Application of Derivatives Test

Result

Application of Derivatives Test - 52

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Point on the curve $$f(x)=\dfrac{x}{1-x^{2}}$$ where the tangent is inclined at an angle of $$\dfrac{\pi }{4}$$ ot the x-axis are

A
(0, 0)

B
$$\left ( \sqrt{3},\dfrac{-\sqrt{3}}{2} \right )$$

C
$$\left ( -2 ,\dfrac{2}{3}\right )$$

D
$$\left (- \sqrt{3},\dfrac{\sqrt{3}}{2} \right )$$

Solution
Question 2
1 / -0

The abscissa of points P and Q in the curve $$y = e^{x}+e^{-x}$$ such that tangents at P and Q make $$60^{o}$$ with the x-axis

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

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

C
ln $$\left ( \dfrac{\sqrt{7}+\sqrt{3}}{2} \right )$$

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

Solution
Question 3
1 / -0

The x-intercept of the tangent at any arbitrary point of the curve $$\dfrac{a}{x^{2}}+\dfrac{b}{y^{2}}=1$$ is proportion to

A
square of the abscissa of the point of tangency

B
square root of the abscissa of the point of tangency

C
cube of the abscissa pf the point of tangency

D
cube root of the abscissa of the point of tangency

Solution
Question 4
1 / -0

If the tangent at any point $$P(4m^{2}, 8m^{3})$$ of $$x^{3}-y^{3}=0$$ is also a normal to the curve $$x^{3}-y^{3}=0$$ , then value of m is

A
$$m = \dfrac{\sqrt{2}}{3}$$

B
$$m = -\dfrac{\sqrt{2}}{3}$$

C
$$m = \dfrac{3}{\sqrt{2}}$$

D
$$m = -\dfrac{3}{\sqrt{2}}$$

Solution
Question 5
1 / -0

Consider the following statement is $$S$$ and $$R$$
$$S$$. Both $$\sin x$$ and $$\cos x$$ are decreasing function in the interval $$\left(\dfrac {\pi}{2}, \pi \right)$$
$$R:$$ If a differentiable function decreases in an interval $$(a, b)$$ then its derivative also decreases in $$(a, b)$$, which of the following is true?

A
Both $$S$$ and $$R$$ are wrong

B
Both $$S$$ and $$R$$ are correct but $$R$$ is not the correct explanation of $$S$$

C
$$S$$ is the correct and $$R$$ is the correct explanation of $$S$$

D
$$S$$ is the correct and $$R$$ is the wrong

Solution

From the graph, it is clear that both $$\sin x$$ and $$\cos x$$ in the internal $$(\pi /2, \pi)$$ are the decreasing functions.
Therefore, $$S$$ is correct.
To disprove $$R$$ let us consider the counter example,
$$f(x)=\sin x $$ in $$(0, \pi/2)$$
so that $$f'(x)=\cos x$$
again from the graph, it is clear that $$f(x)$$ is increasing in $$(0, \pi /2)$$, but $$f'(x)$$ is decreasing in $$(0, \pi /2)$$
Therefore, $$R$$ is wrong. Therefore, d is the correct option.
Question 6
1 / -0

The normal to the curve $$x = a (\cos 0 + 0\sin 0), y= a (\sin 0- 0\cos 0)$$ at any point 0 is such that

A
it makes a constant angle with x-axis

B
it passes through the origin

C
it is at a constant distance from the origin

D
none of these

Solution
Question 7
1 / -0

The slope of the tangent to the curve $$y = f(x)$$ at $$\left [ x, f(x) \right ]$$ is 2x + 1. If the curve passes through the point (1, 2)then the area bounded by the curve, the x-axis and the line x = 1 is

A
$$\dfrac{5}{6}$$

B
$$\dfrac{6}{5}$$

C
$$\dfrac{1}{6}$$

D
6

Solution
Question 8
1 / -0

The point(s) on the curve $$y^{3} + 3x^{2} = 12y,$$ where the tangent is vertical, is (are)

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

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

C
(0, 0)

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

Solution
Question 9
1 / -0

A curve passes through $$(2,1)$$ and is such that the square of the ordinate is twice the contained by the abscissa and the intercept of the normal. Then the equation of curve is

A
$$x^2 +y^2=9x$$

B
$$4x^2 +y^2=9x$$

C
$$4x^2 +2y^2=9x$$

D
None of these

Solution
Question 10
1 / -0

The curve for which the ratio of the length of the segment by any tangent on the $$Y-$$axis to the length of the radius vector is constant $$(K)$$, is

A
$$(y+\sqrt {x^2 -y^2})x^{k-1}=c$$

B
$$(y+\sqrt {x^2 +y^2})x^{k-1}=c$$

C
$$(y-\sqrt {x^2 -y^2})x^{k-1}=c$$

D
$$(y+\sqrt {x^2 +y^2})x^{k-1}=c$$

Solution