Discussion

Explanation:

The roots of the given equation are = $\frac{6\pm \sqrt{36-4c}}{2}$ = 3 ± $\sqrt{9-c}$

Now, 3 + √(9-c) is definitely greater than 1 hence cannot lie between 0 and 1.

∴ The roots which can lie between 0 and 1 will be 3 - $\sqrt{9-c}$.

⇒ 0 < 3 - $\sqrt{9-c}$ < 1

⇒ -3 < - $\sqrt{9-c}$ < -2

⇒ 3 > $\sqrt{9-c}$ > 2

⇒ 9 > 9 – c > 4

⇒ 0 >  -c  > -5

⇒ 0 < c < 5

Hence, option (b).

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