CAT 2025 Question Paper | Quant Slot 3

CAT Questions

CAT 2025 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 7 : The sum of all possible real values of \( x \) for which \( \log _ { x - 3 } \left( x ^ { 2 } - 9 \right) = \log _ { x - 3 } ( x + 1 ) + 2 \), is

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Video Explanation

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Explanatory Answer

Given:
\[ \log_{(x-3)}(x^2-9)=\log_{(x-3)}(x+1)+2 \]

Apply domain constraints:
\(x-3>0\) and \(x-3\neq1\), so \(x>3\) and \(x\neq4\).
Also, \(x^2-9>0\) and \(x+1>0\).
Thus, the domain is \(x>3,\ x\neq4\).

Using \(\log a-\log b=\log\frac{a}{b}\):
\[ \log_{(x-3)}\left(\frac{x^2-9}{x+1}\right)=2 \]
\[ \frac{x^2-9}{x+1}=(x-3)^2 \]
Since \(x>3\), divide by \(x-3\):
\[ \frac{x+3}{x+1}=x-3 \]
\[ x+3=(x-3)(x+1) \]
\[ x^2-3x-6=0 \]

Using the quadratic formula:
\[ x=\frac{3\pm\sqrt{33}}{2} \]
The negative root is rejected since \(x>3\).
The valid root is: \[ x=\frac{3+\sqrt{33}}{2} \]

Sum of all possible real values of \(x\):
\[ \frac{3+\sqrt{33}}{2} \]

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The question is " The sum of all possible real values of \( x \) for which \( \log _ { x - 3 } \left( x ^ { 2 } - 9 \right) = \log _ { x - 3 } ( x + 1 ) + 2 \), is "

Hence, the answer is '\( \frac { 3 + \sqrt { 33 } } { 2 } \)'

Choice 3 is the correct answer.

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