Application of Derivatives Test

Result

Application of Derivatives Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

The slope of the tangent to the curve $${r^2} = {a^2}\cos 2\theta$$, where $$x = r\cos \theta ,y = r\sin \theta $$, at the point $$\theta=\frac{\pi}{6}$$ is

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

B
$$-1$$

C
$$1$$

D
$$0$$
Question 2
1 / -0

The line $$\dfrac{x}{a}+\dfrac{y}{b}=1$$ touches the curve $$y=be^{-x/a}$$ at the point.

A
$$(a, b/a)$$

B
$$(-a, b/a)$$

C
$$(0, b)$$

D
None of these

Solution
Question 3
1 / -0

if $$m$$ is the slope of a tangent to the curve $$e^{y}=1+x^{2}$$, then $$m$$ belongs to the interval

A
$$[-1, 1]$$

B
$$[-2, -1]$$

C
$$[1, 2]$$

D
$$[1, 3]$$

Solution
Question 4
1 / -0

IF $$f(x)=\dfrac{x^2}{2-2cos x} ; \ g(x)=\dfrac{x^2}{6x-6sin x}$$ where $$0< \times < 1$$, then

A
both $$ 'f'$$ and$$ 'g' $$are increasing functions

B
$$'f' $$ is decreasing &$$ 'g'$$ is increasing function

C
$$'f' $$ increasing functions &$$ 'g'$$ is decreasing function

D
both $$'f'$$ & $$'g'$$ are decreasing function

Solution
Question 5
1 / -0

A tangent drawn to the curve $$y = f\left( x \right)$$ at $$P\left( {x,y} \right)$$
cuts the x and y axes at A and B, respectively, such that $$AP:PB = 1:3$$. If $$f\left( 1 \right) = 1$$ then the curve passes through $$\left( {k,\frac{1}{8}} \right)$$ where $$k$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Equation of tangent at $$P(x,y)$$ is

$$Y - y = \dfrac{dy}{dt}(X-x)$$

It meets x - axis at $$A$$

$$A = \left(x-\dfrac{dx}{dy}y, 0\right)$$

and y -axis at $$B$$

$$B = (0, y - \dfrac{dy}{dt})$$

And also given $$\dfrac{AP}{PB} = 1 : 3$$

Using section formula

$$\left[\dfrac{mx_1+m_2x_2}{m_1m+m_2}, \dfrac{m_1y_1+m_2y_2}{m_1+m_2}\right]$$

$$\dfrac{3}{4} \left(x - \dfrac{dx}{dy} y\right)v = \dfrac{1}{4} \left(y - x \dfrac{dy}{dx}\right) = (x,y)$$

$$\dfrac{3}{4} \left( x-y \dfrac{dx}{dy}\right) = x$$; $$\dfrac{1}{4} \left(y - x \dfrac{dy}{x} \right) = y$$

$$3 \dfrac{dx}{x} + \dfrac{dy}{y} = 0$$

$$\log(x^3y) =$$ const (on integrative)

Given $$f(1) = 1$$, so here $$cons \, t = 1$$

$$x^3y = 1$$ (k, 1/8)

$$k^3 = 8$$

$$k = 2$$
Question 6
1 / -0

On the curve $${x}^{3} = 12y$$ , then the interval at which the abscissa changes at a faster rate than the ordinate ?

A
$$x\in \left( -2,2 \right)$$

B
$$x\in \left( -2,2 \right) -\left\{ 0 \right\}$$

C
$$x\in \left( -3,3 \right) -\left\{ 0 \right\}$$

D
None of these

Solution

$$\left| dy \right| >\left| dx \right| $$

$$\left| \dfrac { dy }{ dx } \right| >1$$

$${ x }^{ 3 }=12y$$

diff wrto y

$$3{ x }^{ 2 }\dfrac { dx }{ dy } =12$$

$$\dfrac { dx }{ dy } =\dfrac { 12 }{ { 3x }^{ 2 } } =\dfrac { 4 }{ { x }^{ 2 } } $$

$$\left| \dfrac { dy }{ dx } \right| >1$$

$$\left| \dfrac { y }{ { x }^{ 2 } } \right| >1$$

$$\left| { x }^{ 2 } \right| >4$$

$$x>\pm 2$$

$$\left( 2,-2 \right) $$
Question 7
1 / -0

Let $$f$$ and $$g$$ be differentiable function satisfying $$g'(a)=2,g(a)=b$$ and $$fog=I$$ (identity function). Then $$f'(b)$$ is equal to

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

B
$$2$$

C
$$\frac{2}{3}$$

D
None of these

Solution
Question 8
1 / -0

The point on the curve $$y = b e^{\dfrac {-x}{a}}$$ at which the tangent drawn is $$\dfrac {x}{a} + \dfrac {y}{b} = 1$$ is

A
$$(0, b)$$

B
$$\left (a, \dfrac {1}{e}\right )$$

C
$$(0, 1)$$

D
$$(1, 0)$$

Solution
Question 9
1 / -0

If $$V$$ is the set of points on the curve $$y^{3} - 3xy +2 = 0$$ where the tangent is vertical then $$V =$$.

A
$$\phi$$

B
$$\left \{(1 , 0)\right \}$$

C
$$\left \{(1, 1)\right \}$$

D
$$\left \{(0, 0), (1, 1)\right \}$$

Solution
Question 10
1 / -0

Paraboals $$(y-\alpha )^{2}=4a(x-\beta )and (y-\alpha )^{2}=4a'(x-\beta ')$$ will have a common normal (other than the normal passing through vertex ofparabola)if:

A
$$\frac{2(a-a')}{\beta '-\beta }< 1$$

B
$$\frac{2(a-a')}{\beta -\beta' }< 1$$

C
$$\frac{2(a'-a)}{\beta -\beta' }< 1$$

D
$$\frac{2(a'a)}{\beta -\beta' }> 1$$

Solution