Application of Derivatives Test

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Application of Derivatives Test - 20

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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

The point on the hyperbola $$y = \dfrac {x - 1}{x + 1}$$ at which the tangents are parallel to $$y = 2x + 1$$ are

A
$$(0, -1)$$ only

B
$$(-2, 3)$$ only

C
$$(0, -1)$$, $$(-2, 3)$$

D
$$(-2, 3)$$, $$(5, 4)$$

Solution

$${y}'=\dfrac{(x+1)-(x-1)}{(x+1)^{2}}$$
$${y}'=2 (\therefore$$ parallel to $$y =2x+1)$$
$$\dfrac{2}{(x+1)^{2}}=2$$
$$(x+1)^{2}=1$$
$$x=0$$ or $$x=-2$$
Using the equation of hyperbola, when $$x=0$$, $$y=-1$$.
when $$x=-2$$, $$y=3$$
Question 5
1 / -0

The number of tangents to the curve $$x^{3/2}+y^{3/2}=a^{3/2}$$, where the tangents are equally inclined to the axes, is

A
$$2$$

B
$$1$$

C
$$0$$

D
$$4$$

Solution

$$\dfrac{dy}{dx}=-1$$ (where x is equally inclined)
$$\dfrac{3}{2}x^{^{1}/_{2}}+^{3}/_{2}y^{^{1}/_{2}}\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}=-\sqrt{\dfrac{x}{y}}=1$$
$$-\sqrt{\dfrac{x}{y}}=-1$$
$$\sqrt{x}=\sqrt{y}$$
$$x=y$$
$$x^{^{3}/_{2}}+x^{^{3}/_{2}}=a^{^{3}/_{2}}$$
$$2x^{^{3}/_{2}}=a^{^{3}/_{2}}$$
$$x=\dfrac{a}{2^{^{2}/_{3}}}$$
$$y=\dfrac{a}{2^{^{2}/_{3}}}$$
$$\therefore$$ only one tangent.
Question 6
1 / -0

The points on the hyperbola $$x^{2}-y^{2}=2$$ closest to the point (0, 1) are

A
$$(\displaystyle \pm\frac{3}{2},\frac{1}{2})$$

B
$$(\displaystyle \frac{1}{2},\pm\frac{3}{2})$$

C
$$(\displaystyle \frac{1}{2},\frac{1}{2})$$

D
$$(\displaystyle \pm\frac{3}{4}\pm\frac{3}{2})$$

Solution

Let point be $$(\sqrt 2 sec\theta,\sqrt 2 tan\theta)$$ closest to (0, 1)
So distance $$=\sqrt {2 sec^2\theta+(\sqrt 2 tan\theta-1)^2}$$
$$=\sqrt {2 sec^2\theta+2tan^2\theta-2\sqrt 2 tan\theta+1}$$
$$=\sqrt {4 tan^2\theta-2\sqrt 2 tan\theta+3}$$
$$=2\sqrt {tan^2\theta-\frac {tan\theta}{\sqrt 2}+\frac {3}{4}}$$
$$=2\sqrt {(tan\theta-\frac {1}{2\sqrt 2})^2+\frac {3}{4}-\frac {1}{8}}$$
So distance is min when $$tan\theta=\frac {1}{2\sqrt 2}$$
$$sec\theta=\pm \frac {3}{2\sqrt 2}$$
Question 7
1 / -0

If the circle $$x^2 + y^2 + 2gx + 2fy + c =0$$ is touched by y = x at P in the first quadrant, such that $$OP = 6 \sqrt2$$, then the value of $$c$$ is

A
36

B
144

C
72

D
None of these

Solution

Let the point of contact be $$x_{1},y_{1}$$
However it lies on the line $$y=x$$
Hence
$$x_{1}=y_{1}$$
Applying distance formula, we get
$$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$$
$$2x_{1}^2=72$$
$$x_{1}=6$$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.
$$2x+2yy'+2g+2fy'=0$$
Now $$y'=1$$ since $$y'$$ is the slope of the line $$y=x$$.
By substituting, we get
$$x+y=-(g+f)$$
Now $$x_{1}=y_{1}=6$$
Hence
$$12=-(g+f)$$ ...(i)
Substituting $$x=y=6$$ in the equation of the circle, we get
$$36+36+2(6)(g+f)+c=0$$
$$72+12(-12)+c=0$$
$$c=144-72$$
$$c=72$$
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

$$\Delta (x)=\begin{vmatrix}
\sin x & \cos x &\sin 2x+\cos 2x \\
0 &1 &1 \\
1 &0 &-1
\end{vmatrix}$$
$${\Delta }'(x)$$ vanishes at least once in

A
$$\left(0, \dfrac{\pi}2\right)$$

B
$$\left(\dfrac{\pi}{2}, \pi \right)$$

C
$$\left(0, \dfrac{\pi}4\right)$$

D
$$\left(-\dfrac{\pi}{2}, 0\right)$$

Solution

$$\Delta' (x)=\begin{vmatrix}
\cos x &-\sin x &2\cos 2x-2 \sin 2x \\ 0 &1 &1 \\ 1 &0 &-1 \end{vmatrix}$$

$$\Rightarrow \Delta' (x)= -\cos x - \sin x -1(2\cos2x - 2\sin2x)$$
(Expanded along first column)

$$\Rightarrow \Delta' (x)= 2\sin2x -\cos x - \sin x -2\cos2x $$

Clearly,
$$ \Delta' (x)$$ is continuous and $$ \Delta' (0) =-3 < 0 $$ and $$ \Delta' \left(\dfrac{\pi}{2}\right)= 1 > 0$$
Hence,
$$ \Delta' (x)$$ will vanish at least once in $$\left(0, \dfrac{\pi}{2}\right)$$

Hence, option A.
Question 10
1 / -0

For the curve $$ y=3\sin \theta\cos\theta, x=e^{\theta}\sin \theta, 0\leq \theta\leq\pi$$; the tangent is parallel to $$x$$ -axis when $$\theta$$ is

A
$$0$$

B
$$\displaystyle \frac{\pi}{2}$$

C
$$\displaystyle \frac{\pi}{4}$$

D
$$\displaystyle \frac{\pi}{6}$$

Solution

$$\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{d\theta}=d\dfrac{3}{2}(sin2\theta)$$
$$\dfrac{dy}{d\theta}=d\dfrac{3}{4}(cos2\theta)$$ ---- (1)
$$\dfrac{dy}{d\theta}=e^{\theta}sin\theta+e^{\theta}cos\theta$$ ---- (2)
$$\dfrac{dx}{dy}=\dfrac{3}{4}\dfrac{cos2\theta}{e^{\theta}(sin\ \theta+ cos\ \theta)}=0$$
$$cos2\theta=0$$
$$2\theta={\pi}/{2}$$
$$\theta={\pi}/{4}$$

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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) $$-3$$

Application of Derivatives Test - 20

Result

Application of Derivatives Test - 20

Score

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Rank

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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

$$(0, -1)$$ only

$$(-2, 3)$$ only

$$(0, -1)$$, $$(-2, 3)$$

$$(-2, 3)$$, $$(5, 4)$$

Solution

Question 5

$$2$$

$$1$$

$$0$$

$$4$$

Solution

Question 6

$$(\displaystyle \pm\frac{3}{2},\frac{1}{2})$$

$$(\displaystyle \frac{1}{2},\pm\frac{3}{2})$$

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

$$(\displaystyle \pm\frac{3}{4}\pm\frac{3}{2})$$

Solution

Question 7

36

144

72

None of these

Solution

Question 8

AP

GP

HP

AGP

Solution

Question 9

$$\left(0, \dfrac{\pi}2\right)$$

$$\left(\dfrac{\pi}{2}, \pi \right)$$

$$\left(0, \dfrac{\pi}4\right)$$

$$\left(-\dfrac{\pi}{2}, 0\right)$$

Solution

Question 10

$$0$$

$$\displaystyle \frac{\pi}{2}$$

$$\displaystyle \frac{\pi}{4}$$

$$\displaystyle \frac{\pi}{6}$$

Solution

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