Tangents and its Equations Test 45

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Tangents and its Equations Test 45

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Weekly Quiz Competition

Question 1
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 2
1 / -0

The equation of the curves through the point $$(1,0)$$ and whose slope is $$ \dfrac{y -1}{x^{2} + x} $$ is

A
$$ (y - 1)(x + 1) + 2x = 0 $$

B
$$ 2x(y - 1) + x + 1 = 0 $$

C
$$ x(y - 1)(x + 1) + 2 = 0 $$

D
None of these

Solution

Slope $$ = \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{y - 1}{x^{2} + x} $$
$$ \implies \dfrac{dy}{y-1} = \dfrac{dx}{x^{2} + x} $$
$$ \implies \int\dfrac{1}{y-1}dy = \int(\dfrac{1}{x} - \dfrac{1}{x+1})dx + C $$
$$\implies \log(y-1)=\log(\dfrac{x}{x+1})+\log{c}$$
$$ \implies \dfrac{(y-1)(x+1)}{x} = k $$
Putting $$ x=1, y=0 $$, we get $$ k=-2 $$
The equation is $$ (y-1)(x+1) + 2x = 0 $$
Question 3
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Directions For Questions

Let $$y=\displaystyle \int_{u(x)}^{v(x)} f(t)dt, $$ let us define $$\dfrac{dy}{dx}$$ as $$\dfrac{dy}{dx}=v'(x)f^2(v(x))-u'(x)f^2(u(x))$$ and the equation of tangent at $$(a,b)$$ and$$ y-b=\left( \dfrac{dy}{dx}\right)_{(a,b)}(x-a).$$

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If $$ f(x) = \displaystyle \int_{1}^{x} e^{t^2/2}(1-t^2)dt, $$ then $$\dfrac{d}{dx} f(x) $$ at x=1 is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$-1$$

Solution

We have, $$f(x)=\displaystyle \int_{1}^{x}e^{t^2/2}(1-t^2)dt$$
$$f'(x)=[e^{x^2/2}(1-x^2)]^2$$
$$f'(1)=e^{1/2}.0=0$$
Question 4
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Directions For Questions

Let $$y=\displaystyle \int_{u(x)}^{v(x)} f(t)dt, $$ let us define $$\dfrac{dy}{dx}$$ as $$\dfrac{dy}{dx}=v'(x)f^2(v(x))-u'(x)f^2(u(x))$$ and the equation of tangent at $$(a,b)$$ and$$ y-b=\left( \dfrac{dy}{dx}\right)_{(a,b)}(x-a).$$

...view full instructions

If $$y=\displaystyle \int_{x}^{x^2}t^2dt ,$$ then the equation of tangent at x=1 is

A
$$x+y=1$$

B
$$y=x-1$$

C
$$y=x$$

D
$$y=x+1$$

Solution

At $$x=1,y=0,$$ $$\dfrac{dy}{dx}=2x.(x^4)^2-(x^2)^2=1$$
$$\,\therefore $$ Equation of tangent is $$y=x-1.$$
Question 5
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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 6
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
Question 7
1 / -0

The point of the curve $$ y^2 = x$$ where the tangent makes an angle of $$ \frac { \pi}{4} $$ with x-axis is

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

B
$$ ( \frac {1}{4}, \frac {1}{2} ) $$

C
$$ (4,2 ) $$

D
$$ (1,1) $$

Solution

Given curve is $$ y^2 = x$$
Therefore as per given condition
$$ \dfrac{dy}{dx}=\dfrac{1}{2y} = tan \dfrac{\pi}{4}=1 \Rightarrow y=\dfrac{1}{2} \\ \Rightarrow x= \dfrac{1}{4} $$
Therefore, correct answer is $$ B $$.
Question 8
1 / -0

The abscissa of the point on the curve $$ 3y=6x- 5x^3 $$ the normal at which passes through origin is :

A
$$ 1 $$

B
$$ \frac {1}{3} $$

C
$$ 2 $$

D
$$ \frac {1}{2} $$

Solution

Let $$ (x_1,y_1) $$ be the point on the given curve $$ 3y=6x-5x^3 $$ at which the normal passes through the origin. Then we have $$ \left( \frac { dy }{ dx } \right) _{ (x_{ 1 },y_{ 1 }) } =2-5x_1^2 .$$
Again the equation of the normal at $$ (x_1,y_1) $$ passing through the origin gives
$$ 2-5x_1^2= \dfrac{-x_1}{y_1}=\dfrac{-3}{6-5x_1^2} $$
Since. $$ x_1=1 $$ satisfies the equation, therefore, correct answer is $$ (A) .$$
Question 9
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The curve $$ y=x^{\frac{1}{5}} $$ has at $$ (0,0) $$

A
a vertical tangent (parallel to y-axis)

B
a horizontal tangent (parallel to x-axis)

C
an oblique tangent

D
no tangent

Solution

We have, $$ y=x^{1/5} $$
$$ \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{5}x^{\tfrac{1}{5}-1}=\frac{1}{5}x^{-4/5} $$
$$ \therefore \left( \dfrac { dy }{ dx } \right) _{ (0,0) }= \tfrac{1}{5} \times (0)^{-4/5}= \infty $$
So, the curve $$ y=x^{1/5} $$ has vertical tangent at $$ (0,0) $$, which is parallel to y-axis.
Question 10
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The equation of normal to the curve $$ 3x^2-y^2 =8 $$ which is parallel to the line $$ x+ 3y=8 $$ is

A
$$ 3x-y=8 $$

B
$$ 3x+y+8=0 $$

C
$$ x+3y \pm 8 = 0 $$

D
$$ x+ 3y=0 $$

Solution

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Tangents and its Equations Test 45

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Question 1

it makes a constant angle with x-axis

it passes through the origin

it is at a constant distance from the origin

none of these

Solution

Question 2

$$ (y - 1)(x + 1) + 2x = 0 $$

$$ 2x(y - 1) + x + 1 = 0 $$

$$ x(y - 1)(x + 1) + 2 = 0 $$

None of these

Solution

Question 3

$$0$$

$$1$$

$$2$$

$$-1$$

Solution

Question 4

$$x+y=1$$

$$y=x-1$$

$$y=x$$

$$y=x+1$$

Solution

Question 5

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

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

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

None of these

Solution

Question 6

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

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

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

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

Solution

Question 7

$$ ( \frac {1}{2}, \frac {1}{4} ) $$

$$ ( \frac {1}{4}, \frac {1}{2} ) $$

$$ (4,2 ) $$

$$ (1,1) $$

Solution

Question 8

$$ 1 $$

$$ \frac {1}{3} $$

$$ 2 $$

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

Solution

Question 9

a vertical tangent (parallel to y-axis)

a horizontal tangent (parallel to x-axis)

an oblique tangent

no tangent

Solution

Question 10

$$ 3x-y=8 $$

$$ 3x+y+8=0 $$

$$ x+3y \pm 8 = 0 $$

$$ x+ 3y=0 $$

Solution

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