Differentiation Test 1

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Differentiation Test 1

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Weekly Quiz Competition

Question 1
1 / -0

All possible number are formed using the digits $$1,1,2,2,2,2,3,4,4$$ taken all at a time. The number of such number in which the odd digits occupy even places is:

A
$$175$$

B
$$162$$

C
$$160$$

D
$$180$$

Solution

Number of such number $$=^4C_3\times \dfrac{3!}{2!}\times \dfrac{6!}{2!4!}=180$$
Question 2
1 / -0

The derivative of $$\tan^{-1} \left (\dfrac {\sin x - \cos x}{\sin x + \cos x}\right )$$, with respect to $$\dfrac {x}{2}$$, where $$\left (x \epsilon \left (0, \dfrac {\pi}{2}\right )\right )$$ is

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

B
$$\dfrac {2}{3}$$

C
$$1$$

D
$$2$$

Solution

$$f(x) = \tan^{-1} \left (\dfrac {\sin x - \cos x}{\sin x + \cos x}\right )$$
$$= \tan^{-1} \left (\dfrac {\tan x - 1}{\tan x + 1}\right ) = \tan^{-1} \left (\tan \left (x - \dfrac {\pi}{4}\right )\right )$$
$$\because x - \dfrac {\pi}{4}\epsilon \left (-\dfrac {\pi}{4}, \dfrac {\pi}{4}\right )$$
$$\therefore f(x) = x - \dfrac {\pi}{4}$$
$$\Rightarrow$$ its derivative w.r.t $$\dfrac {x}{2}$$ is $$\dfrac {1}{1/2} = 2$$.
Question 3
1 / -0

A group of students comprises of $$5$$ boys and $$n$$ girls. If the number of ways, in which a team of $$3$$ students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is $$1750$$, then $$n$$ is equal to

A
$$25$$

B
$$28$$

C
$$27$$

D
$$24$$

Solution

Number of ways of selecting a team with 1 boy and 2 girls$$=^{5}C_{1}. ^{n}C_{2} $$

Number of ways of selecting a team with 2 boy and 1 girls$$=^{5}C_{2} . ^{n}C_{1} $$

Given $$^{5}C_{1}. ^{n}C_{2} + ^{5}C_{2} . ^{n}C_{1} = 1750$$

$$5\cdot \dfrac {n(n-1)}{2!}+10\cdot n=1750$$

$$n^{2} + 3n = 700$$

$$\therefore n = 25$$.
Question 4
1 / -0

The value of $$(^{21}C_1-^{10}C_1)+(^{21}C_2-^{10}C_2)+(^{21}C_3-^{10}C_3)+(^{21}C_4+^{10}C_4)+......+(^{21}C_{10}-^{10}C_{10})$$ is:

A
$$2^{21}-2^{11}$$

B
$$2^{21}-2^{10}$$

C
$$2^{20}-2^9$$

D
$$2^{20}-2^{10}$$

Solution

$$(^{21}C_1-^{10}C_1)+(^{21}C_2-^{10}C_2)+(^{21}C_3-^{10}C_3)+(^{21}C_4+^{10}C_4)+......+(^{21}C_{10}-^{10}C_{10})$$

$$= ^{21}C_1+^{21}C_2+^{21}C_3+ ..... + ^{21}C_{10} - [2^{10}-1]$$

$$=\cfrac 12 [2 ^{21}C_1 + 2 ^{21}C_2+ 2 ^{21}C_3 + ..... + 2 ^{21}C_{10}] - [2^{10}-1]$$

$$=\cfrac 12 [^{21}C_1 + ^{21}C_2+ .... + ^{21}C_{10} + ^{21}C_{11} + ..... + ^{21}C_{19} + ^{21}C_{20}] - [2^{10}-1]$$

$$=\cfrac 12 [2^{21} - 1 -1 ] - [2^{10}-1]$$

$$=2^{20} - 1 - 2^{10} +1$$

$$=2^{20}-2^{10}$$
Question 5
1 / -0

The number of four -digit numbers strictly grater than $$4321$$ that can be formed using the digits $$0,1,2,3,4,5$$ (repetition of digits is allowed) is:

A
$$288$$

B
$$306$$

C
$$360$$

D
$$310$$

Solution

(1) The number of four-digit numbers Strating with 5 is equal to $$6^3 = 216$$
(2) Starting with 44 and 55 is equal to $$36\times 2 = 72$$
(3) Starting with 433, 434 and 435 is equal to $$6\times 3 = 18$$
(4) Remaining numbers are 4322, 4323, 4324, 4325 is equal to 4
so total number are
$$716 + 72 + 18 + 4 = 310$$
Question 6
1 / -0

Let $$A$$ be the set of all $$3 \times 3$$ symmetric matrices all of whose entries are either $$0$$ or $$1$$. Five of these entries are $$1$$ and four of them are $$0$$.
The number of matrices in $$A$$ is

A
$$12$$

B
$$6$$

C
$$9$$

D
$$3$$

Solution

If two zeros are the entries in the diagonal, then
$$^{3}\mathrm{C}_{2}\times^{3}\mathrm{C}_{1}$$
If all the entries in the principle diagonal is 1, then
$$^{3}\mathrm{C}_{1}$$
$$\Rightarrow$$ Total matrix $$= 12.$$
Question 7
1 / -0

The value of '$$a$$' in order $$f(x)=\sqrt{3}\sin x-\cos x -2ax+b$$ decrease for all real values of $$x$$, is given by

A
$$a>1$$

B
$$a \ge 1$$

C
$$a \ge \overline{2}$$

D
$$a < \overline{2}$$

Solution

Given, $$f(x)=\sqrt{3}\sin x-\cos x-2ax+b$$
$$f'(x)=\sqrt{3}\cos x + \sin x -2a$$
$$f'(x)=2\left (\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x\right)-2a$$
$$f'(x)=2\left (\sin\left (\dfrac{\pi}{3}+x\right)-a\right)$$
if $$a>1$$, then for all real values of $$x$$ $$,f'(x)<0$$.
So, $$a>1$$
Question 8
1 / -0

Area of a triangle formed by the points A(5, 2), B(4, 7) and C(7, -4) is _____.

A
2 sq. units

B
$$2\sqrt2$$ sq. units

C
4 sq. units

D
$$4\sqrt2$$ sq. units

Solution

Area of a triangle ABC = $$ \frac { 1 }{ 2 } \left|[ x_{ 1 }\left( y_{ 2 }-y_{ 3 }

\right) +x_{ 2 }\left( y_{ 3 }-y_{ 1 } \right) +x_{ 3 }\left( y_{ 1 }-y_{ 2 }

\right) \right]|$$
Substituting the given coordinates, we have
=$$ \frac { 1 }{ 2 } \left[ 5 \left(7+4) \right) -\left( 4+2\right) +7\left( 2-7) \right) \right]$$
$$ =\frac { 1 }{ 2 } \left( 5 ( 11) +4\times( -6)
+7\times( -5)\right)$$
$$ =\frac { 1 }{ 2 } \left[(55) - ( 24) - ( 35) \right]$$
$$ =\frac { 1 }{ 2 } \times |-4| = 2$$ square units.
Question 9
1 / -0

Value of 0! is always 1.

A
True

B
False

C
Either

D
Neither

Solution

we know $$1!=1$$
Also
$$n!=n\times (n-1)\times (n-2)........3\times 2\times 1\\ n!=n\times (n-1)!\\ 1!=1(1-1)!\\ 1=1(0)!\\ 0!=1$$
$$0! $$ is always $$1$$
Hence its true that $$0!$$ is always $$1$$
Question 10
1 / -0

If $$y=2 \sin x -3x^4 + 8$$, then $$\dfrac{dy}{dx}$$ is

A
$$2\sin x -12 x^3$$

B
$$2 \cos x-12 x^3$$

C
$$2 \cos x+12 x^3$$

D
$$2 \sin x+12 x^3$$

Solution

If $$y=2 \sin x -3x^4 + 8$$
$$\Rightarrow \cfrac{dy}{dx}=2 \cos x-3(4x^{(4-1)})+0$$
$$\Rightarrow \cfrac{dy}{dx}=2 \cos x-12x^3$$
Therefore, the correct answer is $$B$$.

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Re-Attempt Weekly Quiz Competition

Differentiation Test 1

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Differentiation Test 1

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Rank

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SHARING IS CARING

Question 1

$$175$$

$$162$$

$$160$$

$$180$$

Solution

Question 2

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

$$\dfrac {2}{3}$$

$$1$$

$$2$$

Solution

Question 3

$$25$$

$$28$$

$$27$$

$$24$$

Solution

Question 4

$$2^{21}-2^{11}$$

$$2^{21}-2^{10}$$

$$2^{20}-2^9$$

$$2^{20}-2^{10}$$

Solution

Question 5

$$288$$

$$306$$

$$360$$

$$310$$

Solution

Question 6

$$12$$

$$6$$

$$9$$

$$3$$

Solution

Question 7

$$a>1$$

$$a \ge 1$$

$$a \ge \overline{2}$$

$$a < \overline{2}$$

Solution

Question 8

2 sq. units

$$2\sqrt2$$ sq. units

4 sq. units

$$4\sqrt2$$ sq. units

Solution

Question 9

True

False

Either

Neither

Solution

Question 10

$$2\sin x -12 x^3$$

$$2 \cos x-12 x^3$$

$$2 \cos x+12 x^3$$

$$2 \sin x+12 x^3$$

Solution

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