retagged by
632 views

1 Answer

1 votes
1 votes
Given that, $ 2 \cos (x(x+1)) = 2^{x} + 2^{-x}$

$ \Rightarrow 2 \cos (x(x+1)) = 2^{x} + \frac{1}{2^{x}} \quad \longrightarrow (1)$

For any set of non-negative real numbers, the arithmetic mean of the set is greater than or equal to the geometric mean of the set.

Algebraically, this is expressed as follows:

For a set of non-negative real numbers $ a_{1}, a_{2}, a_{3}, \dots, a_{n},$ the following always holds:

$ \boxed {\frac{ a_{1} + a_{2} + a_{3} + \dots + a_{n} }{n} \geq \sqrt[n]{a_{1} a_{2} a_{3} \dots a_{n}}}$

Now, from the equation $(1)$, take the right hand side term,

$ 2^{x} + \frac {1}{2^{x}} \geq 2 \sqrt {2^{x} \cdot \frac {1}{2^{x}}}$

$ \Rightarrow 2^{x} + \frac {1}{2^{x}} \geq 2 $

Put the value in the equation $(1),$ we get

$ 2 \cos (x(x+1)) = 2 \quad \longrightarrow (2)$

$ \Rightarrow \cos (x(x+1)) = 1 $

$ \Rightarrow \cos (x(x+1)) = \cos \; 0^{\circ} $

$ \Rightarrow (x(x+1)) = 0 $

$ \Rightarrow \boxed{x=0\;\text{(or)}\; x = -1}$

Now, put  $x=0,$ in the equation $(1),$ we get

$ 2 \cos (0(0+1)) = 2^{0} + \frac{1}{2^{0}}$

$ \Rightarrow 2  \cos 0 = 1+1$

$ \Rightarrow 2 (1) = 2$

$ \Rightarrow \boxed{2 = 2}$ (Satisfied)

And, put $x=-1,$ in the equation $(1),$ we get

$ 2 \cos (-1(-1+1)) = 2^{-1} + \frac{1}{2^{-1}}$

$ \Rightarrow 2 \cos 0 = \frac{1}{2} + 2$

$ \Rightarrow 2(1) = 2 + \frac{1}{2}$

$ \Rightarrow \boxed{2 \neq \frac{5}{2}}$ ( Not satisfied)

So, only $ \boxed{x=0},$ is the real root of the equation.

Correct Answer $: \text{B}$
edited by
Answer:

Related questions

2 votes
2 votes
1 answer
1
go_editor asked Mar 8, 2020
788 views
If $m$ and $n$ are integers such that $(\sqrt{2})^{19}3^{4}4^{2}9^{m}8^{n}=3^{n}16^{m}(\sqrt[4]{64})$ then $m$ is$-20$$-12$$-24$$-16$
1 votes
1 votes
1 answer
2
go_editor asked Mar 8, 2020
635 views
The income of Amala is $20\%$ more than that of Bimala and $20\%$ less than that of Kamala. If Kamala's income goes down by $4\%$ and Bimala’s goes up by $10\%$, then t...
1 votes
1 votes
1 answer
3
go_editor asked Mar 8, 2020
667 views
In a class, $60\%$ of the students are girls and the rest are boys. There are $30$ more girls than boys. If $68\%$ of the students, including $30$ boys, pass an examinati...
1 votes
1 votes
1 answer
4
go_editor asked Mar 8, 2020
617 views
On selling a pen at $5\%$ loss and a book at $15\%$ gain, Karim gains Rs. $7$. If he sells the pen at $5\%$ gain and the book at $10\%$ gain, he gains Rs. $13$. What is t...
1 votes
1 votes
1 answer
5
go_editor asked Mar 8, 2020
828 views
Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is $5:6$$4:5$$3:4$$2:3$