Tangents and its Equations Test 11

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Tangents and its Equations Test 11

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Weekly Quiz Competition

Question 1
1 / -0

lf the chord joining the points where $$x= p,\ x =q$$ on the curve $$y=ax^{2}+bx+c$$ is parallel to the tangent drawn to the curve at $$(\alpha, \beta)$$ then $$\alpha=$$

A
$$2pq$$

B
$$\sqrt{pq}$$

C
$$\displaystyle \frac{p+q}{2}$$

D
$$\displaystyle \frac{p-q}{2}$$

Solution

$$y'=2ax+b$$
$$y'|_{x=b}=(2a\alpha +b)$$
$$2a\alpha+b=\left ( \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )$$
$$2a\alpha+b=\dfrac{ap^{2}+bp-aq^{2}-bq}{(p-q)}$$
$$=\dfrac{a(p-q)(p+q)+b(p-a)}{(p-q)}$$
$$2a\alpha+b=a(p+q)+b$$
$$\alpha=\left ( \dfrac{p+q}{2} \right )$$
Question 2
1 / -0

The arrangement of the following curves in the ascending order of slopes of their tangents at the given points.
$$A) \displaystyle y=\frac{1}{1+x^{2}}$$ at $$x=0$$

$$B) y=2e^{\frac{-x}{4}},$$ where it cuts the y-axis
$$C) y= cos(x)$$ at $$\displaystyle x=\frac{-\pi}{4}$$
$$D) y=4x^{2}$$ at $$x=-1$$

A
DCBA

B
ACBD

C
ABCD

D
DBAC

Solution

(A) $$y^{1}=-\frac{2x}{(1+x^{2})}$$
$$y^{1}|_{x=0}=0$$

(B) $$y^{1}=-\frac{2x}{(4}$$
$$y^{1}|_{x=0}=-\frac{1}{2}$$

(C) $$y^{1}=-sin\ x$$
$$y^{1}|_{x=^{-pi}/_4}=\frac{1}{\sqrt{2}}$$

(D) $$y^{1}=8x$$
$$y^{1}|_{x=-1}=-8$$

$$(C)>(A)>(B)>(D)$$
Question 3
1 / -0

Observe the following lists for the curve $$y=6+x-x^{2}$$ with the slopes of tangents at the given points; I, II, III, IV
Point
Tangent slope
I: $$(1, 6)$$
a) $$3$$
II: $$(2, 4)$$
b) $$5$$
III: $$(-1, 4)$$
c) $$-1$$
IV: $$(-2, 0)$$
d) $$-3$$

A
$$a ,b, c ,d$$

B
$$b, c, d ,a$$

C
$$c, d ,b ,a$$

D
$$c, d ,a ,b$$

Solution

$$y^{'}=\ \ \ 1-2x$$
(A) $$m_{1}=-1$$
(B) $$m_{2}=-3$$
(C) $$m_{3}=3$$
(D) $$m_{4}=5$$
Question 4
1 / -0

lf the tangent to the curve $$2y^{3}=ax^{2}+x^{3}$$ at the point $$(a,\ a)$$ cuts off intercepts $$\alpha$$ and $$\beta$$ on the coordinate axes such that $$\alpha^{2}+\beta^{2}=61$$, then $$a$$ is equal to

A
$$\pm 30$$

B
$$\pm 5$$

C
$$\pm 6$$

D
$$\pm 61$$

Solution

$$2\times 3\times y^{2}y'=2ax+3x^{2}$$
$$y' = \dfrac{2ax+3x^2}{6y^2}$$
$$m = \left.y'\right|_{(a,a)}=\dfrac{5}{6}$$

Equation of the tangent at $$(a,a)$$ with slope, $$m = \dfrac56$$
$$(y-a)=\dfrac{5}{6}(x-a)$$

Intersection points with coordinate axes are, $$\left(0,\dfrac{a}{6}\right)$$ and $$\left(-\dfrac{a}{5},0\right)$$
So, $$\alpha = \dfrac a6$$ and $$\beta = -\dfrac a5$$
Given that,
$${\alpha}^2 + {\beta}^2 = 61$$

$$\Rightarrow \dfrac{a^2}{36} + \dfrac{a^2}{25} = 61$$

$$\Rightarrow a^2 = {30}^2$$
$$\Rightarrow a = \pm 30$$

Question 5

1 / -0

Match List-I with List-II and select the correct answer using the code given below. A B C D

List-I	List-II
a) Equation of tangent to the curve $$y=be^{-x/a}$$ at $$x=0$$	1) $$x-2y=2$$
b) Equation of tangent to the curve $$y=x^{2}+1$$ at $$(1, 2)$$	2) $$y = 2x$$
c) Equation of normal to the curve $$y=2x-x^{2}$$at $$(2, 0)$$	3) $$x-y =\pi$$
d) Equation of normal to the curve $$y= \sin x$$ at $$x=\pi $$	4) $$ \displaystyle {\frac{x}{a}+\frac{y}{b}=1} $$

4 1 2 3

4 2 1 3

1 2 3 4

1 4 3 2

Solution

(A) $$\displaystyle y'=\frac{-b}{a}e^{-x/a}$$
$$\displaystyle m=-\frac{b}{a}$$
$$\displaystyle (y-b)=\frac{-b}{a}(x-0)$$
$$\displaystyle \frac{x}{a}+\frac{y}{b}=1$$

(B) $$y'=2x$$
$$y'|_{x=1}=2$$
$$(y-2)=2(x-1)$$
$$y=2x$$

(C) $$y'=2-2x$$
$$y'|_{x=2}=-2$$
Slope of normal $$\displaystyle =\frac{1}{2}$$
$$\displaystyle (y-0)=\frac{1}{2}(x-2)$$
$$2y=x-2$$

(D) $$y'=\cos x$$
$$y'|_{x={\pi}}=-1$$
Slope of normal =1
$$(y-0)=1(x-\pi)$$
$$x-y=\pi$$

Question 6
1 / -0

Area of the triangle formed by the tangent, normal at $$(1, 1)$$ on the curve $$\sqrt{x}+\sqrt{y}=2$$ and the y axis is (in sq. units)

A
$$1$$

B
$$2$$

C
$$\displaystyle \frac{1}{2}$$

D
$$4$$

Solution

$$\dfrac{1}{\sqrt{x}}+\dfrac{y^{1}}{\sqrt{y}}=0$$
$$y'=-\sqrt{\dfrac{y}{x}}$$
$$y'=-1\ \ at\ (n=1,\ y=1)$$
$$(y-1)=-1(x-1)$$
$$y=2$$
$$(y-1)=1(x-1)$$
$$y=0$$
$$Area=\dfrac{1}{2}\times 2 \times 1$$
$$=1$$
Question 7
1 / -0

Assertion (A): The points on the curve $$y=x^{3}-3x$$ at which the tangent is parallel to $$x$$-axis are $$(1, -2)$$ and $$(-1, 2).$$
Reason (R): The tangent at $$(x_{1}, y_{1})$$ on the curve $$y=f(x)$$ is vertical then $$\displaystyle \frac{dy}{dx}$$ at $$(x_{1}, y_{1})$$ is not defined.

A
Both A and R are true and R is the correct explanation for A

B
Both A and R are true but R is not the correct explanation for A

C
A is true but R is false

D
A is false but R is true

Solution

$$y'=3x^{2}-3$$
$$y'=0$$
$$\therefore$$ tangent is parallel to x-axis
$$x^{2}=1$$
$$x =^+_-1$$
$$y=-2$$
$$y=2$$
y tangent is vertical then $$\dfrac{dy}{dx}$$ is not defined.
Question 8
1 / -0

lf the tangent to the curve $$f(x)=x^{2}$$ at any point $$(c, f(c))$$ is parallel to the line joining points $$(a, f(a))$$ and $$(b,f(b))$$ on the curvel then $$a,\ c,\ b$$ are in

A
AP

B
GP

C
HP

D
AGP

Solution

$$\partial^{'}M=2s$$ at $$(c,\ \partial(c))=2c$$
$$\dfrac{\partial(a)-\partial(L)}{a-b}=2c$$
$$\dfrac{a^{2}=b^{2}}{a-b}=2c$$
$$a+b=2c$$
$$\therefore\ a,c,b$$ are in AP
Question 9
1 / -0

I. lf the curve $$y=x^{2}+ bx +c$$ touches the straight line $$y=x$$ at the point $$(1, 1)$$ then $$b$$ and $$c$$ are given by $$1, 1.$$
II. lf the line $$Px+ my +n=0$$ is a normal to the curve $$xy=1$$, then $$P> 0,\ m <0$$.
Which of the above statements is correct

A
only I

B
only II

C
both I and II

D
Neither I nor II

Solution

(I) $$1+b+c=1$$
$$b+c=0$$
$$y'=2x+b$$
$$y'=1$$
$$2+b=1$$
$$(b=-1)$$
$$c=1$$
(II) $$y'=-\dfrac{1}{x^{2}}$$
$$m=-\dfrac{1}{x^{2}}$$
$$-\dfrac{P}{m}=\dfrac{x^{2}}{x}$$
$$-P=x^{2}m$$
p and m will be of opposite num.
Question 10
1 / -0

lf the parametric equation of a curve given by $$x=e^{t}\cos t,\ y=e^{t}\sin t$$, then the tangent to the curve at the point $$t=\dfrac{\pi}{4}$$ makes with axis of $$x$$ the angle.

A
$$0$$

B
$$\dfrac{\pi}{4}$$

C
$$\dfrac{\pi}{3}$$

D
$$\dfrac{\pi}{2}$$

Solution

$$at\ t={\pi}/{4}$$
$$y=e^{t}sint$$
$$\dfrac{dy}{dl}=e^{t}(sin\ t+cos\ t)$$
$$\dfrac{dx}{dl}=d^{t}(cos\ t+sin\ t)$$
$$\dfrac{dy}{dx}=\dfrac{sin \ t+cos\ t}{cos\ t-sin\ t}$$
$$= \alpha$$
$$\therefore tan^{-1}(\infty)={\pi}/{2}$$

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Re-Attempt Weekly Quiz Competition

Point	Tangent slope
I: $$(1, 6)$$	a) $$3$$
II: $$(2, 4)$$	b) $$5$$
III: $$(-1, 4)$$	c) $$-1$$
IV: $$(-2, 0)$$	d) $$-3$$

Tangents and its Equations Test 11

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Tangents and its Equations Test 11

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SHARING IS CARING

Question 1

$$2pq$$

$$\sqrt{pq}$$

$$\displaystyle \frac{p+q}{2}$$

$$\displaystyle \frac{p-q}{2}$$

Solution

Question 2

DCBA

ACBD

ABCD

DBAC

Solution

Question 3

$$a ,b, c ,d$$

$$b, c, d ,a$$

$$c, d ,b ,a$$

$$c, d ,a ,b$$

Solution

Question 4

$$\pm 30$$

$$\pm 5$$

$$\pm 6$$

$$\pm 61$$

Solution

Question 5

4 1 2 3

4 2 1 3

1 2 3 4

1 4 3 2

Solution

Question 6

$$1$$

$$2$$

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

$$4$$

Solution

Question 7

Both A and R are true and R is the correct explanation for A

Both A and R are true but R is not the correct explanation for A

A is true but R is false

A is false but R is true

Solution

Question 8

AP

GP

HP

AGP

Solution

Question 9

only I

only II

both I and II

Neither I nor II

Solution

Question 10

$$0$$

$$\dfrac{\pi}{4}$$

$$\dfrac{\pi}{3}$$

$$\dfrac{\pi}{2}$$

Solution

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