Application of Derivatives Test

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Application of Derivatives Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The curve given by the equation $$y-e^{xy}+x=0$$ has a vertical tangent at the point

A
$$(0, 1)$$

B
$$(1, 1)$$

C
$$(- 1, 1)$$

D
$$(1, 0)$$

Solution

$$y-e^{xy}+x=0$$

$$\therefore \dfrac{dy}{dx}-e^{xy}\left ( x\dfrac{dy}{dx}+y \right )+1=0$$

$$\Rightarrow \dfrac{dy}{dx}=\dfrac{-1+ye^{xy}}{1-xe^{xy}}$$

$$\therefore $$ if the tangent is vertical at $$(x,y),xe^{xy}=1$$ 1 which is possible only if $$x=1,y=0.$$
Question 2
1 / -0

The sum of the intercepts made on the axes of coordinates by any tangent to the curve $$\sqrt{x}+\sqrt{y}=2$$ is equal to

A
$$4$$

B
$$2$$

C
$$8$$

D
None of these

Solution

Curve equation $$\sqrt{x}+\sqrt{y}=2 \Rightarrow x > 0, y>0$$
$$\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}=-\sqrt{\dfrac {y}{x}}$$
and since in question its asking in general not for a particular point on curve. Hence it we find it for any point on curve then it will be true for any point.
So as use can see (1, 1) lies on curve
So, $$\dfrac{dy}{dx} \int_{1, 1}=-1$$
So tangent equation at (1, 1) is $$y= (-1)x+c$$
and it will pass through (1, 1)
So, $$c=2$$
tangent equation : $$y= -x +2$$
$$x+y=2$$
$$\dfrac{x}{2}+\dfrac{y}{2} =1$$
So x - intercept $$=2$$
y- intercept $$=2$$
So sum $$=4$$
Question 3
1 / -0

lf $$\alpha = \cos 10^{\circ} - \sin 10^{\circ}, \beta = \cos 45^{\circ} - \sin 45^{\circ}, \gamma = \cos 70^{\circ} - \sin 70^{\circ}$$ then the descending order of $$\alpha, \beta, \gamma$$ is

A
$$\alpha, \beta, \gamma$$

B
$$\gamma, \beta, \alpha$$

C
$$\alpha, \gamma, \beta$$

D
$$\beta, \alpha, \gamma$$

Solution

let $$f(x) = \cos {x} - \sin {x}$$
$$f'(x) = -(\sin {x} + \cos {x}) < 0$$ for $$ 0 < x < \dfrac {\Pi}{2}$$
so $$f(x)$$ is a decreasing function.
Question 4
1 / -0

lf the curve $$y=px^{2}+qx+r$$ passes through the point (1, 2) and the line $$y=x$$ touches it at the origin, then the values of $$p,\ q$$ and $$r$$ are

A
$$p=1,q=-1,r=0$$

B
$$p=1,q=1,r=0$$

C
$$p=-1,q=1,r=0$$

D
$$p=1,q=2,r=3$$

Solution

$$2=p+q+r$$
$$r=0$$
$$\dfrac{dx}{dy}|_{x=0}=1$$
$$\dfrac{dy}{dx}=2px+q$$
$$q=1$$
$$2=p+1+0$$
$$p=1$$
Question 5
1 / -0

If there is an error of $$k%$$ in measuring the edge of a cube, then the percent error in estimating its volume is

A
$$k$$

B
$$3k$$

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

D
none of these

Solution

Volume of cube$$V=x^{3}$$
$$\displaystyle \frac{dV}{dx}=3x^{2}$$
Percentage error in measuring side $$= k%$$
$$\displaystyle \Rightarrow \frac{\delta x}{x}=\frac{k}{100}$$
$$\displaystyle {\delta{x}}=\frac{xk}{100}$$
Approximate error in estimating volume $$=dV=(\frac{dV}{dx}){\delta x}=3x^{2}\frac{xk}{100}$$
Percentage error in estimating volume$$=\frac{dV}{V}\times 100=3k$$
Question 6
1 / -0

At what points of curve $$y= \displaystyle \frac {2}{3} x^3 + \displaystyle \frac{1}{2} x^2 $$, the tangent makes equal angle with the axis

A
$$\left( \displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 5 }{ 24 } \right) $$ and $$\left( -1,-\displaystyle \frac { 1 }{ 6 } \right) $$

B
$$\left(\displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 4 }{ 9 } \right)$$ and $$\left( -1,0 \right) $$

C
$$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 1 }{ 7 } \right) $$and $$\left( -3,\displaystyle \frac { 1 }{ 2 } \right) $$

D
$$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 4 }{ 47 } \right)$$ and $$\left( -1,-\displaystyle \frac { 1 }{ 3 } \right) $$

Solution

Given equation of curve is
$$y= \dfrac {2}{3} x^3 + \dfrac{1}{2} x^2 $$
$$\dfrac{dy}{dx}=2x^{2}+x$$
Since ,the tangent makes equal angle with x-axis,
$$\dfrac{dy}{dx}=1$$
$$\Rightarrow 2x^{2}+x-1=0$$
$$\Rightarrow x=-1,\dfrac{1}{2}$$
At $$ x=-1\Rightarrow y=-\dfrac{1}{6}$$
At $$ x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{24}$$
Question 7
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 8
1 / -0

The number of tangents to the curve $$x^{3/2} + y^{3/2}= 2a^{3/2}$$, $$a>0$$, which are equally inclined to the axes, is

A
$$2$$

B
$$1$$

C
$$0$$

D
$$4$$

Solution

For a tangent to be equally inclined to the axes, the slope of the tangent should be 1.
Thus
$$y'=1$$.
Now
$$x^{\frac{3}{2}}+y^{\frac{3}{2}}=2a^{\frac{3}{2}}$$.

$$\dfrac{3}{2}(\sqrt{x})+\dfrac{3}{2}(\sqrt{y})=0$$

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

$$x=y$$
$$x_{1}=y_{1}$$

Substituting int he above equation, we get
$$2x^{\frac{3}{2}}=2a^{\frac{3}{2}}$$

$$x=y=a$$.
Hence the tangent touches the curve at $$(a,a)$$.
Thus we get only one tangent, satisfying that criteria.
Question 9
1 / -0

If $$m$$ is the slope of a tangent to the curve $$e^y= 1+x^2$$, then

A
$$\left | m \right | > 1$$

B
$$m>1$$

C
$$m>-1$$

D
$$\left | m \right | \leq 1$$

Solution

$$e^{y}=1+x^{2}$$
$$\dfrac{dy}{dx}=\dfrac{2x}{1+x^{2}}$$
So $$|m|=|\dfrac{2x}{x^{2}+1}| $$
Now, since $$x^{2}\ge 0$$
$$\Rightarrow x^{2}+1 \ge 1$$
$$\Rightarrow \dfrac{1}{ x^{2}+1} \le 1$$
$$\Rightarrow \dfrac{2x}{ x^{2}+1} \le 1$$
$$ |m| \le 1$$
Question 10
1 / -0

If at each point of the curve $$y=x^3-ax^2 +x+1$$, the tangent is inclined at an acute angle with the positive direction of the $$x$$-axis, then

A
$$a>0$$

B
$$a\leq \sqrt 3$$

C
$$-\sqrt 3 \leq a \leq \sqrt 3 $$

D
none of these

Solution

$$y=x^3-ax^2 +x+1$$
$$\dfrac{dy}{dx}=3x^{2}-2ax+1$$
Since, the tangent is inclined at an acute angle with the positive direction of x-axis.
$$\dfrac{dy}{dx}>0$$
$$\Rightarrow 3x^{2}-2ax+1>0$$
$$\Rightarrow D<0$$
$$\Rightarrow 4(a^2-3)<0$$
$$\Rightarrow (a-\sqrt{3})(a+\sqrt{3})<0$$
$$\Rightarrow -\sqrt{3}<a<\sqrt{3}$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 21

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Application of Derivatives Test - 21

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SHARING IS CARING

Question 1

$$(0, 1)$$

$$(1, 1)$$

$$(- 1, 1)$$

$$(1, 0)$$

Solution

Question 2

$$4$$

$$2$$

$$8$$

None of these

Solution

Question 3

$$\alpha, \beta, \gamma$$

$$\gamma, \beta, \alpha$$

$$\alpha, \gamma, \beta$$

$$\beta, \alpha, \gamma$$

Solution

Question 4

$$p=1,q=-1,r=0$$

$$p=1,q=1,r=0$$

$$p=-1,q=1,r=0$$

$$p=1,q=2,r=3$$

Solution

Question 5

$$k$$

$$3k$$

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

none of these

Solution

Question 6

$$\left( \displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 5 }{ 24 } \right) $$ and $$\left( -1,-\displaystyle \frac { 1 }{ 6 } \right) $$

$$\left(\displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 4 }{ 9 } \right)$$ and $$\left( -1,0 \right) $$

$$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 1 }{ 7 } \right) $$and $$\left( -3,\displaystyle \frac { 1 }{ 2 } \right) $$

$$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 4 }{ 47 } \right)$$ and $$\left( -1,-\displaystyle \frac { 1 }{ 3 } \right) $$

Solution

Question 7

36

144

72

None of these

Solution

Question 8

$$2$$

$$1$$

$$0$$

$$4$$

Solution

Question 9

$$\left | m \right | > 1$$

$$m>1$$

$$m>-1$$

$$\left | m \right | \leq 1$$

Solution

Question 10

$$a>0$$

$$a\leq \sqrt 3$$

$$-\sqrt 3 \leq a \leq \sqrt 3 $$

none of these

Solution

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