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