Write ‘True’ or ‘False’ and justify your answer. The value of 2 be (a+1 a ), where a is a positive number, and a 1.

False.Consider 'a' and'1' a as positive numbers and a 0 Arithumetic mean (AM) of a and 1 a = (a+1 a ) 2 Geometric mean (GM) of a and 1 a = a 1 a =1 AM>GM (a+1 a ) 2 >1 (a+1 a )>2 Let a+1 a =2 2 >2 >1 Which can never be possible. Hence, our consideration that a+1 a =2 is false. Alternate Answer a is positive and a 1 i.e., a can be above 0 all real and values except 1. Let 0 2 for any value of a. Let a=0.2 +1 a =0.2+1 0.2 =5.2 a=0.9 +1 a =0.9+1 0.9 =0.9+1.111=2.011 Put a+1 a =2 2 >2 >1 Which is impossible so 2 +1 a Hence, the given statement is false. If we take any value of a more than one, then the value of a+1 a is always greater then 2 which repeats the result.

Write ‘True’ or ‘False’ and justify your answer. The value of 2 b…

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Write ‘True’ or ‘False’ and justify your answer.
The value of $2\sin\theta\text{can be}\Big(\text{a}+\frac{1}{\text{a}}\Big),$ where a is a positive number, and $\text{a}\neq1.$

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False.Consider 'a' and$\frac{'1'}{\text{a}}$ as positive numbers and $\text{a}\neq0$
Arithumetic mean (AM) of a and $\frac{1}{\text{a}}=\frac{\Big(\text{a}+\frac{1}{\text{a}}\Big)}{2}$
Geometric mean (GM) of a and $\frac{1}{\text{a}}=\sqrt{\text{a}\times\frac{1}{\text{a}}}=1$
$\because\ \text{AM}>\text{GM}$
$\therefore\ \frac{\Big(\text{a}+\frac{1}{\text{a}}\Big)}{2}>1$
$\Rightarrow\ \Big(\text{a}+\frac{1}{\text{a}}\Big)>2$
Let $\text{a}+\frac{1}{\text{a}}=2\sin\theta$
$\Rightarrow\ 2\sin\theta>2$
$\Rightarrow\ \sin\theta>1$
Which can never be possible.
Hence, our consideration that $\text{a}+\frac{1}{\text{a}}=2\sin\theta$ is false.
Alternate Answer
a is positive and $\text{a}\neq1$ i.e., a can be above 0 all real and values except 1.
Let 0 < a < 1 then $\frac{1}{\text{a}}$ will be more then 1 so $\text{a}+\frac{1}{\text{a}}>2$ for any value of a.
Let $\text{a}=0.2\Rightarrow\text{a}+\frac{1}{\text{a}}=0.2+\frac{1}{0.2}=5.2$
$\text{a}=0.9\Rightarrow\text{a}+\frac{1}{\text{a}}=0.9+\frac{1}{0.9}=0.9+1.111=2.011$
Put $\text{a}+\frac{1}{\text{a}}=2\sin\theta$
$\therefore\ 2\sin\theta>2$
$\Rightarrow\ \sin\theta>1$
Which is impossible so $2\sin\theta\neq\text{a}+\frac{1}{\text{a}}$
Hence, the given statement is false. If we take any value of a more than one, then the value of $\text{a}+\frac{1}{\text{a}}$ is always greater then 2 which repeats the result.

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