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Mathematics · 2024 · 2 marks
CBSE 2024 · Region 1 · Set 1 · Q25
Prove that $\displaystyle 5-2 \sqrt{3}$ is an irrational number. It is given that $\displaystyle \sqrt{3}$ is an irrational number.Show that the number $\displaystyle 5 \times 11 \times 17+3 \times 11$ is a composite number.
Prove that $\displaystyle 5-2 \sqrt{3}$ is an irrational number. It is given that $\displaystyle \sqrt{3}$ is an irrational number.
Show that the number $\displaystyle 5 \times 11 \times 17+3 \times 11$ is a composite number.
Marking-scheme solution
Assuming \(\displaystyle 5-2 \sqrt{3}\) to be a rational number.
Let \(\displaystyle 5-2 \sqrt{3}=\frac{a}{b}\) where \(\displaystyle a\) and \(\displaystyle b\) are integers & \(\displaystyle b \neq 0\)
\[\Rightarrow \sqrt{3}=\frac{5 b-a}{2 b}
\]
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, \(\displaystyle 5-2 \sqrt{3}\) is an irrational number.
\[\begin{aligned}
5 \times 11 \times 17+3 \times 11 & =11 \times(5 \times 17+3) \\
& =11 \times 88 \text { or } 11 \times 11 \times 2^{3}
\end{aligned}
\]
It means the number can be expressed as a product of two factors other than $\displaystyle 1$, therefore the given number is a composite number.
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CBSE Class 10 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.