Tangents and its Equations Test 47

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Tangents and its Equations Test 47

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Weekly Quiz Competition

Question 1
1 / -0

The line $$y = mx + 1$$ is a tangent to the curve $$y^{2} = 4x$$ if the value of m is .......

A
$$1$$

B
$$2$$

C
$$3$$

D
$$\dfrac{1}{2}$$

Solution

Given, $$y^{2} = 4x \Rightarrow 2y \dfrac{dy}{dx} = 4 \Rightarrow \dfrac{dy}{dx} = \dfrac{2}{y}$$
but slope of tangent is $$m = \dfrac{2}{y} \Rightarrow y = \dfrac{2}{m}$$
When $$ y= \dfrac{2}{m}, x = \dfrac{y^{2}}{4} = 4/m^{2}_{/4} = \dfrac{1}{m^{2}}$$
$$x = \dfrac{1}{m^{2}}, y = \dfrac{2}{m}$$ lie on the line
$$y = mx + 1$$
$$\dfrac{2}{m} = m \times \dfrac{1}{m^{2}} + 1 \Rightarrow \dfrac{2}{m} = \dfrac{1}{m} + 1$$
$$\dfrac{2}{m} = 1 \Rightarrow m = 1$$
Correct answer (A).
Question 2
1 / -0

The slope of the normal to the curve $$ y = 2x ^{2} + 3 \sin x $$ at $$ x = 0 $$ is

A
$$3$$

B
$$1/3$$

C
$$-3$$

D
$$-1 /3$$

Solution

Given, $$ y = 2x ^{2} + 3 \sin x $$
slope $$ = \dfrac{dy}{dx} = 4x + 3 \cos x $$
$$\dfrac{dy}{dy}|_{x=0} = 4 ( 0 ) = 3 $$
slope of normal = $$ -1/ \dfrac{dy}{dx} = -1/3 $$
Question 3
1 / -0

The line $$ y = x + 1 $$ is a tangent to the curve $$y^{2} = 4 x $$ at the point

A
$$ ( 1 , 2 ) $$

B
$$ ( 2 , 1 ) $$

C
$$ ( 1 , -2 ) $$

D
$$ ( -1 , 2 )$$

Solution

Given,
$$y^{2} = 4 x \\ \Rightarrow 2y \dfrac{dx}{dx} = 4 \\ \Rightarrow \dfrac{dx}{dy} = 2/y $$
but given that $$ y = x + 1 $$ is a tangent to the curve its slope = $$ 1 ( y = mx +c ) $$
$$\dfrac 2y =1\implies y=2$$
$$x=2-1=1$$
The line $$ y = x + 1 $$ is a tangent to the curve $$y^{2} = 4 x $$ at the point $$(1,2)$$.
Question 4
1 / -0

The points on the curve $$9 y^{2} = x^{3}$$, where the normal to the curve makes equal intercepts with the axes are ...........

A
$$\left ( 4, \pm \dfrac{8}{3} \right )$$

B
$$\left ( 4, \dfrac{-8}{3} \right )$$

C
$$\left ( 4, + \dfrac{8}{3} \right )$$

D
$$\left (\pm 4, \dfrac{8}{3} \right )$$
Question 5
1 / -0

The coordinates of a point P(x, y) lying in the first quadrant of the ellipse $$\displaystyle \frac{x^{2}}{8}+\frac{y^{2}}{18}=1$$ so that the area of the triangle formed by the tangent at P and the axes is the smallest are

A
(3,2)

B
(-2,3)

C
(2,-3)

D
(2,3)

Solution

Let point be $$(2\sqrt 2 sin\theta, 3\sqrt 2 cos\theta)$$
$$y'=\dfrac {-3\sqrt 2}{2\sqrt 2}tan\theta$$
$$y'=\dfrac {-3}{2}tan\theta$$
$$(y-3\sqrt 2cos\theta)=\dfrac {-3}{2}tan\theta (x-2\sqrt 2 sin\theta)$$
$$x=\dfrac {2}{sin\theta}$$ (x intercept)
$$y=\dfrac {3}{cos\theta}$$ (y intercept)
so area $$=\dfrac {1}{2}\dfrac {\times 2\times 3}{sin\theta cos\theta}$$
area is min when $$sin 2\theta=1$$
$$\theta=\dfrac {\pi}{4}$$
so point (2,3)
Question 6
1 / -0

Directions For Questions

$$\displaystyle \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1$$ is the equation ot an ellipse. Tangents are drawn to the ellipse and it's auxillary circle at the points where a common ordinate cuts them.

...view full instructions

The greatest inclination between the tangents is

A
$$\displaystyle \tan^{-1} \left ( \dfrac{\mathrm{a}+\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$$

B
$$\displaystyle \tan^{-1} \left (\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$$

C
$$\tan^{-1}\sqrt{\dfrac{\mathrm{a}}{\mathrm{b}}}$$

D
$$\tan^{-1}\sqrt{\dfrac{\mathrm{b}}{\mathrm{a}}}$$

Solution

The slope of the tangent to the ellipse $$=-\displaystyle \dfrac{\mathrm{b}}{\mathrm{a}}\cot\theta$$
The slope of tangent to the circle is $$-\cot\theta$$
$$\therefore$$ the angle $$\phi$$ between the tangents is given by
$$\tan\phi=\dfrac{(-\dfrac{\mathrm{b}}{\mathrm{a}}+1)\cot\theta}{1+\dfrac{\mathrm{b}}{\mathrm{a}}\cot^{2}\theta}=\dfrac{(\mathrm{a}-\mathrm{b})}{\mathrm{a}\tan\theta+\mathrm{b}\cot\theta}=\dfrac{\mathrm{a}-\mathrm{b}}{(\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta})^{2}+2\sqrt{\mathrm{a}\mathrm{b}}}$$
This is maximum when $$\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta}=0$$
$$\therefore$$ the maximum angle $$=\displaystyle \tan^{-1}(\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}})$$
Question 7
1 / -0

A curve passes through $$(2, 0)$$ and the slope of the tangent at any point $$(x, y)$$ is $$x^2 -2x$$ for all values of $$x$$. The point of minimum ordinate on the curve where $$x > 0$$ is $$(a, b)$$'
Then find the value of $$a + 6b$$.

A
$$2$$

B
$$4$$

C
$$-2$$

D
$$-4$$

Solution

$$\dfrac {dy}{dx} = x^2 - 2x$$
$$y = \dfrac {x^3}{3} - x^2 + C$$
Passing through (2, 0)
$$\Rightarrow C = \dfrac {4}{3}$$
$$\therefore y = \dfrac {x^3}{3} - x^2 + \dfrac {4}{3}$$
$$y' = x^2 - 2x = x(x - 2)$$
For maxima or minima, $$y'=0$$
$$ \Rightarrow x= 0,2$$
$$y''>0$$ at $$x=2$$
At x = 2, y takes the minimum value.
$$\therefore $$ minimum value of y is $$= \dfrac {8}{3} - 4 +\dfrac {4}{3} = 0$$
$$\therefore a = 2, b = 0$$
$$a + 6b = 2$$
Question 8
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 9
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 that point the tangent to the graph is $$y=3x-5$$, then the function is

A
$$(x-1)^{2}$$

B
$$(x+1)^{2}$$

C
$$(x+1)^{3}$$

D
$$(x-1)^{3}$$

Solution

$$f''(x)=6(x-1)$$
$$ \Rightarrow f'(x) = 3x^2 -6x+k$$...(1)
Slope of tangent at $$(2,1)$$ is $$3$$.
$$ \Rightarrow 3=3 \times 2^2 -6 \times 2 +k$$
$$ \Rightarrow k=3$$...(2)
From (1) and (2)
$$f(x)=x^3-3x^2+3x+c$$
Using $$f(x)=y=1$$ when $$x=2$$, we get
$$1=2+c \Rightarrow c=-1$$
Hence, the function is
$$f(x)=x^3-3x^2+3x-1$$
$$ \Rightarrow f(x)=(x-1)^3$$
Question 10
1 / -0

Suppose $$a,b,c$$ are such that the curve $$y = ax^2 + bx + c$$ is tangent to $$y = 3x -3$$ at $$(1, 0)$$ and is also tangent to $$y = x + 1$$ at $$(3, 4)$$ then the value of $$(2a -b -4c)$$ equals

A
$$7$$

B
$$8$$

C
$$9$$

D
$$10$$

Solution

$$y = ax^2 + bx + c $$
Since $$(1,0)$$ lies on the curve
$$ \Rightarrow a + b + c = 0 $$ ...... (i)
Slope of curve $$\displaystyle =\frac {dy}{dx} = 2ax + b$$
Slope of curve at (1,0)$$=2a+b$$
Slope of curve at (3,4)$$=6a+b$$
Slope of tangent $$y=3x-3$$ is 3
Slope of tangent $$y=x+1$$ is 1
$$(\dfrac {dy}{dx})_{x=1} = 3$$ and $$(\dfrac {dy}{dx})_{x=3} = 1$$
$$2a + b = 3$$ .... (ii)
$$6a + b = 1$$ .... (iii)
on solving we get a $$= -\dfrac {1}{2}, b = 4, c = -\dfrac {7}{2}$$
$$\therefore 2a - b - 4c = -1 -4 + 14 = 9$$

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Tangents and its Equations Test 47

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SHARING IS CARING

Question 1

$$1$$

$$2$$

$$3$$

$$\dfrac{1}{2}$$

Solution

Question 2

$$3$$

$$1/3$$

$$-3$$

$$-1 /3$$

Solution

Question 3

$$ ( 1 , 2 ) $$

$$ ( 2 , 1 ) $$

$$ ( 1 , -2 ) $$

$$ ( -1 , 2 )$$

Solution

Question 4

$$\left ( 4, \pm \dfrac{8}{3} \right )$$

$$\left ( 4, \dfrac{-8}{3} \right )$$

$$\left ( 4, + \dfrac{8}{3} \right )$$

$$\left (\pm 4, \dfrac{8}{3} \right )$$

Question 5

(3,2)

(-2,3)

(2,-3)

(2,3)

Solution

Question 6

$$\displaystyle \tan^{-1} \left ( \dfrac{\mathrm{a}+\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$$

$$\displaystyle \tan^{-1} \left (\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$$

$$\tan^{-1}\sqrt{\dfrac{\mathrm{a}}{\mathrm{b}}}$$

$$\tan^{-1}\sqrt{\dfrac{\mathrm{b}}{\mathrm{a}}}$$

Solution

Question 7

$$2$$

$$4$$

$$-2$$

$$-4$$

Solution

Question 8

$$0$$

$$2$$

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

$$5$$

Solution

Question 9

$$(x-1)^{2}$$

$$(x+1)^{2}$$

$$(x+1)^{3}$$

$$(x-1)^{3}$$

Solution

Question 10

$$7$$

$$8$$

$$9$$

$$10$$

Solution

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