$The number of value(s) of x for which {log}_{3} (- 3 x^{2} + 6 x) = 1 is/are:$

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1$
Given: ${log}_{3} (- 3 x^{2} + 6 x) = 1$
Now, for logarithm to be defined:
$- 3 x^{2} + 6 x > 0 \Rightarrow x^{2} - 2 x < 0 \Rightarrow x (x - 2) < 0 \Rightarrow x \in (0, 2) \dots (A)$

Now, ${log}_{3} (- 3 x^{2} + 6 x) = 1$
$\Rightarrow - 3 x^{2} + 6 x = 3 \Rightarrow x^{2} - 2 x + 1 = 0 \Rightarrow x = 1 \dots (B)$
Now, $x = 1 \in A \cap B$
Thus, $x = 1$ is the only solution.
Hence the correct answer is option (b).

Suggest Corrections

Similar questions

\lim_{x \to 1} \frac{x^{3} + 3 x^{2} - 6 x + 2}{x^{3} + 3 x^{2} - 3 x - 1}

\int \frac{2 x - 3}{x^{2} + 6 x + 13} d x

\int

The range of values of x which satisfies

the inequality {log}_{(x + 3)} (x^{2} - x) < 1 is

Join BYJU'S Learning Program