CAT 2023 Question Paper | Quant Slot 3

CAT 2023 Quant was dominated by Arithmetic followed by Algebra. 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 5 : A quadratic equation \(x^2+b x+c=0\) has two real roots. If the difference between the reciprocals of the roots is \(\frac{1}{3}\), and the sum of the reciprocals of the squares of the roots is \(\frac{5}{9}\), then the largest possible value of \((b+c)\) is

Video Explanation

Explanatory Answer

let \( \alpha \& \beta \) be the roots of \( x ^ { 2 } + b x + c = 0 \),
\( \frac { 1 } { \alpha } - \frac { 1 } { \beta } = \frac { 1 } { 3 } \)
\( \frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } = \frac { 5 } { 9 } \)
\( \left( \frac { 1 } { \alpha } - \frac { 1 } { \beta } \right) ^ { 2 } = \frac { 1 } { 9 } \)
\( \frac { 4 } { 9 } = \frac { 2 } { \alpha \beta } \)
\( \alpha \beta = 9 / 2 \)
\( \frac { 1 } { \alpha ^ { 2 } + \beta ^ { 2 } } = \frac { \alpha ^ { 2 } + \beta ^ { 2 } } { \alpha ^ { 2 } \beta ^ { 2 } } = \frac { ( \alpha + \beta ) ^ { 2 } - 2 \alpha \beta } { ( \alpha \beta ) ^ { 2 } } = \frac { 5 } { 9 } \)
\( ( \alpha + \beta ) ^ { 2 } = \frac { 5 } { 9 } \left( \frac { 9 } { 2 } \right) \left( \frac { 9 } { 2 } \right) + 2 \left( \frac { 9 } { 2 } \right) \)
\( ( \alpha + \beta ) ^ { 2 } = \frac { 45 } { 4 } + 9 = 20 + \frac { 1 } { 4 } = 20.25 \)
\( \alpha + \beta = \pm 4.5 \)
\( \alpha + \beta = \frac { - b } { 1 } \)
\( b = - ( \alpha + \beta ) \)
\( \alpha \beta = \frac { c } { 1 } \)
\( c = \alpha \beta = 4.5 \)
\( b + c = - ( \alpha + \beta ) + \alpha \)
to maximize \( ( b + c ) \), we pick \( ( \alpha + \beta ) = - 4.5 \)
\( b + c = - ( - 4.5 ) + 4.5 = 9 \)
So, the largest possible value of (b + c) is 9.

The question is " A quadratic equation \(x^2+b x+c=0\) has two real roots. If the difference between the reciprocals of the roots is \(\frac{1}{3}\), and the sum of the reciprocals of the squares of the roots is \(\frac{5}{9}\), then the largest possible value of \((b+c)\) is "

The answer is '9'

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