2022/042) Solve for x $(1+ \frac{1}{x})^{x+1} = (1+ \frac{1}{n})^n$

 we have

$(1+ \frac{1}{x})^{x+1} = (\frac{x+1}{x})^{x+1}$

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

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

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

comparing with the RHS we get -(x+1) = n  or x = -(n+1) 

2022/041) Find the digit X such that 9986860883748524X5070273447265625 equals $1995^{10}$

Because 1995 is divisible by 3 so all the powers $≥2$ of 1995 shall be divisible by 9.

So sum of digits of the result must be divisible by 9.

We get sum of digits of given number = 160 + X.

Smallest multiple of 9 above 160 is 162 which gives X= 2 and next multiple of 9 gives X = 11.

so X = 2 

 

2022/040) If $x = 3 + \sqrt[3]3 + \sqrt[3]{3^2}$ then show that $x^3 - 9x^2+ 18x -12 = 0$

 From the given condition we have

$(x-3) = \sqrt[3]3 + \sqrt[3]{3^2}$

squaring both sides

$(x-3)^3 = (\sqrt[3]3)^3 + ( \sqrt[3]{3^2})^2 + 3  \sqrt[3]{3} \sqrt[3]{3^2}(  \sqrt[3]{3} +  \sqrt[3]{3^2})$ using $(a+b)^3 = a ^3 + b^3 + 3ab(a+)$

or $(x-3)^3 = 3 + 9 + 9 ( x-3)$ using $(  \sqrt[3]{3^2} +  \sqrt[3]{3^2} = x- 3$

or $(x^3 - 9x^2 + 27 x - 27 = 12 + 9x - 27 = 9x + 12$

or $x^3 - 9x^2+ 18x -12 = 0$

2022/039) Find x integer such that $\frac{8^x - 2^x }{6^x - 3^x} = 2$

 We have numerator = $8^x - 2^x = (2^3)^x - 2^x = (2^(3x) - 2^x = 2^x(2^{2x} - 1)= 2^x(2^x + 1)(2^x-1)$

Denominator = $6^x - 2^x = 3^x(2^x-1)$

So $\frac{8^x - 2^x }{6^x - 3^x} = \frac{2^x(2^x+1)}{3^x} = 2$

Or $2^x(2^x+1) = 2 * 3^x$

x cannot be zero as in the original question x = 0 is not possible

So we must ahve $2^x = 2$ and $2^x+ 1 = 3^x$ by equating even on both sides and odd on both sides

1st equuation gives x = 1 and it satisfies 2nd equation as well.

so Solution x = 1 

2022/038) The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, find the length of smallest sides

Let the sides be a -b , a , a+b

as the tringle is right angled so $(a-b)^2 + a^2 = (a+b)^2$

or $a^2-2ab + b^2 + a ^2 = a^2 + 2ab + b^2$

or   $a^2= 4ab$

or $a= 4b$

as the triangls is right angled so aea = $\frac{1}{2} a(a-b) =  \frac{1}{2} 4b(4b-b) = 6b^2 = 24$ or b= 2

the smallest sde = 3b = 6. 

2022/037) If a, b, c are in H.P., how do you prove that $\frac{b+c-a}{a}$, $\frac{c+a-b}{b}$, $\frac{a+b-c}{c}$ are in A.P.​?

 a,b,c are in HP so $\frac{1}{a}$, \frac{1}{b}$,$\frac{1}{c}$ are in AP

or $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$

multiply both LHS and RHS by a + b + c we get

$\frac{a+b+c}{b} - \frac{a+b+c}{a} = \frac{a+b+c}{c} - \frac{a+b+c}{b}$

Or 

$(\frac{a+b+c}{b} - 2) - (\frac{a+b+c}{a}-2)  = (\frac{a+b+c}{c} -2)  - (\frac{a+b+c}{b} -2)$

or $(\frac{a+c-b }{b}) - (\frac{b+c-a }{a})  = (\frac{a+b-c }{c})  - (\frac{c + a - b}{b}$)

Hence  $\frac{b+c-a}{a}$, $\frac{c+a-b}{b}$, $\frac{a+b-c}{c}$ are in A.P

2022/036) Let $t_n= \sin^n \theta + \cos^n \theta$ show that $6t_{10} − 15t_8 + 10t_6 = 1$ .

Let us put $p = \sin ^2 \theta \cos^2 \theta$  

We have $t_2 = \sin ^2 \theta + \cos^2 \theta = 1 \cdots(1)$

and $t_4 = (\sin^4x + \cos^4 x) = (\sin ^2 x + \cos^2 x) - 2 \sin ^2 x +\cos^2 x = 1 - 2p$ 

Now we an sert 

$t_n t_2 = ( \sin ^n \theta + \cos^n \theta)(\sin ^2 \theta + \cos^2 \theta) = \sin  ^{(n+2)}\theta +  \cos^{(n+2)} + \sin ^n \theta \cos^2 \theta +    \sin ^2 \theta \cos^n \theta$

or $t_n = t_{(n+2)} + t_{(n-2)}p$

or $t_(n+2) = t_n - t(n-2)p$

Using this we have

$t_6 = t_4 - t_2p = t_4 - p = 1- 3p$

 $t_8 = t_6 - t_4p = 1-3p -  (1-2p ) p^2 = 1- 4p + 2p^2$

$t_{10} = t_8 - t_6p  = (1-4p +2p^2) - (1-3p)p = 1 - 5p + 5p^2$

so $6t_{10} − 15t_8 + 10t_6 = 6(1-5p + 5p^2 ) - 15(1- 4p + 2p^2 ) + 10(1-3p)  = 1$

  

2022/035) Find the smallest prime factor of $2019^8 + 1$

let the prime be p

we have $2019^8 \equiv -1 \pmod p$ 

or $2019^{16} \equiv 1 \pmod p$

so $2019^{16k} \equiv 1 \pmod p$

as per Formats Little theorem p shall be of the form 16k + 1.

any number less than 16 is not possible because  $2019^8 \equiv -1 \pmod p$ and $16=2^4$ and $8=2^3$ so any factor of 16 shall not give a power =1.

so p = 16 *2 + 1 = 33 (not a prime) or 16 * 3 + 1= 49( not a prime) or 16 * 4 + 1 = 65( not a prime) or 16 * 5 + 1 = 81 (not a prime) or 16 *6  + 1 = 97.

so we check for 97 

now $2019 \equiv - 18 \pmod {97}$ 

so $2019^2 \equiv 324 \pmod {97}$

or   $2019^2 \equiv 33  \pmod {97}$

so $2019^4 \equiv 33^2  \pmod {97}$

or $2019^4 \equiv 22   \pmod {97}$

or $2019^8 \equiv 484  \pmod {97}$

or $2019^8 \equiv -1  \pmod {97}$

or or $2019^8 + 1\equiv 0   \pmod {97}$

so smallest prime factor = 97 

2022/034) How many real roots does $x^ 4 + 12x − 5$ have?

 let $f(x) = x^4+ 12x--5$

As there change of sign once in f(x) so there is one positve real toot as per descartes' rule of signs

$f(- x) = x^4- 12x--5$

As there change of sign once in f(-x) so there is one  real toot as per descartes' rule of signs

so there are 2 real roots

2022/033) Let p be a prime and r is a number less than p. Show that $\frac{(p-1)!}{r!(p-r)!}$ is an integer

Number of ways of choosing r objects from p objects is $\frac{(p!}{r!(p-r)!}$. so $\frac{(p!}{r!(p-r)!}$ is an integer

or $\frac{p(p-1)!}{r!(p-r)!}$ is integer . Now because p does not divide the denominator  because p is prime dividing by p we get the result