, then the relative error in $$f$$ is

A
$$\dfrac {2}{\alpha}$$

B
$$\alpha \left (\dfrac {1}{u}+\dfrac {1}{v}\right )$$

C
$$\alpha \left (\dfrac {1}{u}-\dfrac {1}{v}\right )$$

D
none of these

Solution

Given, $$\displaystyle \dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$$
$$\Rightarrow \displaystyle -\dfrac {\Delta v}{v^2}+\dfrac {\Delta u}{u^2}=-\dfrac {2\Delta f}{f^2}$$
Since, given equal errors in measuring u and v i.e.$$\Delta u=\Delta v=\alpha$$
$$\Rightarrow {\alpha}\left(\dfrac{1}{u}-\dfrac{1}{v}\right)\left(\dfrac{1}{u}+\dfrac{1}{v}\right)=-\dfrac {2\Delta f}{f^2}$$
But, $$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{2}{f}$$
$$\Rightarrow\displaystyle \frac{ \Delta f}{f}={\alpha}\left(\dfrac{1}{u}+\dfrac{1}{v}\right)$$
Hence, correct option is B
Question 8
1 / -0

The tangent of the acute angle between the curves $$y=|x^2-1| $$ and $$y=\sqrt {7-x^2}$$ at their points of intersection is

A
$$\displaystyle \frac {5\sqrt 3}{2}$$

B
$$\displaystyle \frac {3\sqrt 5}{2}$$

C
$$\displaystyle \frac {5\sqrt 3}{4}$$

D
$$\displaystyle \frac {3\sqrt 5}{4}$$

Solution

The given curves are $$y=|x^{2}-1|$$ and $$y=\sqrt{7-x^2}$$
To find the point of intersection, let us equate both the curves.
$$|x^{2}-1|=\sqrt{7-x^2}$$
Squaring on both sides gives us $$x^{4}-x^{2}-6=0$$
$$(x^2-3)(x^2+2) =0 $$
On solving above quadratic equation, we get $$x^{2}=3$$
The point of intersection is $$\left(\pm \sqrt{3},2\right)$$
The slope of the tangent to $$y=|x^{2}-1|$$ at the point $$\left( \sqrt{3},2\right)$$is
$$m_{1}=\displaystyle\dfrac{dy}{dx}=2x=2\sqrt{3}$$
The slope of tangent to $$y=\sqrt{7-x^2}$$ at the point $$\left(\sqrt{3},2\right)$$is
$$m_{2}=\displaystyle\dfrac{dy}{dx}=\dfrac{-x}{\sqrt{7-x^2}}=\dfrac{-\sqrt{3}}{2}$$
Let $$\theta$$ be the acute angle between the two tangents.
$$\displaystyle\tan\theta = \dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}$$
$$\Rightarrow \displaystyle\tan\theta = \dfrac{-5\sqrt{3}}{4}$$
$$\therefore$$The tangent of the acute angle between the curves is $$\displaystyle\dfrac{5\sqrt{3}}{4}$$ .
Hence, option C is correct.
Question 9
1 / -0

Consider the function $$f(x)= \begin{cases} x \sin \displaystyle \frac {\pi}{x}, for \ x>0\\ 0, for \ x=0 \end{cases}$$. Then, the number of points in $$(0,1)$$ where the derivative $$f'(x)$$ vanishes is

A
$$0$$

B
$$1$$

C
$$2$$

D
infinite

Solution
Question 10
1 / -0

The area of a triangle is computed using the formula $$S=\dfrac {1}{2}$$ bc sin A. If the relative errors made in measuring b, c and calculating S are respectively $$0.02$$, $$0.01$$ and $$0.13$$ the approximate error in A when $$A=\pi /6$$ is

A
$$0.05$$ radians

B
$$0.01$$ radians

C
$$0.05$$ degree

D
$$0.01$$ degree

Solution

Error formula for given equation is
$$\dfrac{\Delta s}{s}=\dfrac{\Delta b}{b}+\dfrac{\Delta c}{c}+\dfrac{\Delta sinx}{sinx}$$
$$0.13=0.02+0.01+\dfrac{\Delta sinx}{1/2}$$
$$\Delta sinx=0.05$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 56

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Application of Derivatives Test - 56

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SHARING IS CARING

Question 1

$$1$$

$$-1$$

$$2$$

$$0$$

Solution

Question 2

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

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

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

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

Solution

Question 3

$$0$$

$$2$$

$$\dfrac{5}{2}$$

$$5$$

Solution

Question 4

$$\left (\displaystyle \frac{1}{2},\displaystyle \frac{1}{4} \right )$$

$$\left ( \displaystyle \frac{1}{2}, \displaystyle \frac{1}{2} \right )$$

$$(2,4)$$

$$\left ( \displaystyle \frac{1}{4}, \displaystyle \frac{1}{2} \right )$$

Solution

Question 5

tangent and normal intersect at the point $$(2,1)$$

normal at $$t=\displaystyle \frac{\pi}{4}$$ is parallel to the $$y$$ axis

tangent at $$t=\displaystyle \frac{\pi}{4}$$ is parallel to the line $$y=x$$

tangent at $$t=\displaystyle \frac{\pi}{4}$$ is parallel to the $$x$$ axis

Solution

Question 6

$$0.005$$

$$-0.005$$

$$0.003$$

$$-0.003$$

Solution

Question 7

$$\dfrac {2}{\alpha}$$

$$\alpha \left (\dfrac {1}{u}+\dfrac {1}{v}\right )$$

$$\alpha \left (\dfrac {1}{u}-\dfrac {1}{v}\right )$$

none of these

Solution

Question 8

$$\displaystyle \frac {5\sqrt 3}{2}$$

$$\displaystyle \frac {3\sqrt 5}{2}$$

$$\displaystyle \frac {5\sqrt 3}{4}$$

$$\displaystyle \frac {3\sqrt 5}{4}$$

Solution

Question 9

$$0$$

$$1$$

$$2$$

infinite

Solution

Question 10

$$0.05$$ radians

$$0.01$$ radians

$$0.05$$ degree

$$0.01$$ degree

Solution

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