Infinite series with a finite solution

This question was uploaded on 01/02/22 on social media accounts.

Solution:

Each term in f(x) is increasing but it has a finite value given so the absolute value of x must be less than 1.

Given series:

\[\color{blue} {f(x)=\frac{\color{red}{(1-x)}(1+x)}{\color{red}{(1-x)}}(1+x^2)(1+x^4)(1+x^8)......}\]

\[\Rightarrow\color{blue} {f(x)=\frac{\color{red}{(1-x^2)}(1+x^2)}{\color{red}{(1-x)}}(1+x^4)(1+x^8)......}\]

Similarly:

\[\Rightarrow\color{blue} {f(x)=\frac{1-x^{\infty}}{1-x}=\frac1{1-x}}\]

Here, higher power of x will tends to zero.

Now:

\[\color{green} {2022=\frac1{1-x}}\]

\[\Rightarrow\color{green} {x=\frac{2021}{2022}}\]

Infinite converging series question from JEE Advanced question paper

This question was uploaded on 29/01/22 on social media accounts.

Question from JEE Advanced question paper.

Solution:

Consider 8 as 2³ in all places

We know that:

\[\color{black} {a^3-b^3=(a-b)(a^2+ab+b^2)}\]

\[\color{black} {a^3+b^3=(a+b)(a^2-ab+b^2)}\]

Now,

\[\color{black} {\frac{n^3-2^2}{n^3+2^3}=\color{blue}{\left(\frac{n-2}{n+2}\right)}\color{red}{\left(\frac{n^2+2n+4}{n^2-2n+4}\right)}}\]

Solving in two parts:

Let's consider the blue part from all terms in the series:

\[\color{blue}{\left(\frac{3-2}{3+2}\right)\left(\frac{4-2}{4+2}\right)\left(\frac{5-2}{5+2}\right)....=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot\cdot\cdot\cdot}{5\cdot6\cdot7\cdot8\cdot9\cdot\cdot\cdot\cdot}=24}\]

Now consider the red part from all terms in the series:

\[\color{red}{\frac{19\cdot28\cdot39\cdot52\cdot67\cdot\cdot\cdot\cdot}{7\cdot12\cdot19\cdot28\cdot39\cdot\cdot\cdot\cdot}=\frac{1}{7\cdot12}}\]

Now by combining both parts:

\[\color{black} {S=\left(\color{blue}{24}\right)\left(\color{red}{\frac{1}{7\cdot12}}\right)=\frac27}\]

The hyperbolic equation on two variables

This question was uploaded on 26/01/22 on social media accounts.

Solution:

Given equation:

\[\color{blue} {a^2+2ab-3b^2-41=0}\]

\[\Rightarrow \color{blue} {a^2\color{red}{+a^2-a^2}+2ab-3b^2-41=0}\]

\[\Rightarrow \color{blue} {2a^2-(a^2-2ab+b^2)-2b^2-41=0}\]

\[\Rightarrow \color{blue} {2(a^2-b^2)-(a-b)^2-41=0}\]

\[\Rightarrow \color{blue} {2(a+b)(a-b)-(a-b)^2=41}\]

\[\Rightarrow \color{blue} {(a-b)(a+3b)=1\times41}\]

Given that a and b belong to natural numbers. So, both must be positive integers.

Clearly, ( a - b ) < ( a + 3b )

\[\color{red} {a-b=1}\Rightarrow\color{red} {a=1+b}\]

Now put this value in other equation:

\[\color{red} {a+3b=41}\Rightarrow\color{red} {1+4b=41}\]

\[\Rightarrow\color{red} {b=10}\]

\[\Rightarrow\color{red} {a=11}\]

\[\color{fuchsia} {a^2+b^2=11^2+10^2=221}\]

Telescopic multiplication of a series

This question was uploaded on 22/01/22 on social media accounts.

Solution:

From the given equations:

\[\color{blue} {a_1=0}\]

\[\color{blue} {a_2=a_1+2(1)}\]

\[\color{blue} {a_3=a_2+2(2)}\]

\[\color{blue} {a_4=a_3+2(3)}\]

.

.

.

\[\color{blue} {a_n=a_{n-1}+2(n-1)}\]

By adding these all equation:

\[\color{red} {a_n=2\frac{(n-1)(n)}2=(n-1)(n)}\]

Now because of the general equation, we can calculate the value of the multiplication:

\[\color{black} {\left(\frac{a_3}{a_4}\right)\left(\frac{a_5}{a_6}\right)\left(\frac{a_7}{a_8}\right).....\left(\frac{a_{4043}}{a_{4044}}\right)}\]

\[=\color{black} {\left(\frac{(2)(3)}{(3)(4)}\right)\left(\frac{(4)(5)}{(5)(6)}\right)\left(\frac{(6)(7)}{(7)(8)}\right).....\left(\frac{(4042)(4043)}{(4043)(4044)}\right)}\]

\[=\frac{2\cdot3\cdot4\cdot5\cdot\cdot\cdot\cdot4042\cdot4043}{3\cdot4\cdot5\cdot6\cdot\cdot\cdot\cdot4043\cdot4044}\]

\[=\frac{2}{4044}=\frac1{2022}\]

The prime number before a perfect cube

This question was uploaded on 19/01/22 on social media accounts.

Solution:

Given that x³ - 1 is a prime number and x is a natural number.

We know that, x³ - 1 = (x - 1)(x² + x + 1)

If x³ - 1 is a prime number then it should not be a composite number,

for that either (x - 1) or (x² + x + 1) should be 1.

If: x - 1 = 1 then x = 2, so that x³ - 1 = 7 is a prime number

But if: x² + x + 1 = 1 then x = 0 or x = -1

Here, x³ - 1 will be negative so it will not possible.

So that, x = 2

Equations with powers three and ten

This question was uploaded on 16/01/22 on social media accounts.

Solution:

Given:

\[\color{red} {x^3=4-2x}\]

Now,

\[\color{blue} {x^{10}=x(\color{red}{x^3})^3=x(\color{red}{4-2x})^3}\]

\[\Rightarrow\color{blue} {x^{10}=-8x^4+48x^3-96x^2+64x}\]

\[\Rightarrow\color{blue} {x^{10}=-8x(\color{red}{x^3})+48(\color{red}{x^3})-96x^2+64x}\]

\[\Rightarrow\color{blue} {x^{10}=-8x(\color{red}{4-2x})+48(\color{red}{4-2x})-96x^2+64x}\]

\[\Rightarrow\color{blue} {x^{10}=-80x^2-64x+192}\]

\[\Rightarrow\color{blue} {x^{10}=-16x(5x+4)+192}\]

Now, after putting this value in second equation:

\[\color{blue} {-\color{fuchsia}{16x(5x+4)}+192+\color{fuchsia}{16x(5x+4)}=192}\]

Cubic and quadratic equation on a single variable

This question was uploaded on 13/01/22 on social media accounts.

Solution:

Given equation:

\[\color{blue} {x^3=32(x+2)=32x+64}\]

\[\Rightarrow \color{blue} {x^3-16x=16x+64}\]

\[\Rightarrow \color{blue} {x(x^2-16)=16(x+4)}\]

\[\Rightarrow \color{blue} {x(x-4)\color{Red}{(x+4)}=16\color{red}{(x+4)}}\]

\[\Rightarrow \color{blue} {x(x-4)=\color{#228b22}{x^2-4x=16}}\]

In this, (x+4) term is canceled,

\[\Rightarrow \color{#ff1694}{x+4=0}\Rightarrow \color{#ff1694}{x=-4}\]

\[\Rightarrow \color{#228b22}{x^2-4x=32}\]

\[\Rightarrow \color{#228b22}{x^2-4x=\left\{16,32\right\}}\]

Question From Indian Math Olympiad Exam

This question was uploaded on 10/01/22 on social media accounts.

Solution:

\[\color{green} {a+b=1}\]

\[\color{purple} {a^2+b^2=2}\]

\[\color{red} {a^2+b^2=(a+b)^2-2ab}\]

\[\Rightarrow\color{red} {2=1-2ab}\]

\[\Rightarrow\color{red} {ab=-\frac12}\]

\[\color{blue} {a^4+b^4=(a^2+b^2)^2-2(ab)^2}\]

\[\Rightarrow\color{blue} {a^4+b^4=4-\frac12=\frac72}\]

\[\color{fuchsia} {a^8+b^8=(a^4+b^4)^2-2(ab)^4}\]

\[\Rightarrow\color{fuchsia} {a^8+b^8=\frac{49}4-\frac18=\frac{97}8}\]

Question from Jee Advanced examination

This question was uploaded on 07/01/22 on social media accounts.

This question is from IIT JEE Advanced examination.

Solution:

From the first equation:

\[\color{red} {3\sin^2A+2\sin^2B=1}\]

\[\Rightarrow\color{red} {3\sin^2A=1-2\sin^2B=\cos2B}\]

\[\Rightarrow\color{red} {3\sin A=\frac{\cos2B}{\sin A}}\]

From the second equation:

\[\color{blue} {3\sin2A=2\sin2B = 3(2\sin A \cos A)}\]

\[\Rightarrow\color{blue} {3\sin A=\frac{\sin2B}{\cos A}}\]

From both equations:

\[\color{red} {\frac{\cos2B}{\sin A}}=\color{blue} {\frac{\sin2B}{\cos A}}\]

\[\Rightarrow\color{blue}{\cos A}\color{red}{\cos2B}-\color{red}{\cos A}\color{blue}{\sin2B}=0\]

\[\Rightarrow\cos(A+2B)=0\]

\[\Rightarrow A+2B=\frac\pi2\]

Algebraic equations

This question was uploaded on 04/01/22 on social media accounts.

Solution 1:

Given equations:

\[\color{red} {a+b=4}\]

\[\color{blue} {a^2+b^2=10}\]

We know that,

\[(\color{red} {a+b})^2=\color{blue} {a^2+b^2}+2\color{fuchsia} {ab}\]

\[\Rightarrow (\color{red} {4})^2=\color{blue} {10}+2\color{fuchsia} {ab}\]

\[\Rightarrow \color{fuchsia} {ab=3}\]

Now,

\[(\color{blue} {a^2+b^2})^2=a^4+b^4+2(\color{fuchsia} {ab})^2\]

\[\Rightarrow(\color{blue} {10})^2=a^4+b^4+2(\color{fuchsia} {3})^2\]

\[\Rightarrow a^4+b^4=82\]