2015/119) Prove that $(sin\theta+i\cos\theta)^8=\cos8\theta-i\sin8\theta$

$(\sin\theta+i\cos\theta)^8 = i^8(\cos\theta - i \sin \theta)^8$
$=(e^{-i\theta})^8= e^{-i8\theta}$
$=\cos 8\theta - i \sin 8\theta$

2015/118) Without using logs and calculator find the number of digits in $2^{100}$

$2^{10} = 1024 > 1000 = 10^3$
so $2^{100} > 10^{30}$
further
$2^{10} < 1025 < 10^3 * \frac{41}{40}=  10^3 *( 1+ \frac{1}{40})$
so $2^{100} < 10^{30} * ( 1+ \frac{1}{40})^{10} < 10^{30} * ( 1+ \frac{1}{40})^{40}<  10^{30} * e <  10^{30} * 3$
so $10^{30} < 2^{100} < 3 * 10^{30}$
hence 31 digits

2015/117) If $m\tan(a-30^\circ)=n\tan(a+120^\circ)$ show that $\cos2a=\frac{m+n}{2(m-n)}$

We have $\tan (a+ 120^\circ) = - \cot(a + 30^\circ)$ using $\tan (x+90^\circ) = - \cot x$
so $m\tan(a-30^\circ)= -n \cot ( a+ 30^\circ)$
so $\tan (a+30^\circ) \tan (a-30^\circ) = \dfrac{-n}{m}$
or $\dfrac{\tan a + \tan 30^\circ}{1- \tan a \tan 30^\circ} * \dfrac{\tan a -\tan 30^\circ}{1 + \tan a \tan 30^\circ)} = \dfrac{-n}{m}$
or $\dfrac{tan ^2 a - tan ^2 30^\circ}{1- tan ^2 a tan ^2 30^\circ} = \dfrac{n}{m}$
or $\dfrac{tan ^2 a - \frac{1}{3}}{1- tan ^2 a \frac{1}{3}} = \dfrac{n}{m}$
or $\dfrac{3 tan ^2 a -1}{3 - tan ^2 a} = \dfrac{n}{m}$
or $\dfrac{1-3 tan ^2 a}{3- tan ^2 a} = \dfrac{n}{m}$
use componendo dividendo to get
$\dfrac{4 - 4 \tan ^2 a}{- 2 - 2 \tan ^2 a} = \dfrac{n+m}{n-m}$
or $2\dfrac{1 - tan ^2 a}{1 + 2 tan ^2 a} = \dfrac{n+m}{m-n}$
or $\dfrac{1 - tan ^2 a}{1 + 2 tan ^2 a} = \dfrac{n+m}{2(m-n)}$
or $\cos 2a = \dfrac{n+m}{2(m-n)}$

2015/116) Find the limit of the product as k goes to infinite $\prod_{n=1}^{k}\frac{n^2}{n^2-1}$

We propose $\prod_{n=1}^{k}\frac{n^2}{n^2-1}=\frac{2k}{k+1}$

for n =2 we have product = $\frac{4}{3} = \frac{2 * 2}{2+1}$

let it be true of n = k

so we have $\prod_{n=1}^{k}\frac{n^2}{n^2-1}=\frac{2k}{k+1}$

multiply by (k+1)st term to get   $f(k+1) = \prod_{n=1}^{k}\frac{n^2}{n^2-1} * \frac{(k+1)^2}{(k+1)^2 - 1}$
= $\frac{2k}{k+1} * \frac{(k+1)^2}{k(k+2)} = \frac{2 * (k +1)}{(k+2)}$
so if it is true for k it is true for k + 1
Now that we have found the closed form as k goes to infinite above product goes to 2 (converges to 2)

2015/115) If $a+b+c=\pi$

Prove that
$\sin^3 a+\sin^3 b+ \sin^3 c=3\cos(\frac{a}{2})\cos(\frac{b}{2})\cos(\frac{c}{2})$
           $+ \cos(\frac{3a}{2})\cos(\frac{3b}{2})\cos(\frac{3c}{2})$

Solution
we have $\sin(3x) = 3\sin(x) - 4\sin^3(x)$
hence
$\sin ^3 x = (\frac{3}{4}\sin x - \frac{1}{4}\sin (3x))$
so 
$\sin^3 A+\sin^3 B + \sin^3 C = \frac{3}{4}( \sin A + \sin B + \sin C) - \frac{1}{4}( \sin 3A + \sin 3B + \sin 3C)\cdots(1)$
 we have if $A + B+ C = \pi$
$\sin \frac{A+B}{2} = sin (\frac{\pi}{2}- \frac{C}{2}) = \cos \frac{C}{2}$
and $\cos \frac{A+B}{2} = \cos (\frac{\pi}{2} - \frac{C}{2}) = \sin  \frac{C}{2}$

Now
$\sin A + \sin B + \sin C= 2 \sin \frac{A+B}{2} \cos\frac{A-B}{2} + 2 \sin \frac{C}{2} \cos \frac{C}{2}$
= $2 \cos\frac{C}{2} \cos\frac{A-B}{2} + 2 \cos\frac{A+B}{2}  \cos \frac{C}{2}$
= $2 \cos   (\cos \frac{A-B}{2} + \cos \frac{A+B}{2})$
= $4 \cos \frac{A}{2}cos \frac{B}{2} \cos \frac{C}{2}$
hence $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2}cos \frac{B}{2} \cos \frac{C}{2}\cdots(2)$
By similar argument if
$3A+3B+3C = 3\pi$
then
$\sin \frac{3A + 3B}{2} = \sin \frac{3\pi- 3c}{2} = \cos \frac{3C}{2}$
and
$\cos \frac{3A + 3B}{2} = \cos \frac{3\pi- 3c}{2} = -\sin \frac{3C}{2}$
using this we get
$\sin 3A + \sin 3B + \sin 3C = - 4 \cos \frac{3A}{2}cos \frac{3B}{2} \cos \frac{3C}{2}\cdots(3)$
using (1), (2),(3) we get the result

2015/114) Solve in real $a^3 + b^3 = 8 – 6ab$

$a^3 + b^3 - 8 + 6ab = 0$

or

$a^3 + b^3 + (-2)^3 -3(-2)(a)(b) = 0$

=> $a + b – 2 = 0$ or $a=b= - 2$

using the fact

$x^3 + y^3 + z^3 – 3xyz = \dfrac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

2015/113) There is a number n between to successive squares.this is k larger than the smaller number and l smaller than the larger number . Prove that n-kl is a perfect square.

Let the number n be between $a^2$ and $(a+1)^2$
As per given condition
$n- a^2 = k\cdots(1)$
$(a+1)^2 –n = l\cdots(2)$
Adding (1) and (2)
$k + l = 2a + 1$
or $l = 2a + 1 - k$
now
$n- kl = (a^2+k) – k(2a+1-k) = a^2 + k – 2ak –k + k^2 = a^2 – 2ak + k^2 = (a-k)^2$
Hence proved

2015/112) show that $\cos2A + \cos6A + \cos8A = \dfrac{\sqrt{13)}-1}{4}$ where $A = \dfrac{pi}{13}$

Let
$x = \cos2A + \cos6A + \cos8A \cdots(1)$
by seeing that $\sqrt{13}$ on right
square both sides of (1) to get
$x^2 = \cos^2 2A + \cos^2 6A + \cos^2 8A + 2 \cos 2A \cos 6A + 2 \cos 2A \cos 8A + 2 \cos 6A \cos 8A$
multiply by 2 to get
$2 x^2 = 2 \cos^2 2A + 2 \cos^2 6A + 2 \cos^2 8A$
        + $2(2 \cos 2A \cos 6A + 2 \cos 2A \cos 8A + 2 \cos 6A \cos 8A)$
= $\cos 4A + 1 + \cos 12 A + 1 + \cos 16 A + 1$ 
                    + $2( \cos 8A +\cos 4 A + \cos 10 A + \cos 6A + \cos 14 A + \cos 2 A)$
= $3 + \cos 4A + \cos 12 A + \cos 16 A$
                      + 2 $( \cos 8A + \cos 6A + \cos 2A + \cos 4A + \cos 10 A + \cos 14A)$

Now $\cos 16 A = \cos 10 A$ as $26 A = 2\pi$
$\cos 14 A = \cos 12 A$ as $26 A = 2\pi$
So we continue
= $3 + \cos 4A + \cos 12 A + \cos 10 A + 2(x+ \cos 4A + \cos 10 A + \cos 12 A)$
= $3 + 2x + 3 (\cos 4A + \cos 12 A+ \cos 10 A)$
Now $\cos 2A + \cos 4A + \cos 6A + \cos 8A + \cos 10 A + \cos 12A = \dfrac{-1}{2}$

So $\cos 4A + \cos 12 A+ \cos 10 A = (\dfrac{- 1}{2}-x)$

So $2x^2 = 3 + 2x + 3(\dfrac{-1}{2}- x)$
Or $4x^2 = 6 + 4x -3 – 6x$
Or $4x^2 + 2x -3 = 0$

This has one positive solution $\dfrac{\sqrt{13)}-1}{4}$ and one negative solution

As $\cos2A + \cos6A + \cos8A = \cos 2A + \cos 6A – \cos 5A$ and $\cos6 A > 0$ and $\cos 2A > \cos 5A$ so this is > 0
So this is $\dfrac{\sqrt{13)}-1}{4}$


I have solved at http://in.answers.yahoo.com/question/index;_ylt=ApwbfmdwbadqGq5DHY.KZsiRHQx.;_ylv=3?qid=20120607073236AAFCjFo

2105/111) Given 4 positive integers a,b,c and d such that $a^5=b^4, c^3=d^2$ and $c−a=19$ what is $d−b$

as $c^3 = d^2$ so c will be a square let $c = x^2$

as $a^5 = b^4$ so $a = y^4$

now
$c-a = 19$
=> $x^2 - y^4 = 19$
=> $(x-y^2)(x+y^2) = 19$
hence $x - y^2 =1$ and $x+y^2 = 19$ as 19 is a prime
so $x = 10$ and $y = 3$

so $a = y^4$ or $a^5 = y^20 = b^ 4$ or $b= y^5 = 243$

$c= x^2 => c= 100$ and hence $d^2 = 10^6$ and so $d = 1000$
$d-b = 1000 - 243 = 757$

2015/110) If 'a' and 'b' are the roots of $x^2-3x+1=0$ then

find the value of $\frac{a^{2014} + b^{2014} + a^{2016} + b^{2016}}{ a^{2015} + b^{2015}}$

Solution

a is root of $x^2 – 3x + 1=0$

so $a^2 - 3a + 1 = 0$

or

$a^2 + 1 = 3a$

so

$\dfrac{a^{2014} + a^{2016}}{a^{2015}} = \dfrac{1+a^2}{a} = 3 \cdots(1)$

Similarly

$\dfrac{b^{2014} + b^{2016}}{b^{2015}} = \dfrac{1+b^2}{b} = 3 \cdots(2)$

using

if $\dfrac{x}{y} = \dfrac{z}{w}$ then both are $\dfrac{x+z}{y+w}$

we get

$\dfrac{a^{2014} + b^{2014} + a ^{2016} + b^{2016}}{a^{2015} + b^{2015}} =  3$

2015/109) A triangle with sides 10, 24, and 26 is inscribed in a circle. What is the radius of the circle?

This is a right angled triangle because $10^2 + 24^2 = 26^2$ ( as u can check)
Therefore the hypotneuse forms the diameter and therefore the radius is $\frac{26}{2} =13$

2015/108) If $a + b = 1$ and $a^2 + b^2 = 2$ what is the value of $a^3 + b^3$ ?

we have

$a+b =1 \cdots(1)$
$a^2+b^2 = 2 \cdots(2)$
from 1st we get
$(a+b)^2 = a^2 +b^2 + 2ab = 2 + 2ab = 1$
or $ab = \dfrac{- 1}{2}$

hence
$a^3+b^3 = (a+b)^3 - 3ab(a+b) = 1 - 3 * \dfrac{-1}{2} * 1 = \dfrac{5}{2}$

2015/107) find the sum of $\frac{1}{3}+\frac{4}{9}+\frac{7}{27}+\frac{10}{81}+\cdots$.

The nth term = $\dfrac{3n+1}{3^{n+1}}$ n is from 0 to infinite
= $\dfrac{n}{3^n}+ \dfrac{1}{3}\cdot\dfrac{1}{3^n}$
The second term is GP and the sum upto infinite terms is $\dfrac{1}{3}\cdot\dfrac{1}{1-\frac{1}{3}} = \dfrac{1}{2}$
let $f(x) = 1 + x + x^2 + \cdots = \dfrac{1}{1-x}$
then differentiate both sides wrt x to get
$f'(x) = 1 + 2x + 3x^2 + \cdots. = \dfrac{1}{(1-x)^2}$
multiply by x to get
$xf'(x) = x + 2x^2 + 3x^3 \cdots = \dfrac{x}{(1-x)^2}$
$x = \dfrac{1}{3}$ gives the 1st sum as $\dfrac{\frac{1}{3}}{(\frac{2}{3})^2}= \dfrac{9}{4}\cdot\dfrac{1}{3} = \dfrac{3}{4}$
so sum = $\dfrac{3}{4}+ \dfrac{1}{2} = \dfrac{5}{4}$

2015/106) Factorize: 2x^2-5xy-3y^2+3x+19y-20

The factor is of the form $(ax+by+c)(dx+ey+f)$

if we ignore c and f and multiply we get

$(ax+by)(dx+ey)= adx^2 + (ae+bd)xy + bey^2$

so we can first factor $2x^2-5x-3y^2$ and then evaluate the other parts

$2x^2-5xy-3y^2= (2x+y)(x-3y)$ factored by quadratic method

so

 $2x^2-5xy-3y^2+3x+19y-20= (2x+y+e)(x-3y+f)$

      $= 2x^2-3y^2-5xy+(2f+e)x + (f-3e)y + ef$

Comparing coefficients of x , y and constant separately we get
$2f+e = 3, f- 3e = 19$ and $ef = - 20$
solving 1^st 2 equations we get $e = -5$ and $f = 3$ and 3^rd equation also meets criteria
so we get 

$2x^2-5xy-3y^2+3x+19y-20= (2x+y-5)(x-3y+4)$

2015/105) if $a=xy^2+yz^2+zx^2$ and $b=x^2y+y^2z+z^2x$ then express $(x^3-y^3)(y^3-z^3)(z^3-x^3)$ in terms of a and b

$(x^3-y^3)(y^3-z^3)(z^3-x^3)$
= $(x^3-y^3)(y^3z^3-y^3x^3-z^6+z^3x^3)$
=$x^3y^3z^3-y^3x^6-z^6x^3+z^3x^6-y^6z^3+y^6x^3+y^3z^6-x^3z^3x^3$
=  $-y^3x^6-z^6x^3+z^3x^6-y^6z^3+y^6x^3+y^3z^6$
= $(z^3x^6+y^3 z^6 + x^3y^6) - (y^3x^6 + x^3z^6 + y^6z^3)$
HENCE
$(x^3-y^3)(y^3-z^3)^(z^3-x^3)$
  $= ((xy^2)^3 + (yz^2)^3+(zx^2)^3) -  ((x^2y)^3 +(y^2z)^3 + (z^2x)^3)\cdots(1)$

we have
$a^3+b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)$
so  $((xy^2)^3 + (yz^2)^3+(zx^2)^3) = (xy^2+yz^2+zx^2)^3- 3(xy^2+yz^2)(yz^2+zx^2)(zx^2+xy^2)$
= $(xy^2+yz^2+zx^2) - 3y(xy+z^2)z(x^2+yz)x(xz+y^2)$
= $b^2-3xyz(x^2+yz)(y^2+xz)(z^2+xy)\cdots(2)$
similarly
$((x^2y)^3 + (y^2z)^3+(z^2x)^3) =a^2-3xyz(x^2+yz)(y^2+xz)(z^2+xy)\cdots(3)$
from (1), (2) and (3) we get

$(x^3-y^3)(y^3-z^3)^(z^3-x^3))=b^3 - a^3$

2015/104) Solve x^x^x^x^ to the infinity=10

x^x^x^x^ to the infinity=10

so $x^{10} = 10$

$x = \sqrt[10]{10}$

2015/103) If $xy + yz + zx =0$ then $\dfrac{1}{x^2 - yz} + \dfrac{1}{y^2 - zx} +\dfrac{1}{z^2 - xy}$ is equal to

$xy + yz + zx = 0$

so $yz = -x(y+z)$
$x^2-yz = x(x+y+z)$

so $\dfrac{1}{x^2-yz} = \dfrac{1}{x(x+y+z)}$
similarly$\dfrac{1}{y^2-xz} = \dfrac{1}{y(x+y+z)}$
and
$\dfrac{1}{z^2-xy} = \dfrac{1}{z(x+y+z)}$

adding all 3 we get your expression

$\dfrac{1}{x^2 - yz} + \dfrac{1}{y^2 - zx} +1\dfrac{1}{z^2 - xy} = \dfrac{1}{x+y+z}(\dfrac{1}{x}+\dfrac{1}{y} + \dfrac{1}{z})$

$\dfrac{1}{x} +\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{yz+xz+xy}{xyz} = 0$

so
$\dfrac{1}{x^2 - yz} + \dfrac{1}{y^2 - zx} +\dfrac{1}{z^2 - xy} = 0$

2015/102) Two roots of the polynomial $x^3 + ax^2 + 15x -7 = 0$ are equal and rational. Find "a"

If 2 roots are rational then 3rd must be rational.
possible roots are -7 , -1, 1, 7
now product of 2 roots (as same) so roots shall be +1 or - 1 as 7 or -7 cannot be a double root

so 3rd root has to be 7 ( as product has to be +ve)

so $f(x) = x^3 + ax^2 + 15x -7 = 0 = f(7)$

or $343 + 49 a + 105 - 7 = 0$

or $a = 9$

check:
$x^3 - 9x^2 + 15x - 7 = (x-1)^2(x-7)$

2015/101) factor $a(b^2+c^2-a^2) + ...$(the full question below)

 $a(b^2+c^2-a^2) + b(c^2+a^2-b^2) + c(a^2+b^2-c^2) - 2abc$
Solution 

$a(b^2+c^2-a^2) + b(c^2+a^2-b^2) + c(a^2+b^2-c^2) - 2abc$
=$(ab^2+ac^2-a^3 + bc^2+ a^2b -b^3) + c(a^2+b^2-c^2) - 2abc$
= $(c^2(a + b) + a(b^2-a^2)   + b(a^2-b^2) + c(a^2+b^2-c^2) - 2abc$
= $(c^2(a+b) + (a-b) (b^2 - a^)) + c(a^2+b^2-c^2) - 2abc$
= $(c^2(a+b) -  (a-b)^2(a+b))+ c(a^2+b^2-c^2) - 2abc$
= $(a+b)(c^2 - (a-b)^2) + c(a^2 + b^2 -2ab -c^2)$
= $(a+b)(c^2 - (a-b)^2) + c((a-b)^2 -c^2)$
= $(c^2 - (a-b)^2)((a+b)-c)$
=  $(c+a-b)(c-a+b)(a+b-c)$
=$(a+b-c)(b+c-a)(c+a-b)$

2015/100)n any triangle ABC, prove that

$a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C -\cos^2 A) + c^2(\cos^2 A - \cos^2 B) = 0$

proof:

$a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C - \cos^2 A) + c^2(\cos^2 A - \cos^2 B)$
= $a^2(\sin^2 C - \sin^2 B) + b^2 (\sin^2 A - \sin^2 C) + c^2( \sin^2 B - \sin^2 A)$

using law of sines we have

let $\dfrac{a}{\sin A} = \dfrac{b}{sin B} = \dfrac{c}{\sin C} = k$ (say)

we get

$a^2(\sin^2 C - \sin^2 B) + b^2 (\sin^2 A- \sin^2 C) + c^2( \sin^2 B - \sin^2 A)$

=$k^2 \sin ^2 A(\sin^2 C - \sin^2 B) + k^2 \sin ^2 B (\sin^2 A -\sin^2 C) + k^2 \sin ^2 C( \sin^2 B - \sin^2 A)$

= $k^2( \sin ^2 A \sin ^2 C – \sin ^2 A \sin ^2 B + \sin ^2 B \sin ^2 A – \sin ^2B \sin^2 C + \sin ^2 C \sin ^2 B – \sin ^2C \sin ^2 A)$

= 0

21015/099) Solve the system of equations

$a(b+c+d+e+f)=184$

$c(a+b+d+e+f)=301$

$e(a+b+c+d+f)=400$

$b(a+c+d+e+f)=225$

$d(a+b+c+e+f)=225$

$f(a+b+c+d+e)=525$

Solution

We note that all the rest are of the form $n(g-n)$
where $g=a+b+c+d+e+f$ and observe that smaller the value, the smaller n and hence b = d

Factoring, we see that
$a(g-a) = 2^2 * 3 * 7$
$b(g-b) = d(g-d) = 3^2 * 5*2$
$e(g-e) = 2^4 * 5^2$
$f(g-f) = 3 * 5^2 * 7$
and
a < b = d < e < f

Because b and d must be greater than a,

so we have the set for a 2,3,4,7

b = 3,5,

d= 3,5

e= 2,5,10

f = 3,5,15 so on

we can quickly find values that satisfy that and the last four equations: they all work out if g=50, which implies that c=7.

This produces the solution
$a=4, b=d=5, c=7, e=8, f=21$

 

2015/098) Prove $(4 \cos\, 20^\circ + 1) \tan 20^\circ = \sqrt3$

we have $tan\, 60^\circ – \tan\, 20^\circ$
= $\dfrac{\sin\, 60^\circ}{\cos\,60^\circ} – \dfrac{\sin\, 20^\circ}{\cos\,20^\circ}$
= $\dfrac{\sin\, 60 ^\circ\cos\, 20^\circ – \cos\, 60^\circ \sin\, 20^\circ}{\cos\, 60^\circ\, \cos\, 20^\circ}$
= $\dfrac{\sin\, 40^\circ}{\cos\, 60^\circ \cos\, 20^\circ}$
=  $2\dfrac{\sin\, 40^\circ}{\cos\, 20^\circ}$
= $2 \dfrac{2 sin\, 20^\circ \cos\, 20^\circ}{\cos\, 20^\circ} = 4 \sin\, 20^\circ$
hence $4\sin \, 20^\circ + \tan\, 20^\circ = \sqrt3$
or $4 \cos\, 20^\circ \tan\, 20^\circ + \tan \,20^\circ = \sqrt3$
or $(4 \cos\, 20^\circ + 1) \tan\, 20^\circ = \sqrt3$  

2015/097) Find the equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

Let the slope of line be m

clearly m is -ve as if m is positive or zero then it shall not cut x and y both in 1st quadrant
so the equation of line is

 $y-5 = m(x-3)$


as it goes via $(3,5)$
 
now x intercept when x = 0 is $y=5-3m$

y intercept when y = 0 is given by $x = \dfrac{3m-5}{m}$    
 so area of the triangle in 1st quadrant is $\dfrac{xy}{2}$
so we need to minimize $xy = \dfrac{(3m-5)^2}{m}$

let m = - p where p > 0

$xy=\dfrac{(5+3p)^2}{p}=(\frac{5}{\sqrt{p}}+3\sqrt{p})^2$
or  $xy=(\frac{5}{\sqrt{p}}-3\sqrt{p})^2+60$

clearly it is lowest when  $\frac{5}{\sqrt{p}}-3\sqrt{p}=0$


 so $p=\frac{5}{3}$

so equation of line is
$y=5-\frac{5}{3}(x-3)$
or $3x+5y=30$

Ths problem appeared as Problem of the week in http://mathhelpboards.com/potw-university-students-34/problem-week-115-june-9th-2014-a-10908.html

where you can find some solution. I answered the question correctly but my solution is not provided there.

2015/096) $ax^2 + bx + c = 0$ has imaginary roots and $a + c \lt b$ then prove that $4a + c \lt 2b$

$f(x) = ax^2 + bx + c$ has imaginary roots so this expression is positive or -ve for all x

$f(-1)= a - b + c < 0$ as$a + c < b$

so

$f(-2) = 4a - 2b +c < 0$ or $4a + c < 2b$

2015/095) If $x^{a}= (x^\frac{b}{2}) (z^\frac{b}{2}) =z^c$ then a,b,c are in?

A) H.P, B) G.P, C) H.P ,D)non of these

we have $x^a = z^c$
so $x =z^\frac{c}{a}\cdots(1)$
further
$x^{a-\frac{b}{2}}=z^{\frac{b}{2}}$
or $x^{2a-b} = z^{b}$
from 1
$z^{(2a-b)*^\frac{c}{a}} = z^{b}$
or
$(2a-b)c= ab$
or
$\dfrac{2}{b} =  \dfrac{1}{a} + \dfrac{1}{c}$
hence they are in HP

2015/094) One roots of a quadratic equation $ax^2+bx+c=0$ is three times the other.

Prove that $3b^2=16ac$

Solution

we are given

$ax^2 + bx+ c = 0 \cdots(1)$

let the roots be t and 3t

so we have $(x-t)(x-3t) = 0$

or $x^2 – 4tx + 3t^2 = 0\cdots(2)$

the roots of(1) and (2) are same so coefficients are proportionate

so

$\dfrac{a}{1} = \dfrac{-b}{4t} = \dfrac{c}{3t^2}$

we need to eliminate t

$4at = -b\cdots(3)$

and $-b = \dfrac{4c}{3t}\cdots(4)$

 

multiplying (3) with (4) we get

$b^2 = \dfrac{16ac}{3}$ or $3b^2 = 16ac$

 

2015/093) Show that $\dfrac{\sec\,A +\tan\, A - 1}{\tan\, A- \sec\, A +1} = \dfrac{\cos\, A}{1-\sin\,A}$

we have
$\dfrac{\sec\,A +\tan\, A - 1}{\tan\, A- \sec\, A +1}$
= $\dfrac{\sec\,A +\tan\, A - (\sec^2A-\tan ^2 A}{\tan\, A- \sec\, A +1}$
=  $\dfrac{\sec\,A +\tan\, A - (\sec\,A+\tan\, A)(\sec\,A-\tan\, A)}{\tan\, A- \sec\, A +1}$
= $\dfrac{(\sec\,A +\tan\, A)(1 - \sec\,A+\tan\, A)}{\tan\, A- \sec\, A +1}$
= $(\sec\,A +\tan\, A)$
= $\dfrac{1}{\cos\, A }+  \dfrac{\sin\, A}{\cos\, A }$
=  $\dfrac{1+ \sin\, A}{\cos\, A }$
=  $\dfrac{(1+ \sin\, A)(1-\sin\, A)}{\cos\, A(1-\sin\, A) }$
= $\dfrac{1-\sin^2A}{\cos\, A(1-\sin\, A) }$
=  $\dfrac{\cos^2A}{\cos\, A(1-\sin\, A) }$
= $\dfrac{\cos\,A}{1-\sin\, A}$

2015/092) Solve $\dfrac{x-a}{b+c} + \dfrac{x-b}{a+c} + \dfrac{x-c}{a+b} = 3$ given a,b,c positive

$\dfrac{x-a}{b+c} + \dfrac{x-b}{a+c} + \dfrac{x-c}{a+b} = 3$
or $(\dfrac{x-a}{b+c}-1) + (\dfrac{x-b}{a+c}-1) + (\dfrac{x-c}{a+b}-1) =0$
or $\dfrac{x-a-b-c}{b+c} + \dfrac{x-b-a-c}{a+c} + \dfrac{x-c-a-b}{a+b} =0$
or  $(x-a-b-c)(\dfrac{1}{b+c} + \dfrac{1}{a+c} + \dfrac{1}{a+b}) =0$
so $x-a-b-c) =0$ or  $\dfrac{1}{b+c} + \dfrac{1}{a+c} + \dfrac{1}{a+b} =0$
as a,b,c are positive so 2nd expression >0 and hence x = a+b+c 

2015/091) Let p be a prime number. Prove that 6[(p-4)!] = 1 (mod p )

we have 6 = 2 * 3
p is prime
now (p-2) mod p = -2
(p-3) mod p = - 3

so 6(p-4)! = - (p-2)(p-3) (p-1)(p-4)! = - (p-1)!

so 6(p-4)! mod p = - 1(p-1)! mod p = -1 * (-1) = 1 mod p

2015/090) Find $\frac{1+a}{1-a}$ if $a = \cos\beta + i\sin\beta$.

$a = \cos\beta + i\sin\beta$
so
 $1+ a = 1+ \cos\beta + i\sin\beta= 2\cos^2\frac{\beta}{2}+2i\cos\frac{\beta}{2}\sin\frac{\beta}{2}= 2 \cos\frac{\beta}{2}(\cos\frac{\beta}{2} +i\sin\frac{\beta}{2})$

further
$1 - a = 1 - \cos\beta -i\sin\beta = ( 2 \sin ^2 \frac{\beta}{2} - 2 i \cos\frac{\beta}{2} \sin\frac{\beta}{2}) = 2 \sin \frac{\beta}{2}(\sin \frac{\beta}{2}- i \cos\frac{\beta}{2})$
= $- 2i \sin \frac{\beta}{2}(\cos\frac{\beta}{2} + i \sin \frac{\beta}{2})$

so $\frac{1+a}{1-a}= \frac{\cos\frac{\beta}{2}}{- i sin \frac{\beta}{2}} = i\cot \frac{\beta}{2}$ 

2015/089) lf $z= \frac{a+3i}{2+ai}$ , a is real.

Show that there is only one value of a for which arg z = $\frac{pi}{4}$  and find it
Solution
$arg(z) = \frac{pi}{4}$

z must be located at first quadrant...


now $\frac{a+3i}{2+ai} = \frac{(a+3i)(2-ai)}{(4+a^2)}$

= $\dfrac{5a + (6-a^2)i}{4+ a^2}$

Arg z = 1 when $5a = 6- a^2$
or $a^2 + 5a – 6 = 0$
$(a+6)(a-1) = 0$
a = -6 or 1
a = -6 means $arg(z) = \frac{3pi}{4}$ and hence this is invalid value and so $a = 1$ is the correct ans.
Refer to https://in.answers.yahoo.com/question/index?qid=20111204094456AAQhinr

2015/088)If a,b,c are real and distinct, then show that

$a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\gt6abc$

taking 6abc to the left and reordering we get

$(a^2b^2-2bac + c^2) + (a^2c^2-2abc + b^2) + (b ^2c^2- 2abc + a^2)$

= $(ab-c)^2 + (ac-b)^2 + (bc-a)^2 \ge 0$

the expression can be zero only if $ab= c, ac = b , bc = a$ or $abc =$1 or $zero$

if a,b,c all are different then above cannot be zero

2015/087) If two vertices of an equilateral triangle are $(0,0),(3,\sqrt{3})$ then find the third vertex

one point is origin $(0,0)$ and another is $(3,\sqrt3)$

the length of the side = $\sqrt{12} = 2\sqrt3$

So it is at 30 degrees with x axis.

For it to be equilateral triangle the 2^nd side shall be at 60 degrees more or 60 degrees less
that is there are 2 triangle one is at 30 degrees more that is at 90 degrees which gives co-ordinates $(0, 2\sqrt3)$
another triangle at 30 degrees less that is at -30 degrees which gives co-ordinates $(3, -\sqrt{3})$

2015/086) Let $S=y^2+20y+12$ y is a positive integer.

What is the sum of all possible values of y for which S is a perfect square.

Solution

let $S = x^2$

so $x^2 = (y+10)^2 – 88$

or $88 = (y+10)^2-x^2 = (y+10 + x) (y+10-x)$

both the terms on RHS has to be even as one add and one even shall give fractional x and y

so $(y+10+x) = 44, (y + 10 – x) = 2$ giving $y = 13, x = 21$

or $(y+10+x) = 22, (y+10-x) = 4$ giving $y=3 , x = 9$

so sum of all possible values of $y = 16$

2015/085) Prove that

$(x+y)^3 + (y+z)^3 + (z+x)^3 -3(x+y)(y+z)(z+x)=2(x^3+y^3+z^3 – 3xyz)$
Solution 
knowing
$a^3+b^3 + c^3 – 3abc = \frac{1}{2}(a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)\cdots (1)$
putting $a = x + y , b= y + z, c= z+x$ we get

$(x+y)^3 + (y+z)^3 + (z+x)^3 -3(x+y)(y+z)(z+x)$

= $\frac{1}{2}(2x+2y+ 2z)((x-z)^2 + (y-x)^2 + (z-y)^2)$

= $2 * \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2 + (z-x)^2)$

= $2(x^3+y^3+z^3 – 3xyz)$ using (1)

2015/084) Prove that the locus of the center of the circle

$\frac{1}{2}(x^2 + y^2) + x \cos(\theta) + y \sin(\theta) - 4 = 0$ is $x^2 + y^2 = 1$

Solution

we have $(x^2 + 2 x \cos \theta) + (y^2 + 2y \sin \theta ) = 8$
or $(x^2 + 2 x \cos \theta+1) + (y^2 + 2y \sin \theta+ 1 ) = 10$

or $( x + \cos \theta)^2 + (y + \sin \theta ) ^2 = 10$
So the locus of the centre
$x = - \cos \theta$
$y = - \sin \theta$
to eliminate $\theta$
or $x^2 + y ^2 = 1$
above is the locus
so above is true

 

2015/083) The L.C.M and H.C.F of two numbers are 1760 and 32 respectively. If one of the number is 160 find the other.

$1760 = 160 * 11 = 32 * 5 * 11$
$160 = 32 * 5$

32 is the HCF and and one number is 32 * 5
the other number has to be 32m when 32m* 5 = 1760 and m is coprime to 5
so m = 11
so other number = 32 * 11 = 352

This can also be solved by using the rule that product of HCF and LCM is the product of numbers

2015/082) Evaluate $(\dfrac{n}{n+1})^{1 + n}$ as $n\rightarrow \infty$

$(\dfrac{n}{n+1})^{1 + n}$

Let's remember one thing first: lim $x\rightarrow\infty (1 + \frac{r}{x})^x = e^r$
$(\dfrac{n}{n+1})^{1 + n}$

=  $(1- \dfrac{1}{n+1})^{1 + n}$

substitute x = n + 1

Now, as n goes to infinity, then n + 1 will also go to infinity, and x will go to infinity

lim $n\rightarrow\infty (1 - \dfrac{1}{n + 1})^{n + 1}$ 

= lim $x\rightarrow\infty (1 - \dfrac{1}{x})^x$ 

 
Which is $e^{-1}$, or $\dfrac{1}{e}$

2015/081) If $a(y + z) = x, b(z + x) = y, c(x + y) = z$ show that $bc + ca + ab + 2abc = 1$

$a(y + z) = x$

Hence $\dfrac{1}{a} = \dfrac{y+z}{x}$

add 1 to both sides

$\dfrac{a+1}{a} = \dfrac{x + y +z}{x}$

so $\dfrac{a}{a+1} = \dfrac{x}{x+y+z}$

similarly

 

$\dfrac{b}{b+1} = \dfrac{y}{x+y+z}$

$\dfrac{c}{c+1} = \dfrac{z}{x+y+z}$

adding the above we get

$\dfrac{a}{a+1}+\dfrac{b}{b+1} +\dfrac{c}{c+1}= 1$

or $a(b+1)(c+1) + b(a+1)(c+1) + c(a+1)(b+1) = (1+a)(1+b)(1+c)$

or $(abc + ab + ac + a) + (abc + bc + ca + b) + (abc + ca + cb + c) = 1 + a + b+ c + ab + bc+ ca +abc$

hence $2abc + ab + bc + ca = 1$

 

2015/080) If $log_4 5 =a$ and $log_5 6 =b$, then $log_3 2 =$

From the given condition

$5= 4^a$
$6 = 5^b = (4^a)^b = (2^2)^{ab} = 2^{2ab}$

so $3 *2 = 2^{2ab}$

so $3 = \frac{2^{2ab}}{2} = 2^{2ab-1}$
or $2 = 3^{\frac{1}{2ab-1}}$

hence $log_3 2$=$\dfrac{1}{2ab-1}$

2015/079) if $a^{x-1} = bc, b^{y-1} = ca, c^{z-1} = ab$ show that $xy + yz + zx = xyz$

$a^{x-1} = bc$

hence $a^x = abc$

or $a = (abc)^{\frac{1}{x}}\cdots (1)$

similarly

$b = (abc)^{\frac{1}{y}}\cdots (2)$

$c = (abc)^{\frac{1}{z}}\cdots (3)$ 

hence $abc = (abc)^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

hence 

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$

or $yz + zx + xy = xyz$

proved

2015/078) Find the equation of the sphere passing $(a,0,0), (0,b,0), (0,0,c)$ and $(0,0,0)$

the general equation of sphere

$x^2+y^2+z^2+mx +ny + rz = p^2$

as it passes through $(0,0,0)$ put $(x,y,z) = (0,0,0)$ to get $p^2 = 0$ or $p = 0$

as it passes through $(a,0,0)$ put $(x,y,z) = (a,0,0)$ to get $a^2 + ma = 0$ or $m = -a$

similarly $n = -b$ and $r = -c$ and we get equation as

$x^2+y^2+z^2 - ax - by - cz = 0$

2015/077) Find the formula for the number pattern : $2,5,10,17 \cdots$

The 1st order difference $3,5,7$
The second order difference $2,2,2,$
So the term is quadratic

$t_n = an^2 + bn + c$

put $n = 1$ to get

$a + b+ c = 2\cdots(1)$

put $n =2$ to get

$4a + 2b + c = 5 \cdots(2)$

put $n = 3$ to get

$9a + 3b + c = 10 \cdots(3)$

subtract (1) from (2) to get
 

$3a + b = 3 \cdots(4)$

subtract (2) from (3) to get
 

$5a + b = 5 \cdots(5)$

from (4) and (5) $2a = 2$ or $a = 1$ so $b = 0$ and hence $c = 1$
so

$t_n = n^2 + 1$

 

2015/076) Prove that: $\tan\,x\tan(60^\circ-x)\tan(60^\circ+x) = \tan3x$

LHS = $\tan\,x\tan(60^\circ-x)\tan(60^\circ+x)$
= $\tan\,x\dfrac{\tan\,60^\circ-\tan\,x}{1+\tan\,60^\circ \tan\,x}\dfrac{\tan\,60^\circ+\tan\,x}{1-\tan\,60^\circ \tan\,x}$
=  $\tan\,x\dfrac{\tan^260^\circ-\tan^2x}{1- \tan^260^\circ \tan^2x}$
=  $\tan\,x\dfrac{3-\tan^2x}{1- 3 \tan^2x}$
=  $\dfrac{3\tan\,x-\tan^3x}{1- 3 \tan^2x}$
= $\tan 3x$ = RHS

2015/075) find $x^{63} + x^{44} + x^{37} + x^{31} + x^{26} + x^9 +6$

what is the value of the above expression for $(x + \dfrac{1}{x})^2 = 3$

we have
$x^2 + 2 + \dfrac{1}{x^2}= 3$
or $x^4 -x^2 + 1=0$
or $(x^4-x^2+1) (x^2+1)= 0$ as $x^2+1$ is not zero
or $x^6+1=0$

so $x^6= - 1$ and $x^{12} = 1$

so   $x^{63} + x^{44} + x^{37}+ x^{31} + x^{26} + x^9 +6$
= $x^3 + x^8 + x + x^7 + x^2 + x^9 + 6$ taking mod 12 of exponent
= $x^3(x^6+1) + x^2(x^6+1) + x ( 1 + x^6) + 6$
= 6 
This I have picked from https://in.answers.yahoo.com/question/index?qid=20150801013839AAEmdz0

2015/074) If $ax^2+bx+c=0$ has complex roots and $a + c \lt b$, then prove $4a+c\lt 2b$

because $f(x) = ax^2+bx+c=0$ has complex roots so this is positive for all x or -ve for all x

$f(-1) = a – b + c$
from given condition $a + c – b \lt 0$ so $f(-1) \lt 0$
so $f(-2) = 4a – 2b + c \lt 0$ or $4a + c \lt 2b$
proved

2015/073) Show that the following is true for all real values of x.

$\dfrac{a^2(x-b)(x-c)}{(a-b)(a-c) } + \dfrac{b^2(x-c)(x-a)}{(b-c)(b-a)} + \dfrac{c^2(x-a)(x-b)}{(c-a)(c-b)} = x^2$

Solution 2015/073)

This can be done in 2 ways

method 1

expand

we have 1^st term

= $\dfrac{a^2(x-b)(x-c)}{(a-b)(a-c)}$
= $\dfrac{-a^2(x-b)(x-c)}{(a-b)(c-a)}$

=$\dfrac{-a^2(x-b)(x-c)(b-c)}{(a-b)(b-c)(c-a)}$

= $\dfrac{x^2(- a^2(b-c)) +xa^2(b^2-c^2) -a^2bc(b-c)}{(a-b)(b-c)(c-a)}$

similarly we have 2^nd term

= $\dfrac{b^2(x-c)(x-a)}{(b-c)(b-a)}$

= $\dfrac{x^2(- b^2(c-a)) +xb^2(c^2-a^2) -b^2ca(c-a)}{(a-b)(b-c)(c-a)}$

3^rd term

= $\dfrac{c^2(x-a)(x-b)}{(c-a)(c-b)}$

= $\dfrac{x^2(- c^2(a-b)) +x(c^2(a^2-b^2) -c^ab(a-b)}{(a-b)(b-c)(c-a)}$

so the sum = 
$\frac{(x^2(- a^2(b-c)-b^2(c-a) -c^2(a-b)) +x(a^2(b^2-c^2) + b^2(c^2-a^2) + c^2(a^2-b^2)) – (a^2bc(b-c) + b^2ca(c-a) + c^2ab(a^2-b^2)}{(a-b)(b-c)(c-a))}$

= $\dfrac{x^2(- a^2(b-c)-b^2(c-a) -c^2(a-b)}{(a-b)(b-c)(c-a)}$

numerator = $(x^2(- a^2(b-c)-b^2(c-a) -c^2(a-b))$

= $x^2(-a^2(b-c) – b^2c+ ab^2 -ac^2 +bc^2)$

= $x^2(-a^2(b-c) – b^2c+ bc^2 -ac^2 +ab^2)$

= $x^2(-a^2(b-c) – bc(b-c) + a(b^2-c^2))$

= $x^2(-a^2(b-c) – bc(b-c) + a(b+c)(b-c))$

= $x^2(b-c)(-a^2-bc+ab+ac)$

= $x^2(b-c)(a-b)(c-a)$

so we have ratio = $x^2$

hence proved

Aliter (Method 2)

This can be proved as below

because this is quadratic this cannot have more than 2 root.

But x= a, x= b, x= c satisfy the condition.

Hence this has to be identity.

Proved

2015/072) show that $\tan 3x - \tan x = \dfrac{2 \sin\, x}{cos 3x}$

we have $\tan 3x - \tan\, x = \dfrac{\sin 3x}{\cos 3x} - \dfrac{\sin\, x}{\cos\, x}$
= $\dfrac{\sin 3x \cos\, x - \cos 3x \sin\, x}{\cos 3x \sin\, x}$
= $\dfrac{\sin 2x}{\cos3x \sin\, x}$
= $\dfrac{2 \sin\, x \cos\, x}{\cos 3x \sin\, x}$
= $2 \dfrac{\sin\, x}{\cos 3x}$

2015/071) Factor $y^4-y^2+16$

This can be factored by completing the square. By changing the $y^2$ term

$y^4 -y^2 +16$
= $y^4 -y^2+8y^2-8y^2 +16$
= $y^4 +8y^2 +16 -9y^2$
= $(y^2+4)^2 -(3y)^2$
= $(y^2-3y+4)(y^2+3y+4)$

2015/070) Find the point on the line $2 x + 4 y + 1 = 0$ which is closest to the point $(-1, -3 )$

There are different methods to solve this problem. I shall use one of them.
 
Because the point is closest to $(-1,-3)$ the line joining the point and $(-1,-3)$ shall be perpendicular to $(2x + 4y + 1) = 0$

slope of $2x + 4y + 1 = 0$ is $\dfrac{- 1}{2}$
so slope of perpendicular line is 2
so the equation of line
$y = 2x + c$
as it passes through $(-1,-3)$
so $-3 = - 2 + c$ or  $c = -1$
so $y = 2x -1$
solving this and $2x + 4y + 1 = 0$ we get the result
$2x + 4 ( 2x-1) + 1 = 0$
or $10x - 3 = 0$
$x = \dfrac{3}{10}$ and $y = \dfrac{- 2}{5}$

So  the point is $(\dfrac{3}{10}, \dfrac{- 2}{5})$
 
 

2015/069) $P(x)$ is unknown and when divided by $(x+1)(x-3)$ the remainder becomes $2x+7$

What is the remainder when $P(x)$ is divided by $(x-3)$ ?

Solution
Because $P(x)$ divided by $(x+1)(x-3)$ the remainder becomes $2x+7$ so there exists $Q(x)$ such that

$P(x) = Q(x) (x+1)(x-3) + 2x + 7$
so remainder when divided by $x-3$ is using remainder theorem
$P(3) = Q(3) (x+1) 0 + 2 * 3 + 7 = 13$

2015/068) find parametric form of Pythagorean triplet for $(3,4,5), (5,12,13) \cdots$ that is difference between hypotenuse and one leg is 1

that means we need to find $(x,y,y+1)$ where

$x^2 + y^2 = y^2 + 2y + 1$

or $x^2 = 2y + 1$

now x has to be odd say $2n + 1$

so $x^2 = 4n^2 + 4n + 1 = 2y + 1$

or $y = 2n^2 + 2n$

so the sides are $( 2n+ 1, 2n^2+2n, 2n^2+2n+ 1)$

by putting n = 1,2 ,3 we get the sequence

2015/067) Integrate $\dfrac{1}{e^x(1+e^x)}$

1st we convert it into partial fraction
$\dfrac{1}{e^x} - \dfrac{1}{e^x +1}$
= $e^{-x} - \dfrac{e^{-x}}{1+ e^{-x}}$
integral of $e^{-x}$ is $-e^{-x}$

and putting $1+e^{-x} = t$ we get $-e^{-x} dx = dt$
so integral of $\dfrac{e^{-x}}{1+ e^{-x}}$ = $ln | 1 + e^{-x} |$
but as $1 + e^{-x} > 1$ and hence > 0 we get the integral of given expression

$- e^{- x} + ln ( 1+ e^{-x}) + C$

2015/066) A line passes through A (1,1) and B (100,1000). How many other points with both coordinates integers are on this line segment between A and B.

One of the points is $A(1,1)$ and second point $B(100,1000)$ The slope is $\dfrac{999}{99}$ or in lowest form $\dfrac{111}{11}$. The equation of line is

$(y-1) = \dfrac{111}{11}(x-1)$

or $11(y-1) = 111(x-1)$

so $y-1 = 111t$ and $x-1 = 11t$

$y = 111 t + 1$ and $x = 11 t + 1$

for x and y to be integer both $111 t$ and $11t$ have to be integer and hence $t$ is integer

the 1^st point is for t = 0 and 2^nd point for t = 9

there are 8 values of t( from 1 to 8) so number of points = 8

2015/065) What is the value of f(14400) from the following case?

• Function f from the positive integers to the positive integers satisfies the following conditions:
  1. $f(xy)=f(x)+f(y)-1$ for any pair of positive integers x and y.
  2. $f(x)=1$ holds for only finitely many x.
  3. $f(30)=4$
  
  Solution
  
  From (1) we find that
  $f(30)=f(2*15)=f(2)+ f(15)-1$
  = $f(2)+ f(3*5)-1$
  = $f(2) + f(3) + f(5) – 2$
  
  With (3) this yields:
  $f(2)+f(3)+f(5)-2=4$
  $f(2)+2f(3)+f(5)=6 (4)$
  Lemma (1)
  
  
  Neither f(2), nor f(3), nor f(5) can be 1.
  
  Proof
  Suppose  one of them is 1, say $f(2)$, then $f(2^k)=kf(2)-k+1=1$
  That means that infinitely many numbers x have $f(x)=1$
  This is a contradiction with (2).
  Combining lemma with [1] tells us that f(2)=f(3)=f(5)=2.
  Lemma (2)
  
  $f(a^n) = n f(a) – (n-1)$
  
  
  
  the above can be proved by induction
  
  
  
  It follows from (1) that:
  $f(14400)=f(144 * 100)$
  = $f( 2^ 4 * 3^ 2 * 2^2 * 5^2)$
  = $f(2^6 * 3^2 * 5^2)$
  = $6f(2) + 2f(3) + 2f(5)- 9 = 6 * 2 – 5 + 2 * 2 -1 + 2 * 2 – 1 - 2 = 10*2 - 9 = 11$
  I have solved the same at https://in.answers.yahoo.com/question/index?qid=20131011072645AAoFbsS
  
  
  
  
  
  
  
  

2015/064) If $\alpha$ and $\beta$ are the solutions of the equation $a \tan \theta + b \sec \theta = c$ , then show that $\tan (\alpha + \beta)= 2\frac{ac}{a^2-c^2}$

$b \sec \theta = (c- a \tan \theta)$

so $b^2 \sec^2 \theta = (c-a \tan \theta)^2$

or $b^2 ( 1 + \tan ^2 \theta) = c^2 + a^2 \tan ^2 \theta - 2ac \tan \theta$

or $(a^2-b^2) \tan ^2 \theta - 2ac \tan \theta + (c^2-b^2) = 0$

or $tan ^2 \theta - \dfrac{2ac}{a^2-b^2} + \dfrac{c^2-b^2}{a^2-b^2} = 0 \cdots(1)$

if $\alpha$ and $\beta$ are the solutions of the equation $a \tan \theta + b \sec \theta = c$

then $\tan \alpha$ and $\tan \beta$ are the solutions of (1)

so $tan\, \alpha + \tan\, \beta = \dfrac{2ac}{a^2-b^2}\cdots (2)$

$tan\, \alpha \cdot \tan\, \beta = \dfrac{c^2-b^2}{a^2-b^2}\cdots (3)$ 



so $\tan (\alpha  + \beta) = \dfrac{ \tan\,\alpha + \tan\, \beta}{1- \tan\, \alpha \tan\,\beta}$
 = $\dfrac{\frac{2ac}{a^2-b^2}}{1- \frac{c^2-b^2}{a^2-b^2}}$
= $\dfrac{2ac}{a^2-c^2}$

PROVED

2015/063) What are the real roots of $x^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1 = 0$

The equation is very close to $(x-1)^6$ except coefficient of $x^3$ which is $-30 x^3$ instead of $-20 x^3$

so equation is

$(x-1)^6 - 10x^3 = 0$

let $\sqrt[3]10 = t$

so we get $(x-1)^6 - (tx)^3 = 0$

x is not zero

so divide by $x^3$
$(\dfrac{(x-1)^2}{x})^6 = t$

OR $(x - 1)^2 - \sqrt[3]{10}x  = 0$
ie $(x^2 - (2+ \sqrt[3]{10})x +1   = 0$
ie $x= \dfrac{1}{2}(2+\sqrt[3]{10})\pm\sqrt{4+4\sqrt[3]{10}+\sqrt[3]{100}-4}$
or  $x= \dfrac{1}{2}(2+\sqrt[3]{10})\pm\sqrt{4\sqrt[3]{10}+\sqrt[3]{100}}$

2015/062) A problem in AP

There are 2 sets of numbers each consisting of 3 terms in A.P & sum of each set is 15. The common difference of the $1^{st}$ set is greater than the common difference of the $2^{nd}$ set by 1 and the ratio of product of the $1^{st}$ set is to the product of $2^{nd}$ set is 7 to 8.Find the numbers. 

Solution
 
As sum of 3 terms is 15 so middle term is 5


let the common difference in $1^{st}$ set be a, so common difference in $2^{nd}$ set is (a-1)

then numbers in 1^st set are 5-a, 5, 5+ a and in 2^nd set are the numbers are 6-a, 5, 4+ a

now as per given condition $\dfrac{(5-a)5(5+a)}{((6-a) 5 (4+a))} = \dfrac{7}{8}$

or $8(25-a^2)= 7(6-a)(4+a) = 7(24+ 2a - a^2)$

or $200- 8a^2 = 168 + 14a - 7a^2$ or $a^2 + 14 a - 32 = 0$

$(a-2)(a+16) = 0$

a= 2 gives a- 1 = 1 gives 1st series = 3,5,7 and second series = 4,5,6

a = - 16 gives a- 1= - 17 the 1st series = 21,5, -11 and second series = 22,5, - 12.

refer to http://in.answers.yahoo.com/question/index;_ylt=Ar2kcwS_d26mHg5f2W.h9F.RHQx.;_ylv=3?qid=20130222042246AAW7rSk

2015/061) Simplify $((-1)+\frac{i\sqrt{2}}{3})^2 + ((-1)-\frac{i\sqrt{2}}{3})^2$

we know

$(a+b)^2 + (a-b)^2 = 2 (a^2+b^2)$

gives $((-1)+\frac{i\sqrt{2}}{3})^2 + ((-1)-\frac{i\sqrt{2}}{3})^2$

  =$2((-1)^2+(\frac{i\sqrt{2}}{3})^2)$

 $= 2 ( 1- \frac{2}{9}) = \frac{14}{9}$