ত্রিভুজের দুটি বাহু ও তাঁদের বিপরীত কোণের সম্পর্ক - অষ্টম শ্রেণী

 

1. Which one is larger in between ${50}^{50}$ and ${49}^{51}$ ?

Ans:  $$\frac{{50}^{50}}{{49}^{51}}=\frac{{50}^{50}}{{49}^{50}\times 49}=\frac{(49+1{)}^{50}}{{49}^{50}\times 49}={\left(\frac{49+1}{49}\right)}^{50}\times \frac{1}{49}={(1+\frac{1}{49})}^{50}\times \frac{1}{49}<e\times \frac{1}{49}$$

$={(1+\frac{1}{49})}^{49}\times (1+\frac{1}{49})\times \frac{1}{49}<e\times (1+\frac{1}{49})\times \frac{1}{49}[As\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=e]$

$<e\times \frac{50}{49}\times \frac{1}{49}=2.703\times \frac{50}{49}\times \frac{1}{49}$

$\frac{135}{2401}<1\Rightarrow {50}^{50}<{49}^{51}$

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: $148=2\times 2\times 37$

 $2^a+2^b+2^c=148$

$\Rightarrow 2^a+2^b+2^c=2^2\times 37$

$\Rightarrow 2^{a-2}+2^{b-2}+2^{c-2}=1+36$

$\Rightarrow 2^{a-2}+2^{b-2}+2^{c-2}=1+4+32$

$\Rightarrow 2^{a-2}+2^{b-2}+2^{c-2}=2^0+2^2+2^5$

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: $\frac{{333}^{222}}{{222}^{333}}$

 $\Rightarrow \frac{{\left({333}^{111}\right)}^2}{{\left({222}^{111}\right)}^3}$

$\Rightarrow \frac{{(3^{111}\times {111}^{111})}^2}{{(2^{111}\times {111}^{111})}^3}$

$\Rightarrow \frac{{\left(3^2\right)}^{111}}{{\left(2^3\right)}^{111}}\times \frac{1}{{111}^{111}}$

$\Rightarrow {\left(\frac{9}{8}\right)}^{111}\times \frac{1}{{111}^{111}}$

$\Rightarrow {\left(\frac{9}{888}\right)}^{111}$

$<1$

i.e. ${333}^{222}<{222}^{333}$

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, $(a\log (a)-2\log (2))=0$

$\Rightarrow a\log (a)=2\log 2$

By comparing both sides

a = 2 i.e. $x=a^2=2^2=4$

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N.B Generally I have found the questions from Internet

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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 $$\psi (x)=x+{\int }_0^x(x-s)sin(s)ds$$ $$\psi (x)=x+[-xcoss{]}_0^x-{\int }_0^{x}ssin(s)ds$$ $$\psi (x)=x-xcosx-[-scoss{]}_0^x-{\int }_0^x(-coss)ds$$ $$\psi (x)=x-xcos(x)+xcosxthe+sinx$$ $$\psi (x)=x-2xcos(x)-sin(x)$$