Ans. 5

In the given word EXAMINATION, there are 11 letters out of which, A, I, and N appear 2 times and all the other letters appear only once.
The words that will be listed before the words starting with E in a dictionary will be the words that start with A only.
Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left position, and then the remaining 10 letters taken all at a time are rearranged.
Since there are 2 Is and 2 Ns in the remaining 10 letters,
Number of words starting with A = \(={10!\over 2!2!}\) \(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2\times1\times2!}\)
= 907200
Thus, the required numbers of words is 907200.

We have 
L.H.S. =\(sin^2{\pi\over6}+cos^2{\pi\over3}-tan^2{\pi\over 4}\)
\(=({1\over2})^2+({1\over2})^2-(1)^2={1\over4}+{1\over4}-1\)
\(={1+1-4\over4}={-2\over4}={-1\over2}=R.H.S\)

(-∞ , -3)

Write the complex number in the form \(\frac { a+ib }{ c+id } \) and then rationalising the denominator, further simplify it
\(\frac { \left( 3+\sqrt { 5i } \right) \left( 3-\sqrt { 5i } \right) }{ \left( \sqrt { 3 } +\sqrt { 2i } \right) -\left( \sqrt { 3 } -\sqrt { 2i } \right) } \)
\(=\frac { { \left( 3 \right) }^{ 2 }-{ \left( \sqrt { 5i } \right) }^{ 2 } }{ \sqrt { 3 } +\sqrt { 2i } -\sqrt { 3 } +\sqrt { 2i } } \quad \left[ \because \ \left( { z }_{ 1 }+{ z }_{ 2 } \right) \left( { z }_{ 1 }-{ z }_{ 2 } \right) ={ z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 } \right] \)

\(=\frac { 9+5 }{ 2\sqrt { 2i } } =\frac { 14 }{ 2\sqrt { 2i } } =\frac { 7 }{ \sqrt { 2i } } \times \frac { \sqrt { 2i } }{ \sqrt { 2i } } \)
[by rationalising the denominator]
\(=\frac { 7\sqrt { 2i } }{ 2{ i }^{ 2 } } =\frac { 7\sqrt { 2i } }{ -2 } =0-i\frac { 7\sqrt { 2 } }{ 2 } \)
\(=0+i\left( \frac { -7\sqrt { 2 } }{ 2 } \right) \)
Which is in the form of (a+ib).

Given, relation between honest and punctual people is h=p+16
And relation between honest people and the people who progress in life is \(P=\left( \frac { h }{ 8 } \right) +5\)
Now, required relation between number of people who progress in life and punctual is given by
\(P=\left( \frac { h }{ 8 } \right) +5 \quad [\because h=p+16]\)
\(=\left( \frac { p }{ 8 } \right) +2+5=\left( \frac { p }{ 8 } \right) +7\)  
If we are punctual, THisthen we can complete our work at time and the quality of work will also good.This help us to get progress in our life.   

\(\phi \)

We express 98 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.
Write 98 = 100 – 2
Therefore, (98)⁵ = (100 – 2)⁵
= ⁵C₀ (100)⁵ – ⁵C₁ (100)⁴.2 + ⁵C₂ (100)³2²
– ⁵C₃ (100)² (2)³ + ⁵C₄ (100) (2)⁴ – ⁵C₅ (2)⁵
= 10000000000 – 5 x 100000000 x 2 + 10 x 1000000 x 4 – 10 x10000 x 8 + 5 x 100 x 16 – 32
= 10040008000 – 1000800032 = 9039207968.

\(\left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right)\)
\( =\frac{1}{3}+\frac{7}{3} i+4+\frac{1}{3} i+\frac{4}{3}-i \)
\(=\left(\frac{1}{3}+4+\frac{4}{3}\right)+i\left(\frac{7}{3}+\frac{1}{3}-1\right) \)
\(=\frac{17}{3}+i \frac{5}{3} \)

We have
\(\text { L.H.S. } =\frac{\sin (x+y)}{\sin (x-y)}=\frac{\sin x \cos y+\cos x \sin y}{\sin x \cos y-\cos x \sin y}\)
Dividing the numerator and denominator by cos x cos y, we get
\(\frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y}\)

Let A ⊂ B
To show: C – B ⊂ C – A
Let x ∈ C – B
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [A ⊂ B]
⇒ x ∈ C – A
∴ C – B ⊂ C – A

There are as many codes as there are ways of filling 4 vacant places   in succession by the first 10 letters of the English alphabet, keeping in mind that the repetition of letters is not allowed.The first place can be filled in 10 different ways by any of the first 10 letters of the English alphabet following which, the second place can be filled in by any of the remaining letters in 9 different ways. The third place can be filled in by any of the remaining 8 letters in 8 different ways and the fourth place can be filled in by any of the remaining 7 letters in 7 different ways.
Therefore, by multiplication principle, the required numbers of ways in which 4 vacant places can be filled is 10 × 9 × 8 × 7 = 5040
Hence, 5040 four-letter codes can be formed using the first 10 letters of the English alphabet, if no letter is repeated.

Let x be the smaller of the two consecutive odd natural number, so that the other one is x +2. Then, we should have
x > 10 ... (1)
and x + ( x + 2) < 40 ... (2)
Solving (2), we get
2x + 2 < 40
i.e., x < 19 ... (3)
From (1) and (3), we get
10 < x < 19
Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required possible pairs will be
(11, 13), (13, 15), (15, 17), (17, 19)

Given,  f(x)=x²
\(\therefore \frac{\mathrm{f}(1.1)-\mathrm{f}(1)}{(1.1-1)}=\frac{(1.1)^2-(1)^2}{(1.1-1)}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1\)

There are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different. Therefore
The required number of arrangements \(=\frac{12 !}{3 ! 4 ! 2 !}=1663200\)
(i) Let us fix P at the extreme left position, we, then, count the arrangements of the remaining 11 letters. Therefore, the required number of words starting with P
\(=\frac{11 !}{3 ! 2 ! 4 !}=138600\)
(ii) There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object EEEEI for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in \(\frac{8 !}{3 ! 2 !}\) ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, E and I can be rearranged in \(\frac{5 !}{4 !}\) ways. Therefore, by multiplication principle, the required number of arrangements
\(=\frac{8 !}{3 ! 2 !} \times \frac{5 !}{4 !}=16800\)
(iii) The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together.
= 1663200 – 16800 = 1646400
(iv) Let us fix I and P at the extreme ends (I at the left end and P at the right end). We are left with 10 letters. Hence, the required number of arrangements
\(=\frac{10 !}{3 ! 2 ! 4 !}=12600\)

We have \(\left[ 1+\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \right] ^{ 4 }\)
=\(\left[ 1+\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \right] ^{ 4 }\)
=⁴C₀+⁴C₁\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \)+⁴C₂\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) ^2\)+⁴C₃\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) ^3\)+⁴C₄\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) ^4\)
=1+4\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \)+6\(\left( \frac { x^{ 2 } }{ 2 } -\frac { 2 }{ { x }^{ 2 } } -2 \right) \)+4\(\left( \frac { { x }^{ 3 } }{ 8 } -\frac { 8 }{ { x }^{ 3 } } -\frac { 3x }{ 2 } +\frac { 6 }{ x } \right) +\left[ ^{ 4 }C_{ 0 }\left( \frac { x }{ 2 } \right) ^{ 4 }-^{ 4 }C_{ 1 }\left( \frac { x }{ 2 } \right) ^{ 3 }\left( \frac { 2 }{ x } \right) +^{ 4 }C_{ 2 }\left( \frac { x }{ 2 } \right) ^{ 2 }\left( \frac { 2 }{ x } \right) ^{ 2 }-^{ 4 }C_{ 3 }\left( \frac { x }{ 2 } \right) \left( \frac { 2 }{ x } \right) ^{ 3 }+^{ 4 }C_{ 4 }\left( \frac { x }{ 2 } \right) ^{ 4 } \right] \)
=1+\(\left( 2x-\frac { 8 }{ x } \right) +\left( \frac { 3 }{ 2 } { x }^{ 2 }+\frac { 24 }{ { x }^{ 2 } } -12 \right) +\left( \frac { { x }^{ 3 } }{ 2 } -\frac { 32 }{ { x }^{ 3 } } -6x+\frac { 24 }{ x } \right) +\left( \frac { { x }^{ 4 } }{ 16 } -{ x }^{ 2 }+6\frac { 16 }{ { x }^{ 2 } } +\frac { 16 }{ { x }^{ 4 } } \right) \)
=-5-4x + \(\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 2 } +\frac { { x }^{ 4 } }{ 16 } +\frac { 16 }{ x } +\frac { 8 }{ { x }^{ 2 } } +\frac { 32 }{ { x }^{ 3 } } -\frac { 16 }{ { 4 }^{ 4 } } \)

Here sin x = \(3\over 5\)
Now sin² x + cos² x = 1
\(\Rightarrow ({3\over 5})^2+cos^2x=1\)
\(\Rightarrow cos^2 \ x=1-{9\over25}\)
\(\Rightarrow cos^2x={16\over25}\)
\(\Rightarrow cos x=\pm{4\over 5}\)
But x lies in second quadrant.
\(\therefore cos x =-{4\over 5}\)
Now cosec x=\({1\over sin \ x}={5\over3}\)
and secx=\({1\over cos x}=-{5\over4}\)
\(tan x={sin \ x\over cos \ x }={3/5 \over -4/5}={-3\over 4}\)
and cot x=\({cos x\over sin x}={-4/5\over 3/5}={-4\over 3}\)

We have, (x-iy)(3+5i) is the conjugate of -6-24i =-6-24i.
\(\Rightarrow \) (x - iy) (3 + 5i) = -6 + 24i
[conjugate of -6-24i = -6 + 24i]
\(\Rightarrow \) 3x + 3iy + 5ix + 5i²y = -6+24i
\(\Rightarrow \) (3x + 5y) + i(5x - 3y) = -6+24i      [i²=-1]  ...(i)
On equating real and imaginary parts both sides of Eq. (i) we get
3x + 5y = -6 ....(ii)
5x - 3y = 24
On multiplying Eq, (i) by 3 and Eq. (ii) by 5, then adding the result, we get
9x + 15y + 25x - 15y = -18 + 120 \(\Rightarrow \) 34x = 102 \(\Rightarrow \) x = 3
On substituting x = 3 in Eq,...(ii) we get
 9 + 5y = -6\(\Rightarrow \) 5y = -15 \(\Rightarrow \)y = -3
Hence, the required values of x and y are respectively 3 and -3       

{-1,1}