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