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  $x= \dfrac{1}{2}(2+\sqrt[3]{10})\pm\sqrt{4\sqrt[3]{10}+\sqrt[3]{100}}$