Differentiation Test 22

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Differentiation Test 22

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Weekly Quiz Competition

Question 1
1 / -0

The triangle with vertices A(4, 4), B(-2, -6) and C(4, -1) is shown in the diagram. The area of $$\Delta$$ ABC is _______

A
5 sq. units

B
12 sq. units

C
15 sq. units

D
20 sq. units

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$= \cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$

$$=\dfrac{1}{2}|4(-6-(-1))-2(-1-4)+4(4-(-6))|$$

$$=\dfrac{1}{2}|4(-5)-2(-5)+4(10)|$$

$$=\dfrac{1}{2}|30|$$

$$=15 $$ sq. units
Question 2
1 / -0

Three points A, B and C have coordinates $$(a, b + c), \ (b, c + a)$$ and $$(c, a + b)$$, respectively. The area of the triangle ABC will be:

A
$$\displaystyle a^{2}+b^{2}+c^{2}$$

B
$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}2$$

C
$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}4$$

D
$$0$$

Solution

Area of triangle $$=\dfrac{ 1 }{ 2 }\left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] $$

Given the points: $$A(a, b+c) , B(b, c+a) , C(c, a+b)$$
Therefore, area:

$$A =\dfrac{ 1 }{ 2 }\left[ a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a) \right]$$

$$A =\dfrac{ 1 }{ 2}{\left[ a(c-b)+b(a-c)+c(b-a) \right] }$$

$$A=\dfrac{ 1 }{ 2}{\left[ ac-ab+ba-bc+cb-ac \right] }=0$$
Question 3
1 / -0

The number of positive fractions $$\dfrac{m}{n}$$ such that $$\dfrac{1}{3}< \dfrac{m}{n}<1 $$ and having the property that the fraction remains the same by adding some positive integer to the numerator and multiplying the denominator by the same positive integer is:

A
$$1$$

B
$$3$$

C
$$6$$

D
infinite

Solution

Given $$m,n$$ are integers $$\dfrac{1}{3}<\dfrac{m}{n}<1$$
$$\Rightarrow \dfrac{m+x}{nx}=\dfrac{m}{n}$$
$$\Rightarrow m+x=mx$$
$$\Rightarrow x=\dfrac{m}{m-1},$$ $$x$$ must be an integer.
So only possible value of $$m$$ satisfying above equation is $$2.$$
$$\Rightarrow m=2$$
$$\Rightarrow \dfrac{m}{n}>\dfrac{1}{3}\Rightarrow 3m>n\Rightarrow 6>n$$

$$\Rightarrow \dfrac{m}{n}<1\Rightarrow m<n\Rightarrow 2<n$$
$$\Rightarrow n=3,4,5$$
$$\therefore 3$$ Possible fractions.
Hence, the answer is $$3.$$
Question 4
1 / -0

$$\displaystyle \frac{d}{dx}\left ( x^{\log x} \right )$$ is equal to

A
$$\displaystyle 2x^{\log x-1}\log x$$

B
$$\displaystyle x^{\log x-1}$$

C
$$\dfrac 23 (\log x)$$

D
$$\displaystyle x^{\log x-1}.\log x$$

Solution

$$\displaystyle \dfrac { d }{ dx } \left( x^{ \log x } \right) =\dfrac { d }{ dx } \left( { e }^{ \log { x } \log { x } } \right) $$
$$\displaystyle =\dfrac { d }{ dx } \left( { \left( \log { x } \right) }^{ 2 } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$$
$$\displaystyle =2\log { x } \dfrac { d }{ dx } \left( \log { x } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$$
$$\displaystyle ={ e }^{ { \left( \log { x } \right) }^{ 2 } }2\log { x } \left( \dfrac { d }{ dx } \left( \log { x } \right) \right) $$
$$\displaystyle =\dfrac { 1 }{ x } 2\log { x } { e }^{ { \left( \log { x } \right) }^{ 2 } }=2x^{ \log { x } -1 }\log { x } $$
Question 5
1 / -0

If $$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$$, then $$y'(0)$$ is equal to

A
$$-1$$

B
$$1$$

C
$$2$$

D
Non existant

Solution

$$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$$
Take log both sides,
$$\displaystyle \log y = \frac{1}{3}\log(1+3x)+\frac{1}{4}\log(1+4x)+\frac{1}{5}\log(1+5x)-\frac{1}{7}\log(1+7x)-\frac{1}{8}\log(1+8x)$$
Differentiating both side w.r.t $$x$$
$$\displaystyle

\frac {1}{y}\times \frac {dy}{dx}=\frac {1}{(1+3x)}+\frac

{1}{1+4x}+\frac {1}{1+5x}-\frac {1}{1+7x}-\frac {1}{1+8x}$$
at $$x=0, y=1$$,
$$\therefore \displaystyle \frac {dy}{dx}=1+1+1-1-1=1$$
Question 6
1 / -0

$$(3, 1), (-3, 2)$$ and $$\displaystyle (0,2-\sqrt{3})$$ are the vertices of __________ triangle of area ___________.

A
an isosceles, $$81$$ sq. units

B
a scalene, $$\cfrac {-3+6\sqrt{3}}{2}$$ sq. units

C
an equilateral, $$\displaystyle 9\sqrt{3}$$ sq. units

D
a right angled, $$81$$ sq. units

Solution

Distance between two points $$= \sqrt { \left( { x }_{ 2 }-{x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Distance between the points $$A (3,1) $$ and $$B (-3,2) = \sqrt { \left( -3-3 \right) ^{ 2 }+\left( 2 - 1 \right) ^{ 2 } } = \sqrt { 36 + 1 } = \sqrt { 37 }$$
Distance between the points $$B(-3,2) $$ and $$C (0,2-\sqrt {3}) = \sqrt { \left( 0 + 3 \right) ^{ 2 }+\left(2 - \sqrt {3} - 2\right) ^{ 2 } } = \sqrt { 9 + 3 } = \sqrt { 12 } $$
Distance between the points $$A(3,1) $$ and $$ C(0,2-\sqrt {3})$$
$$ = \sqrt { \left( 0-3\right) ^{ 2 }+\left( 2-\sqrt {3} - 1\right) ^{ 2 } } = \sqrt { 9 + 1 + 3 -2\sqrt {3} } = \sqrt { 13 - 2\sqrt {3} } $$
Since the length of the sides between all vertices are different, they are the vertices of a scalene triangle.
Area of a triangle $$= \left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Area of $$\triangle ABC= \left| \cfrac { (3)(2-2+\sqrt {3})+(-3)(2-\sqrt {3} - 1)+0(1-2) }{ 2 } \right| $$
$$ = \left| \cfrac { 3\sqrt {3} -3 + 3\sqrt {3} }{ 2 } \right| $$
$$ = \cfrac{-3 + 6 \sqrt {3}}{2}$$ sq. units
Question 7
1 / -0

If the coordinates of two points A and B are $$(3, 4)$$ and $$(5, -2)$$, respectively, then the coordinates of any point P if $$PA = PB$$ and area of $$\displaystyle \Delta PAB=10$$ is

A
$$(7, 2)$$ or $$(1, 0)$$

B
$$(-7, 2)$$ or $$(3, 0)$$

C
$$(7, -2)$$ or $$(5, 0)$$

D
$$(7, -2)$$ or $$(-1, 0)$$

Solution

Let P $$ = (x, y) $$
Given, $$ PA = PB $$
$$=> {PA}^{2} = {PB}^{2} $$
$$ => {(x-3)}^{2} + {(y-4)}^{2} = {(x-5)}^{2} + {(y+2)}^{2} $$
$$=> {x} ^{2} -6x + 9 + {y} ^{2} -8y + 16 = {x}^{2} -10x + 25 + {y}^{2} + 4y + 4 $$
$$=> 4x - 12y = 4 $$
$$=> x - 3y = 1 $$ .......(i)

Also, Area of $$ \Delta PAB=10 $$
$$\left| \cfrac {x (4+2)+3(-2-y)+5(y-4)}{ 2 } \right|=10 $$
$$ \left| \cfrac { 6x -6-2y +5y -20 }{ 2 } \right| = 10 $$
$$ \cfrac {6x+3y -26}{2} = \pm 10 $$
$$ 6x+2y-26= \pm 20 $$
$$ 6x+2y = 46 $$ ........(ii)
or $$ 6x + 2y = 6 $$ .......(iii)
Solving equation (i) and (ii), we get $$ x = 7, y = 2 $$
Solving equation (i) and (iii), we get $$ x = 1, y = 0 $$.
So, the co-ordinates of P are $$ (7, 2)$$ or $$(1, 0) $$.
Question 8
1 / -0

- nmmn - mmnm - mnnm -

A
nmmn

B
mnnm

C
nnmm

D
nmnm

Solution

The series is n n m m / n n m m / n n m m / n n m m
Question 9
1 / -0

Find pair of consecutive odd natural numbers, both of which are larger than 13 such that their sum is less than 40

A
(15, 17)

B
(13, 17)

C
(19, 21)

D
(21, 17)

Solution

Let the consecutive odd numbers be $$ x, x + 2 $$

Given, $$ x > 13 $$ -- (1)

Also, $$ x + 2 > 13 $$
$$ x > 11 $$ -- (2)

And $$ x + x + 2 < 40 $$
$$ 2x + 2 < 40 $$
$$ 2x < 38 $$
$$ x < 19 $$ -- (3)

From, $$ 1, 2, 3 $$
$$ 13 < x < 19 $$
This means, the numbers are $$ 15, 17 $$
Question 10
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle.Choose the correct answer for the term marked by the symbol (?)

A
$$18$$

B
$$40$$

C
$$21$$

D
$$24$$

Solution

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Differentiation Test 22

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Differentiation Test 22

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Question 1

5 sq. units

12 sq. units

15 sq. units

20 sq. units

Solution

Question 2

$$\displaystyle a^{2}+b^{2}+c^{2}$$

$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}2$$

$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}4$$

$$0$$

Solution

Question 3

$$1$$

$$3$$

$$6$$

infinite

Solution

Question 4

$$\displaystyle 2x^{\log x-1}\log x$$

$$\displaystyle x^{\log x-1}$$

$$\dfrac 23 (\log x)$$

$$\displaystyle x^{\log x-1}.\log x$$

Solution

Question 5

$$-1$$

$$1$$

$$2$$

Non existant

Solution

Question 6

an isosceles, $$81$$ sq. units

a scalene, $$\cfrac {-3+6\sqrt{3}}{2}$$ sq. units

an equilateral, $$\displaystyle 9\sqrt{3}$$ sq. units

a right angled, $$81$$ sq. units

Solution

Question 7

$$(7, 2)$$ or $$(1, 0)$$

$$(-7, 2)$$ or $$(3, 0)$$

$$(7, -2)$$ or $$(5, 0)$$

$$(7, -2)$$ or $$(-1, 0)$$

Solution

Question 8

nmmn

mnnm

nnmm

nmnm

Solution

Question 9

(15, 17)

(13, 17)

(19, 21)

(21, 17)

Solution

Question 10

$$18$$

$$40$$

$$21$$

$$24$$

Solution

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