Differentiation Test 29

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Differentiation Test 29

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Weekly Quiz Competition

Question 1
1 / -0

Ten points lie in a plane so that no three of them are collinear. The number of lines passing through exactly two of these points are dividing the plane into two regions each containing four of the remaining points is

A
1

B
5

C
10

D
Dependent on the configuration of points.

Solution

You can clearly see from the attached figure
Five such lines are possible joining $$AF,BG,CH,DI,EJ$$
Question 2
1 / -0

Area of the triangle formed by the points $$(0,0),(2,0)$$ and $$(0,2)$$ is

A
$$1$$ sq.units

B
$$2$$ sq.units

C
$$4$$ sq.units

D
$$8$$ sq.units

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore,
$$=\dfrac{1}{2}\left|0(0-2)+2(2-0)+0(0-0)\right|$$

$$=\dfrac{1}{2}\left|0+4+0\right|$$

$$=\dfrac{1}{2}\times 4=2\ sq.\ units$$

Hence, this is the answer.
Question 3
1 / -0

If $$y = \dfrac {1}{1 + x + x^{2}}$$, then $$\dfrac {dy}{dx}$$ is equal to

A
$$y^{2} (1 + 2x)$$

B
$$\dfrac {-(1 + 2x)}{y^{2}}$$

C
$$\dfrac {1 + 2x}{y^{2}}$$

D
$$-y (1 + 2x)$$

E
$$-y^{2} (1 + 2x)$$

Solution

Given, $$y = \dfrac {1}{1 + x + x^{2}} \Rightarrow 1 + x + x^{2} = \dfrac {1}{y}$$
On differentiating w.r.t. $$x$$, we get
$$0 + 1 + 2x = \dfrac {1}{y^{2}} \dfrac {dy}{dx}$$
$$\Rightarrow \dfrac {dy}{dx} = -(1 + 2x)y^{2}$$.
Question 4
1 / -0

Out of $$7$$ consonants and $$4$$ vowels, words are formed each having $$3$$ consonants and $$2$$ vowels. The number of such words that can be formed is

A
$$210$$

B
$$25200$$

C
$$2520$$

D
$$302400$$

Solution

Total no. of consonants is $$7$$ and of vowels is $$4$$
Out of $$7$$ consonants, $$3$$ are chosen to form a word. So, this can be done in $${ }^{7}C_{3}$$ ways
Out of $$4$$ vowels, $$2$$ are chosen to form a word. So, that can be done in $${ }^{4}C_{2}$$ ways
So, the no. of select $$3$$ consonants and $$2$$ vowels is $$^{7}C_{3} \times ^{4}C_{2}$$
Now, we can arrange the word containing $$5$$ letters in $$5!$$ ways.
Thus, the total no. of words formed each having $$3$$ consonants and $$2$$ vowels is
$$^{7}C_{3} \times ^{4}C_{2}\times 5! = 25200$$
Hence, the answer is $$25200$$.
Question 5
1 / -0

Find the term independent of $$x$$ in $${ \left( \cfrac { 3 }{ 2 } { x }^{ 2 }-\cfrac { 1 }{ 3x } \right) }^{ 9 }$$.

A
$$\dfrac {6}{15}$$

B
$$\dfrac {7}{18}$$

C
$$\dfrac {7}{8}$$

D
$$\dfrac {4}{9}$$

Solution

We know that the general term in expansion of $$(a+b)^n$$ is given by $$^nC_{r}a^{n-r}b^{r}$$ where r ranges from $$0 $$ to $$ 9$$.
Here $$a= \cfrac{3}{2}x^2, \ b= -\cfrac{1}{3x} \ \And \ n=9$$
For any term to be independent of $$x$$, coefficient of $$x$$ in $$a^{n+1-r}b^{r-1}$$ should be zero.
$$\Rightarrow \ (x^2)^{9-r}(x^{-1})^{r}=x^0$$
$$\Rightarrow 18-2r-r=0$$
$$\Rightarrow r=6$$
Therefore, the term independent of $$x$$ is $$^9C_6 \left (\cfrac{3}{2}\right)^3 \left (-\cfrac{3}{3}\right)^6=\cfrac{7}{18}$$.
Question 6
1 / -0

Consider an incomplete pyramid of balls on a square base having $$18$$ layers; and having $$13$$ balls on each side of the top layer. Then the total number $$N$$ of balls in that pyramid satisfies

A
$$9000 < N < 10000$$

B
$$8000 < N < 9000$$

C
$$7000 < N < 8000$$

D
$$10000 < N < 12000$$

Solution

The top layer has $$(13\times 13)$$ balls the layer below it will have $$(14\times 14)$$ balls

We have $$18$$ layers

So the total number of balls

$$N=(13\times 13)+(14\times 14)+.......(30\times 30)\\ N={ 13 }^{ 2 }+{ 14 }^{ 2 }+.....{ 30 }^{ 2 }$$

Sum of squares of first $$n$$ natural numbers is $$\dfrac { n(n+1)(2n+1) }{ 6 } \\ $$

$$\therefore N=$$ sum of first $$30$$ $$-$$ sum of first $$12$$

$$N=\dfrac { 30\times 31\times 61 }{ 6 } -\dfrac { 12\times 13\times 25 }{ 6 } \\ N=8805\\ $$

$$\Rightarrow 8000<N<9000$$
Hence, option B is correct.
Question 7
1 / -0

If $$f(x)=\left| \log { \left| x \right| } \right| $$, then

A
$$f(x)$$ is continuous and differentiable for all $$x$$ in its domain

B
$$f(x)$$ is continuous for all $$x$$ in its domain but not differentiable at $$x=\pm 1$$

C
$$f(x)$$ is neither continuous nor differentiable at $$x=\pm 1$$

D
None of the above

Solution

It is evident from the graph of $$f(x)=\left| \log { \left| x \right| } \right| \quad $$ that $$f(x)$$ is everywhere continuous but not differentiable at $$x=\pm 1$$
Question 8
1 / -0

Choose $$3, 4, 5$$ points other than vertices respectively on the sides $$AB, BC$$ and $$CA$$ of a $$\triangle ABC$$. The number of triangles that can be formed by using only these points as vertices, is

A
$$220$$

B
$$217$$

C
$$215$$

D
$$210$$

E
$$205$$

Solution

Required number of triangles that can be formed by using only given points as vertices
$$= ^{12}C_{3} - \left \{^{3}C_{3} + ^{4}C_{3} + ^{5}C_{3}\right \}$$
$$= \dfrac {12\times 11\times 10}{3\times 2\times 1} - \left \{1 + 4 + \dfrac {5\times 4}{2\times 1}\right \}$$
$$= 220 - (5 + 10)$$
$$= 220 - 15 = 205$$
Question 9
1 / -0

If $$ y^x = 2^x , $$ then $$ \dfrac {dy}{dx} $$ is equal to :

A
$$ \dfrac {y}{x} \log \left( \dfrac {2}{y} \right) $$

B
$$ \dfrac {x}{y} \log \left( \dfrac {2}{y} \right) $$

C
$$ \dfrac {y}{x} \log \left( \dfrac {y}{2} \right) $$

D
$$ \dfrac {x}{y} \log \left( \dfrac {y}{2} \right) $$

E
$$ \dfrac {y}{x} \log \left( 2y \right) $$

Solution

Given, $$ y^x = 2^x $$
On taking $$\log$$ on both sides, we get
$$ x \log y = x \log 2 $$
On differentiating w.r.t. $$x,$$ we get
$$ x\times \dfrac {1}{y} \times \dfrac {dy}{dx} + \log y = \log 2 $$
$$ \Rightarrow \dfrac {dy}{dx} = \dfrac {y}{x} [ \log 2 - \log y ] $$
$$ \Rightarrow \dfrac {dy}{dx}= \dfrac {y}{x} \left[ \log \dfrac{2}{y} \right] $$
Question 10
1 / -0

If $$\alpha ,\beta , \gamma $$ are three consecutive integers. If these integers are raised to first, second and third positive powers respectively, and added then they form a perfect square, the square root of which is equal to the sum of these integers. Also, $$\alpha < \beta < \gamma $$. Then, $$\gamma$$ is equals to:

A
$$3$$

B
$$14$$

C
$$5$$

D
$$11$$

Solution

Let the numbers be $$n-1,n,n+1$$
As per the given information, we have
$$(n-1)+{ n }^{ 2 }+{ (n+1) }^{ 3 }={ (n-1+n+n+1) }^{ 2 }$$
$$\Rightarrow { n }^{ 3 }-5{ n }^{ 2 }+4n=0$$
$$\Rightarrow n=0,1,4$$
For $$n=0,$$ the numbers are: $$-1,0,1$$ this is out as all the numbers should be positive
For $$n=1,$$ the numbers are: $$0,1,2$$ this is out as all the numbers should be positive (‘0’ can’t be taken as positive)
For $$n=4,$$ the numbers are: $$3,4,5$$
We have got largest number which is $$5$$
Hence, the value of $$\gamma$$ is $$5$$.

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Re-Attempt Weekly Quiz Competition

Differentiation Test 29

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Differentiation Test 29

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SHARING IS CARING

Question 1

1

5

10

Dependent on the configuration of points.

Solution

Question 2

$$1$$ sq.units

$$2$$ sq.units

$$4$$ sq.units

$$8$$ sq.units

Solution

Question 3

$$y^{2} (1 + 2x)$$

$$\dfrac {-(1 + 2x)}{y^{2}}$$

$$\dfrac {1 + 2x}{y^{2}}$$

$$-y (1 + 2x)$$

$$-y^{2} (1 + 2x)$$

Solution

Question 4

$$210$$

$$25200$$

$$2520$$

$$302400$$

Solution

Question 5

$$\dfrac {6}{15}$$

$$\dfrac {7}{18}$$

$$\dfrac {7}{8}$$

$$\dfrac {4}{9}$$

Solution

Question 6

$$9000 < N < 10000$$

$$8000 < N < 9000$$

$$7000 < N < 8000$$

$$10000 < N < 12000$$

Solution

Question 7

$$f(x)$$ is continuous and differentiable for all $$x$$ in its domain

$$f(x)$$ is continuous for all $$x$$ in its domain but not differentiable at $$x=\pm 1$$

$$f(x)$$ is neither continuous nor differentiable at $$x=\pm 1$$

None of the above

Solution

Question 8

$$220$$

$$217$$

$$215$$

$$210$$

$$205$$

Solution

Question 9

$$ \dfrac {y}{x} \log \left( \dfrac {2}{y} \right) $$

$$ \dfrac {x}{y} \log \left( \dfrac {2}{y} \right) $$

$$ \dfrac {y}{x} \log \left( \dfrac {y}{2} \right) $$

$$ \dfrac {x}{y} \log \left( \dfrac {y}{2} \right) $$

$$ \dfrac {y}{x} \log \left( 2y \right) $$

Solution

Question 10

$$3$$

$$14$$

$$5$$

$$11$$

Solution

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