1. Which one is larger in between \(50^{50}\) and \(49^{51}\) ?
Ans: \[ \frac{50^{50}}{49^{51}}=\frac{50^{50}}{49^{50}\times49}=\frac{(49+1)^{50}}{49^{50}\times 49}=\left(\frac{49+1}{49}\right)^{50} \times \frac{1}{49}= \left(1 +\frac{1}{49}\right)^{50} \times \frac{1}{49}< e \times \frac{1}{49} \]
\(=\left(1 +\frac{1}{49}\right)^{49}\times \left(1 +\frac{1}{49}\right) \times \frac{1}{49}< e \times\left(1 +\frac{1}{49}\right)\times \frac{1}{49} [As \lim_{x\to \infty}\left( 1+ \frac{1}{x}\right)^x = e]\)
\(< e \times\frac{50}{49}\times \frac{1}{49}=2.703 \times\frac{50}{49}\times \frac{1}{49}\)
\( \frac{135}{2401} < 1 \Rightarrow 50^{50} < 49^{51} \)
2. Solve \( ab+c = 2020 \) and \( a+bc=2021 \) for a,b,c
Ans: \( a+bc - (ab+c) = 2021-2020 \)
\( \Rightarrow a-ab + bc - c = 1 \)
\( \Rightarrow a(1-b) - c(1-bc) = 1 \)
\( \Rightarrow(1-b)(a-c) = 1 \)
Now we can write \( 1-b = 1 \) which gives b = 0 and ab+c=2020 equation gives c = 2020, a+bc=2021 gives a = 2021.
Again we can write \( 1-b = -1 \) which gives b =2, a-c=-1, and by putting b=2 in given equation we will get
\( 2a+c=2020 \) and \( a+2c=2021 \).
by multiplying 2 with \( a+2c=2021 \) and subtracting \( 2a+c=2020 \) we will get
\( 2a+4c-(2a+c)= 4042-2020 \)
or, \( 3c = 2022 \)
or, c = 674
therefore \( a = 2021-2c = 2021-2 \times 674 = 2020 - 1348 = 673 \)
Therefore the solutions are (2021,0,2022) and (672,2,673)
3. Solve for m where \( 3^m - 2^m = 65 \)
Ans: If m=0 then \( 3^m - 2^m =3^0 - 2^0=1-1=0 \),
If m=1 then \( 3^m - 2^m =3^1 - 2^1=3-2=1 \)
If m=2 then \( 3^m - 2^m =3^2 - 2^2=9-4=5 \)
If m=3 then \( 3^m - 2^m =3^3 - 2^3=27-8=19 \)
If m=4 then \( 3^m - 2^m =3^4 - 2^4=81-16=65 \)
Therefore m = 4
4. Solve a for \( 2^a+2^b+2^c=148 \)
Ans: \( 148= 2\times2\times 37 \)
\( 2^a+2^b+2^c=148 \)
\( \Rightarrow 2^a+2^b+2^c=2^2\times 37 \)
\( \Rightarrow 2^{a-2} + 2^{b-2}+ 2^{c-2} = 1 + 36 \)
\( \Rightarrow 2^{a-2} + 2^{b-2}+ 2^{c-2} = 1 + 4+32 \)
\( \Rightarrow 2^{a-2} + 2^{b-2}+ 2^{c-2} = 2^0 + 2^2+2^5 \)
By comparing both sides,
\( a-2=0\Rightarrow a = 2 \)
\( b-2=2\Rightarrow b = 4 \)
\( c-2=5\Rightarrow c=7 \)
The required solution is (2,4,7)
5. Which is bigger number in between \( {222}^{333} \) and \( {333}^{222} \)?
Ans: \( \frac{{333}^{222}}{{222}^{333}} \)
\( \Rightarrow\frac{\left({333}^{111}\right) ^2}{\left({222}^{111}\right)^3} \)
\( \Rightarrow\frac{\left(3^{111}\times {111}^{111}\right)^2}{\left(2^{111}\times {111}^{111}\right)^3} \)
\( \Rightarrow \frac{\left(3^2\right)^{111}}{\left(2^3\right)^{111}} \times \frac{1}{{111}^{111}} \)
\( \Rightarrow \left(\frac{9}{8}\right)^{111} \times \frac{1}{{111}^{111}} \)
\( \Rightarrow \left(\frac{9}{888}\right)^{111} \)
\( \lt 1 \)
i.e. \( {333}^{222} \lt {222}^{333} \)
6. Solve for \( \sqrt{x}^x = 4^{\sqrt{x}} \)
Ans: \( \sqrt{x}^x = 4^{\sqrt{x}} \)
\( \Rightarrow x^{\frac{x}{2}} = 2^{2x^{\frac{1}{2}}} \)
Let \( \sqrt{x} = a \) i.e. \( x=a^2 \)then
\( (a^2)^{\frac{a^2}{2}} = 2^{2a} \)
\( \Rightarrow a^{a^2}= 2^{2a} \)
\( \Rightarrow \log(a^{a^2})= \log(2^{2a}) \)
\( \Rightarrow a^2\log(a)= 2a\log(2) \)
\( \Rightarrow a^2\log(a)=-2a\log(2) = 0 \)
\( \Rightarrow a(a\log(a)-2\log(2) )= 0 \)
Either \( a = 0 \) or x=0
Or, \( (a\log(a)-2\log(2) )= 0 \)
\( \Rightarrow a\log(a) = 2\log2 \)
By comparing both sides
a = 2 i.e. \( x = a^2 = 2^2 = 4 \)
N.B Generally I have found the questions from Internet
7. Solve Volterra equation
\[ \psi(x) = x + \int_0^x (x-s) \psi(s)ds\]
Solution
Taking differentials \[ \psi(x)^{\prime} = 1 + \int_0^x -\psi(s)ds\] and \[ \psi(x)^{\prime \prime} = -\psi(x) \] which is a shm equation so the solution becomes \[ \psi(x) = c_1 sin(x) + c_2 cos(x) \] assuming \[ \psi(0) = 0 \] which gives \[ c_2 = 0 \] \[ \psi^\prime(0) = 1 \] \[ c_1 = 1\] Therefore : \[ \psi(x) = sin(x) \] now substituting this into giving equation we will get \[ \psi(x) = x + \int_0^x (x-s)sin(s)ds \] \[ \psi(x) = x + [-xcoss]_0^x - \int_0^xssin(s)ds \] \[ \psi(x) = x-xcos x - [-scoss]_0^x - \int_0^x (-coss)ds \] \[ \psi(x) = x - x cos(x) + xcosx the + sinx \] \[ \psi(x) = x - 2xcos(x) - sin(x) \]
