Differentiation Test 27

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Differentiation Test 27

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Weekly Quiz Competition

Question 1
1 / -0

$$(9, 2), (5, -1) $$ and $$ (7, -5)$$ are the vertices of the triangle. Find its area.

A
10 square units

B
11 square units

C
12 square units

D
13 square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 9$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = -1$$, $$x_{3} = 7$$ and $$y_{3} = -5$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[9(-1 + 5) + 5(-5 - 2) + 7(2 + 1)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 35 + 21]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 22\right|$$
$$=$$ $$\left|11 \right|$$
Area always in absolute value.
So, area of the triangle $$= 11$$ square units.
Question 2
1 / -0

What is the area of the triangle whose vertices are: $$(-3, 15), (6, -7) $$ and $$(10, 5)$$?

A
$$94$$ square units

B
$$96$$ square units

C
$$97$$ square units

D
$$98$$ square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is $$A=\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
Here $$x_{1} = -3$$, $$y_{1} = 15$$, $$x_{2} = 6$$, $$y_{2} = -7$$, $$x_{3} = 10$$ and $$y_{3} = 5$$

Substituting the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[-3(-7 - 5) + 6(5 - 15) + 10(15 + 7)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 60 + 220]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 196\right|$$
$$=$$ $$\left|98 \right|$$

Area is always in absolute value.
So, area of the triangle $$= 98$$ square units.
Question 3
1 / -0

Identify the missing integer: $$9, 45,$$ ____$$, 1125, 5625...$$

A
$$220$$

B
$$223$$

C
$$224$$

D
$$225$$

Solution

This is continuous series multiplied by $$5$$.
$$9 \times 5 = 45$$
$$45 \times 5 = 225$$
$$225 \times 5 = 1,125$$
$$1,125 \times 5= 5,625.$$
So, the missing integer is $$225$$.
Question 4
1 / -0

What is the area of the triangle for the following points $$(6, 2), (5, 4)$$ and $$(3, -1)$$?

A
2.3 square units

B
4.5 square units

C
4.1 square units

D
3.6 square units

Solution

Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 6$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = 4$$, $$x_{3} = 3$$ and $$y_{3} = -1$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[6(4 + 1) + 5(-1 - 2) + 3(2 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[30 - 15 - 6]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 9\right|$$
$$=$$ $$\left|4.5 \right|$$
Area always in absolute value.
So, area of the triangle $$= 4.5$$ square units.
Question 5
1 / -0

The area of the triangle whose vertices are $$(0, 1), (1, 4)$$ and $$(1, 2)$$ is ___ square units.

A
1

B
2

C
3

D
4

Solution

Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 0$$, $$y_{1} = 1$$, $$x_{2} = 1$$, $$y_{2} = 4$$, $$x_{3} = 1$$ and $$y_{3} = 2$$
Substitute the values, we get
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[0(4 - 2) + 1(2 - 1) + 1(1 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[0 + 1 - 3]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times -2\right|$$
$$=$$ $$\left|- 1 \right|$$
Area always in absolute value.
So, area of the triangle is $$ 1$$ square units.
Question 6
1 / -0

If $$r=\left[2\phi +\cos^2\left(2\phi +\dfrac{\pi}4\right)\right]^{\tfrac12},$$ then what is the value of the derivative of $$\dfrac{dr}{d\phi}$$ at $$\phi=\dfrac{\pi}4?$$

A
$$2\left(\displaystyle\frac{1}{\pi+1}\right)^{\tfrac12}$$

B
$$2\left(\displaystyle\frac{2}{\pi+1}\right)^{2}$$

C
$$\left(\displaystyle\frac{2}{\pi+1}\right)^{\tfrac12}$$

D
$$2\left(\displaystyle\frac{2}{\pi+1}\right)^{\tfrac12}$$

Solution

$$r=\sqrt { 2\phi +{ cos }^{ 2 }\left( 2\phi +\dfrac { \pi }{ 4 } \right) } $$
$$\phi =\dfrac { \pi }{ 4 } $$

$$\Rightarrow r=\sqrt { 2\times \dfrac { \pi }{ 4 } +{ cos }^{ 2 }\left( 2\times \dfrac { \pi }{ 4 } +\dfrac { \pi }{ 4 } \right) } $$

$$=\sqrt { \dfrac { \pi }{ 2 } +{ cos }^{ 2 }\left( \dfrac { 3\pi }{ 4 } \right) } $$

$$r\left( \dfrac { \pi }{ 4 } \right) =\sqrt { \dfrac { \pi }{ 2 } +\dfrac { 1 }{ 2 } } =\dfrac { \sqrt { \pi +1 } }{ \sqrt { 2 } } $$

$${ r }^{ 2 }=2\phi +{ cos }^{ 2 }\left( 2\phi +\dfrac { \pi }{ 4 } \right) $$

Differentiate w.r.t $$\phi $$

$$\Rightarrow 2r\dfrac { dr }{ d\phi } =2+2\cos { \left( 2\phi +\dfrac { \pi }{ 4 } \right) } \left( -\sin { \left( 2\phi +\dfrac { \pi }{ 4 } \right) } \right) \left( +2 \right) $$

$$2r\left( \dfrac { \pi }{ 4 } \right) \dfrac { dr }{ d\phi } =2+2\cos { \dfrac { 3\pi }{ 4 } \left( -\sin { \left( \dfrac { 3\pi }{ 4 } \right) } \right) } \left( +2 \right) $$

$$=2+2\times \dfrac { -1 }{ \sqrt { 2 } } \times \dfrac { -1 }{ \sqrt { 2 } } \times \left( +2 \right) $$

$$2\times \left( \sqrt { \dfrac { \pi +1 }{ 2 } } \right) \times \dfrac { dr }{ d\phi } =4$$

$$\dfrac { dr }{ d\phi } =\dfrac { 2\sqrt { 2 } }{ \sqrt { \pi +1 } } $$
Question 7
1 / -0

$$f(x) = \log \left (e^{x} \left (\dfrac {x - 2}{x + 2}\right )^{\dfrac {3}{4}} \right ) \Rightarrow f'(0) =$$

A
$$\dfrac {1}{4}$$

B
$$4$$

C
$$\dfrac {-3}{4}$$

D
$$1$$

Solution

$$f(x) = \log \left (e^{x} \left (\dfrac {x - 2}{x + 2}\right )^{\dfrac {3}{4}}\right )$$

$$f(x) = x + \dfrac {3}{4} [\log (x - 2) - \log (x + 2)]$$

$$f'(x) = 1 + \dfrac {3}{4} \left [\dfrac {1}{x - 2} - \dfrac {1}{x + 2} \right ]$$

$$f'(x) = \left [1 + \dfrac {3}{x^{2} - 4}\right ]$$

$$f'(0) = 1 -\dfrac {3}{4} = \dfrac {1}{4}$$
Question 8
1 / -0

If n is an integer with $$0\le n \le 11$$ then the minimum value of $$n!(11-n)!$$ is attained when a value of n =

A
11

B
5

C
7

D
9

Solution

Let $$y=n!(11-n)!$$
Consider $$x= ^{11}C_n=\dfrac{11!}{n!(11-n)!}$$
For maximum value of $$x$$ we must have $$n=6$$ or $$n=5$$
Thus $$x_{max}=^{11}C_6=\dfrac{11!}{5!6!}=\dfrac{11!}{y_{min}}$$
$$\Rightarrow y_{min}=5!6!$$ i.e. $$n=6$$ or $$n=5$$
Question 9
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$$6, 10, 18, 34, 66$$
The first number in the list above is $$6$$. Determine a rule for finding each successive number in the list.

A
Add $$4$$ to the preceding number.

B
Take $$\displaystyle \frac { 1 }{ 2 } $$ of the preceding number and then add $$7$$ to that result.

C
Double the preceding number and then subtract $$2$$ from that result.

D
Subtract $$2$$ from the preceding number and then double that result.

E
Triple the preceding number and then subtract $$8$$ from that result.

Solution

The series is $$6,10,18,34,66$$
In the given series:
$$6+6=12-2=10$$
$$10+10=20-2=18$$
$$18+18=36-2=34$$
$$34+34=68-2=66$$
Hence in this each successive number is obtained by double the preceding number and then subtract $$2$$ from the result.
Question 10
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If $$y = \tan^{-1} \left (\dfrac {1}{1 + x + x^{2}}\right ) + \tan^{-1} \left (\dfrac {1}{x^{2} + 3x + 2}\right ) + \tan^{-1} \left (\dfrac {1}{x^{2} + 5x + 6}\right ) + .... +$$ upto $$n$$ terms then $$\dfrac {dy}{dx}$$ at $$x = 0$$ and $$n = 1$$ is equal to

A
$$\dfrac {1}{2}d$$

B
$$-\dfrac {1}{2}$$

C
$$0$$

D
$$\dfrac {1}{3}$$

Solution

$$y= \tan^{-1} \left (\dfrac {1}{1 + x(x + 1)}\right ) + \tan^{-1}\left (\dfrac {1}{1 + (x + 1)(x + 2)}\right ) +\\ \quad\tan^{-1} \left (\dfrac {1}{1 + (x + 2)(x + 3)}\right ) + ..... + \tan^{-1} \left (\dfrac {1}{1 + (x + n - 1) (x + n)}\right )$$

when $$n = 1$$
$$y = \tan^{-1} \left (\dfrac {1}{1 + x(x + 1)}\right ) = \tan^{-1} \left (\dfrac {(x + 1) - x}{1 + x(x + 1)}\right ) = \tan^{-1} (x + 1) - \tan^{-1} x$$

$$\Rightarrow \dfrac {dy}{dx} = \dfrac {1}{1 + (x + 1)^{2}} - \dfrac {1}{1 + x^{2}} \\ \therefore \dfrac {dy}{dx}]_{x = 0} = \dfrac {1}{1 + 1} - \dfrac {1}{1} = \dfrac {1}{2} - 1 = -\dfrac {1}{2}$$

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Re-Attempt Weekly Quiz Competition

Differentiation Test 27

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Differentiation Test 27

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SHARING IS CARING

Question 1

10 square units

11 square units

12 square units

13 square units

Solution

Question 2

$$94$$ square units

$$96$$ square units

$$97$$ square units

$$98$$ square units

Solution

Question 3

$$220$$

$$223$$

$$224$$

$$225$$

Solution

Question 4

2.3 square units

4.5 square units

4.1 square units

3.6 square units

Solution

Question 5

1

2

3

4

Solution

Question 6

$$2\left(\displaystyle\frac{1}{\pi+1}\right)^{\tfrac12}$$

$$2\left(\displaystyle\frac{2}{\pi+1}\right)^{2}$$

$$\left(\displaystyle\frac{2}{\pi+1}\right)^{\tfrac12}$$

$$2\left(\displaystyle\frac{2}{\pi+1}\right)^{\tfrac12}$$

Solution

Question 7

$$\dfrac {1}{4}$$

$$4$$

$$\dfrac {-3}{4}$$

$$1$$

Solution

Question 8

11

5

7

9

Solution

Question 9

Add $$4$$ to the preceding number.

Take $$\displaystyle \frac { 1 }{ 2 } $$ of the preceding number and then add $$7$$ to that result.

Double the preceding number and then subtract $$2$$ from that result.

Subtract $$2$$ from the preceding number and then double that result.

Triple the preceding number and then subtract $$8$$ from that result.

Solution

Question 10

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

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

$$0$$

$$\dfrac {1}{3}$$

Solution

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