Application of Derivatives Test

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Application of Derivatives Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

If there is an error of $$0.01 cm$$ in the diameter of a sphere then percentage error in surface area when the radius $$= 5 cm$$, is

A
$$0.005\%$$

B
$$0.05\%$$

C
$$0.1\%$$

D
$$0.2\%$$

Solution

Surface area of sphere $$S=4\pi r^{2}$$
$$\displaystyle S=\pi D^{2}$$
$$\Rightarrow S=100\pi$$
Also, $$ \displaystyle \frac{dS}{dD}=2\pi D=20\pi$$ [r=5]
Approximate error in S is
$$\displaystyle dS=(\frac{dS}{dD})\Delta D$$
$$ =20\pi (0.01)$$

then
$$\dfrac{dS}{S}=\dfrac{1}{500} $$
$$dS=0.2$$% of S
Percentage error in $$S=0.2%$$
Question 2
1 / -0

If errors of $$1\%$$ each are made in the base radius and height of a cylinder, then the percentage error in its volume is

A
$$1\%$$

B
$$2\%$$

C
$$3\%$$

D
none of these

Solution

Given, percentage error in r is 1%
$$\Rightarrow \displaystyle \frac{\Delta r}{r}=\frac{1}{100}$$

$$\Rightarrow \displaystyle \Delta r=\frac{r}{100}$$
Also given, percentage error in h is 1%
$$\Rightarrow \displaystyle \frac{\Delta h}{h}=\frac{1}{100}$$

$$\Rightarrow \displaystyle \Delta h=\frac{h}{100}$$

Now, volume of cylinder $$V=\pi r^{2}h$$
$$\dfrac{dV}{dr}=\pi2rh+\pi r^2\dfrac{dh}{dr} $$
$$\Delta V=\pi [r^{2}\Delta h+2rh\Delta r]$$
$$\displaystyle \Delta V=\pi[r^{2}\frac{h}{100}+2rh\frac{r}{100}]$$

$$\displaystyle\Delta V=\pi r^{2}h[\frac{3}{100}]$$
$$\Rightarrow\displaystyle\frac{\Delta V}{V}=\frac{3}{100}$$
Percentage error in V is 3%
Question 3
1 / -0

If $$y = 4x - 5$$ is a tangent to the curve $$\displaystyle y^2 = px^3 + q$$ at $$(2, 3)$$, then

A
$$p = 2, q = -7$$

B
$$p = -2, q = 7$$

C
$$p = -2, q = -7$$

D
$$p = 2, q = 7$$

Solution

Given equation of curve
$$\displaystyle y^2 = px^3 + q\\$$
$$\displaystyle \dfrac{dy}{dx}=\dfrac{3px^{2}}{y}\\$$
Slope of tangent at (2,3) $$= \displaystyle (\dfrac{dy}{dx})_{(2,3)}=2p$$

Given equation of tangent is
$$y=4x-5$$
Slope of tangent =4

$$\Rightarrow 2p=4$$
$$\Rightarrow p=2$$

Since, (2,3) lies on
$$ y^{2} = px^{3} + q$$
$$9=16+q$$
$$\Rightarrow q=-7$$
Question 4
1 / -0

The point(s) at each of which the tangents to the curve $$\displaystyle y = x^3 - 3x^2 - 7x + 6$$ cut off on the positive semi axis $$OX$$ a line segment half that on the negative semi axis $$OY$$, then the co-ordinates of the point(s) is/are give by:

A
$$(-1, 9)$$

B
$$(3, -15)$$

C
$$(1, -3)$$

D
none

Solution

$$y={ x }^{ 3 }-3{ x }^{ 2 }-7x+6\\ \dfrac { dy }{ dx } =3{ x }^{ 2 }-6x-7$$
Let coordinate of X be $$\left( a,0 \right) $$ then coordinate of Y is $$\left( 0,-2a \right) $$
Therefore slope of XY is $$\cfrac { 0+2a }{ a-0 } =2$$
Equating slope $$3{ x }^{ 2 }-6x-7=2\\ 3{ x }^{ 2 }-6x-9=0$$
Solving we get
$$x=-1,y=9\quad \\ x=3,y=-15$$
Hence, option 'B' is correct.
Question 5
1 / -0

If the tangent to the curve $$xy + ax + by = 0$$ at (1, 1) makes an angle $$\displaystyle \tan ^{-1}(2)$$ with x-axis, then $$\displaystyle a + 2b$$ is equal to

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

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

C
$$3$$

D
$$-3$$

Solution

$$xy'+y+a+by'=0$$
$$(x+b)y'+a+y=0$$
$$y'=\dfrac{-(a+y)}{x+b}$$
Now
$$y'_{1,1}=\dfrac{-(a+1)}{b+1}$$
$$=2$$
Or
$$2b+2=-a-1$$
$$2b+a=-3$$
Question 6
1 / -0

If the tangent at each point of the curve $$\displaystyle y=\frac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$$ makes an acute angle with the positive direction of x-axis, then

A
$$a\ge 1$$

B
$$-1\le a\le 1$$

C
$$a\le -1$$

D
none of these

Solution

We have, $$\displaystyle y=\dfrac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$$
$$\displaystyle \Rightarrow \dfrac { dy }{ dx } =2{ x }^{ 2 }-4ax+2.$$
Since, the tangent makes an acute angle with the positive direction of x-axis, therefore,
$$\displaystyle \dfrac { dy }{ dx } \ge 0\Rightarrow 2{ x }^{ 2 }-4ax+2\ge 0$$ for all $$x$$
$$\Rightarrow 16{ a }^{ 2 }-16\le 0$$
$$(\because $$ Disc. $$={ \left( 4a \right) }^{ 2 }-4\left( 2 \right) \left( 2 \right) \le 0)$$
$$\Rightarrow { a }^{ 2 }-1\le 0$$ i.e., $$\left( a-1 \right) \left( a+1 \right) \le 0$$
$$\Rightarrow -1\le a\le 1.$$
Question 7
1 / -0

The set of values of $$a$$ for which the function $$\displaystyle f(x) = (4a - 3) (x + \ln5) + 2(a - 7) \cot \frac {x}{2} \sin^2 \frac {x}{2}$$ does not posses critical points in its domain is

A
$$\displaystyle (- \infty, - \frac {4}{3}) \cup (2, \infty)$$

B
$$\displaystyle (- \infty, 2)$$

C
$$\displaystyle [1, \infty)$$

D
$$\displaystyle (1, \infty)$$

Solution

$$f\left( x \right) =\left( 4a-3 \right) \left( x+\log 5 \right) +2\left( a-7 \right) \cot \left( \dfrac x2 \right) \sin ^{ 2 } \left( \dfrac x2 \right) $$
$$\Rightarrow f\left( x \right) =\left( 4a-3 \right) \left( x+\log 5 \right) +\left( a-7 \right) \sin { x } $$
$$f(x)$$ posses critical points when $$f'\left( x \right) =\left( 4a-3 \right) +\left( a-7 \right) \cos { x } =0$$
$$\Rightarrow \cos { x } =-\dfrac { 4a-3 }{ a-7 } $$
$$\Rightarrow -1\le -\dfrac { 4a-3 }{ a-7 } \le 1$$
$$\Rightarrow -\dfrac { 4 }{ 3 } \le a\le 2$$
Therefore, $$f(x)$$ does not posses critical points when $$a\in (-\infty,-\dfrac 43)\cup (2,\infty)$$

Ans: A
Question 8
1 / -0

Let $$f\left( x \right) =\left\{ \begin{matrix} { x }^{ { 3 }/{ 5 } }\quad \quad \quad x\le 1 \\ -{ \left( x-2 \right) }^{ 3 }\quad x>1 \end{matrix} \right. $$
then the number of critical points on the graph of the function is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$f\left( x \right) =\left\{ \begin{matrix} { x }^{ { 3 }/{ 5 } }\quad \quad \quad x\le 1 \\ -{\left( x-2 \right) }^{ 3 }\quad x>1 \end{matrix} \right. $$
Critical points will at
At $$x=2$$, as$${ f }^{ ' }(x)=0$$
At $$x=0$$, $${ f }^{ ' }(x)$$ is not defined
At $$x=1$$, $${ f }^{ ' }(x)$$ does not exist , therefore its a critical point.
Thus, there are three critical points.
Hence, option 'C' is correct.
Question 9
1 / -0

Let $$f$$ be a decreasing function in $$(a,b]$$, then which of the following must be true?

A
$$f$$ is continuous at $$b$$

B
$$\displaystyle f'(b)<0$$

C
$$\displaystyle \lim_{x\to b}f(x)\leq f(b)$$

D
$$\displaystyle \lim_{x\to b}f(x)\geq f(b)$$

Solution

$$f$$ is a decreasing function in $$(a,b]$$
So $$f'(x) \le 0 $$ not strictly decreasing.
Hence $$f(b) \le f(x)$$ in $$(a,b]$$
Regarding continuity, only LHS known at b, no information about RHS.
Question 10
1 / -0

$$\displaystyle f:(0, \infty) \rightarrow (-\frac {\pi}{2}, \frac {\pi}{2})$$ be defined as, $$\displaystyle f(x) = arc \: \tan( \: x)$$
The above function can be classified as

A
injective but not surjective

B
surjective but not injective

C
neither injective nor surjective

D
both injective as well as surjective

Solution

$$f(x) = \tan^{-1}x$$
$$f'(x) = \cfrac{1}{1+x^2} > 0 \forall x \in R$$
Thus $$f(x)$$ is injective.
Also $$f(x)$$ is surjective as co-domain and range are same.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 24

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Application of Derivatives Test - 24

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SHARING IS CARING

Question 1

$$0.005\%$$

$$0.05\%$$

$$0.1\%$$

$$0.2\%$$

Solution

Question 2

$$1\%$$

$$2\%$$

$$3\%$$

none of these

Solution

Question 3

$$p = 2, q = -7$$

$$p = -2, q = 7$$

$$p = -2, q = -7$$

$$p = 2, q = 7$$

Solution

Question 4

$$(-1, 9)$$

$$(3, -15)$$

$$(1, -3)$$

none

Solution

Question 5

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

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

$$3$$

$$-3$$

Solution

Question 6

$$a\ge 1$$

$$-1\le a\le 1$$

$$a\le -1$$

none of these

Solution

Question 7

$$\displaystyle (- \infty, - \frac {4}{3}) \cup (2, \infty)$$

$$\displaystyle (- \infty, 2)$$

$$\displaystyle [1, \infty)$$

$$\displaystyle (1, \infty)$$

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 9

$$f$$ is continuous at $$b$$

$$\displaystyle f'(b)<0$$

$$\displaystyle \lim_{x\to b}f(x)\leq f(b)$$

$$\displaystyle \lim_{x\to b}f(x)\geq f(b)$$

Solution

Question 10

injective but not surjective

surjective but not injective

neither injective nor surjective

both injective as well as surjective

Solution

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