Show that AB + BC = AC

\(z=\frac { 1-i }{ 1+i } x \frac { 1-i }{ 1-i } =\frac { 1-1-2i }{ 1+1 } =-i\)   = i

\(\sin\theta =\frac { -4 }{ 5 } \ and\ \tan \theta =\frac { 3 }{ 4 } \)

\(cos\left( \frac { 3\pi }{ 4 } +x \right) -cos\left( \frac { 3\pi }{ 4 } -x \right)\)
\(=\left( -\frac { 1 }{ \sqrt { 2 } } cos\quad x\quad -\quad \frac { 1 }{ \sqrt { 2 } } sin\quad x \right) -\left( -\frac { 1 }{ \sqrt { 2 } } cos\quad x\quad +\quad \frac { 1 }{ \sqrt { 2 } } sin\quad x \right) \)
\(=-\sqrt { 2 } sinx.\)

We have,
\((f+g)(x)=x^{2}+2 x+1,(f-g)(x)=x^{2}-2 x-1\)
\((\text { fg })(x)=x^{2}(2 x+1)=2 x^{3}+x^{2},\left(\frac{f}{g}\right)(x)=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \)

\(n({ A }^{ ' }\cap { B }^{ ' })=n(U)-n(A\cup B)\)
Ans.300

Here f(x)=\(\frac { a+b\quad sin\quad x }{ c+d\quad cos\quad x } \)
\(\therefore f'(x)=\frac { d }{ dx } \left[ \frac { a+b\quad sin\quad x }{ c+d\quad cos\quad x } \right] \)
= \(\frac { (c+dcosx)\frac { d }{ dx } (a+b\quad sinx)-(a+b\quad sinx)\frac { d }{ dx } (c+d\quad cosx) }{ c+d\quad cosx^{ 2 } } \)
\(=\frac { (a-b\quad sinx)(-d\quad sinx) }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bc\quad cosx+bd\quad cos^{ 2 }x+\quad ad\quad sinx+bd\quad sin^{ 2 }x }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bc\quad cosx+bd\quad sinx+\quad bd(cos^{ 2 }x+sin^{ 2 }x) }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bd(cos^{ 2 }x+sin^{ 2 }x) }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bc\quad cosx+ad\quad sinx+bd }{ (c+d\quad cosx)^{ 2 } } \)

Total letters in the word PERMUTATIONS = 12.
Here T = 2.
(i) Now first letter is P and last letter is S, which are fixed.
So the remaining 10 letters are to be arranged between P and S
∴ Number of permutations = \(10!\over 2!\)
\(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=1814400
(ii) There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.
∴ Number of permutations of all vowels together = ⁵p₅
\(={5!\over 0!}=5 \times 4 \times 3 \times 2 \times 1 = 120\)
Now consider the 5vowels together as one letter. Sothe number of letters in the word when all vowels are together = 8.
∴ Number of permutations  \(={8!\over 2!}\)
\(={8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=20160
Hence the total number of permutations
= 120 x 20160 = 2419200
(iii) Here P and S are on 1st and 6th places
P and S are on 2nd and 7th places
P and S are on 3rd and 8th places P and S are on 4th and 9th places
P and S are on 5th and 10th places P and S are on 6th and 11th places
P and S are on 7th and 12th places
Now we see that P and S can be put in 7 ways and also P and S can interchange their positions.
∴ Number of permutations = 2 x 7 = 14
Now the remaining 10 places can be filled with remaining 10 letters.
∴ Number of permutations =  \(10!\over 2!\)
\(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=1814400
Thus total number of permutations = 14 x 1814400= 25401600

The given series is
\(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+....\)
Let 'a_n' be nth term of the given series and 'S_n' be the sum of n^th terms of the given series.
\(\therefore a_n=\frac{1}{(n^{th}\quad term\quad of\quad 1,2,3...)(n^{th}\quad term\quad of 2,3,4...)}\)
= \(\frac{1}{[1+(n+1)\times 1][2+(n-1)\times 1]}\)
= \(\frac{1}{n(n+1)}=\frac{1}{n}+(\frac{-1}{n+1})\) [By partial fraction]
Putting n = 1, 2, 3, ....n, we have
\(a_1=\frac{1}{1}-\frac{1}{2};\)
\(a_2=\frac{1}{2}-\frac{1}{3};\)
\(a_3=\frac{1}{3}-\frac{1}{4}.....\)
\(a_n=\frac{1}{n}-\frac{1}{n+1}....\)
Adding vertically, we have:
S_n = a₁ + a₂ + a₃ + .... + a_n
= \(\frac{1}{1}-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)
\(\therefore\)  S_n = \(\frac{n}{n+1}\)

Mean\(\overline { x } =\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 50 } \times 17900=358\)
Mean deviation about mean\(=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-\overline { x } \right| } \)
\(=\frac{1}{50}\times7896 = 157.92\) 

2x + y≥ 6 … (1)
3x + 4y ≤ 12 … (2)
The graph of the lines, 2x + y= 6 and 3x + 4y = 12, are drawn in the figure below.
Inequality (1) represents the region above the line, 2x + y= 6 (including the line 2x + y= 6), and inequality (2) represents the region below the line, 3x + 4y =12 (including the line 3x + 4y =12).
Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.

Let \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ n(n+1) } =\frac { n }{ (n+1) } \)
For n =1
 P(1) = \(\frac { 1 }{ 1(1+1) } =\frac { 1 }{ 1+1 } \Rightarrow \frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) = \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } =\frac { k }{ (k+1) } \)   ...(i)
For n = k+1
∴ P(k+1)= \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } +\frac { 1 }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)
= \(\frac { k }{ k+1 } +\frac { 1 }{ (k+1)(k+2) } =\frac { k(k+2)+1 }{ (k+1)(k+2) } =\frac { { k }^{ 2 }+2k+1 }{ (k+1)(k+2) } \)       [Using (i)]
=  \(\frac { (k+1)^{ 2 } }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)= P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.

We have
L.H.S.= sin x + sin 3x + sin 5x + sin 7x
= (sin 7x + sin x) + (sin 5x + sin 3x)
\(=[2sin({7x+x\over2})cos({7x-x\over2})]+[2sin({5x+3x\over2})cos({5x-3x\over2})]\)
= 2 sin 4x cos 3x + 2 sin 4x cos x
\([\because \ sinC+sin D=2sin{C+D\over2}.cos{C-D\over2}]\)
= 2 sin 4x [cos 3x + cos x]
= 2 sin 4x\([2cos({3x+x\over2})cos({3x-x\over2})]\)
\([\because \ cos C+cosD=2cos {{C+D\over2}.cos{C-D\over2}}]\)
= 2 sin 4x [2 cos 2x cos x]
= 4 cos x cos 2x sin 4x = R.H.S.

We have
L.H.S. = \({cos \ 4x +cos \ 3x +cos \ 2x\over sin \ 4x+sin 3\ x+sin \ 2x}\)
\(={(cos \ 4x +cos \ 2x )+cos \ 2x\over (sin \ 4x+sin 2\ x)+sin \ 2x}\)
\(={2cos{4x+2x\over2}cos{4x-2x\over2}+cos3x\over 2sin{4x+2x\over2}cos{4x-2x\over2}+sin3x}\)
\(\left[\begin{array}{l} \because \sin \mathrm{C}+\sin \mathrm{D}=2 \sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cos \left(\frac{\mathrm{C}-\mathbf{D}}{2}\right) \\ \cos \mathrm{C}+\cos \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right) \end{array}\right]\)
\(={2cos \ 3x \cos \ x + cos \ 3x\over 2sin \ 3x \cos x+ sin \ 3x}\)
\(={cos \ 3x (2 \cos \ x + 1)\over sin \ 3x(2 \cos x+ 1)}={cos \ 3x\over sin \ 3x}\)
= cot 3x = R.H.S.