NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.2
Question 9

Question. 9

The value of \(2\sin\theta\) can be \(a+\dfrac{1}{a}\), where \(a\gt0\) and \(a\ne1\).

Answer:

False.

Detailed Answer with Explanation:

Step 1: We are told that \(a > 0\) and \(a \neq 1\).

Step 2: From the AM–GM inequality (Arithmetic Mean – Geometric Mean), we know that for any positive number \(a\):

\[ a + \dfrac{1}{a} \geq 2, \]

and equality (i.e., exactly equal to 2) happens only when \(a = 1\).

Step 3: Since the problem says \(a \neq 1\), this means:

\[ a + \dfrac{1}{a} > 2. \]

Step 4: Now, let’s check the possible values of \(2\sin\theta\).
We know that \(-1 \leq \sin\theta \leq 1\).
So multiplying the whole inequality by 2 gives:

\[ -2 \leq 2\sin\theta \leq 2. \]

Step 5: This means that \(2\sin\theta\) can never be greater than 2. Its maximum possible value is exactly 2, and minimum is -2.

Step 6: But from Step 3, \(a + \dfrac{1}{a} > 2\).
So \(a + \dfrac{1}{a}\) is always bigger than the maximum possible value of \(2\sin\theta\).

Final Step: Therefore, \(2\sin\theta\) can never equal \(a + \dfrac{1}{a}\) (except at \(a = 1\), which is not allowed).
Hence, the statement is False.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 8: Introduction to Trignometry and Its Applications – Exercise 8.2 | Detailed Answers