CAT 2023 Question Paper | Quant Slot 2

CAT 2023 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 : Let $k$ be the largest integer such that the equation $(x-1)^2+2 k x+11=0$ has no real roots. If $y$ is a positive real number, then the least possible value of $\frac{k}{4 y}+9 y$ is

Video Explanation

Explanatory Answer

$( x - 1 ) ^ { 2 } + 2 k x + 11 = 0$
$x ^ { 2 } - 2 x + 1 + 2 k x + 11 = 0$
$x ^ { 2 } + 2 ( k - 1 ) x + 12 = 0$
Since the equation above has no real roots, the discriminant of the equation should be negative.
$b ^ { 2 } - 4 a c < 0$
$b ^ { 2 } < 4 a c$
$( 2 ( k - 1 ) ) ^ { 2 } < 4 \cdot 1 \cdot 12$
$4 ( k - 1 ) ^ { 2 } < 4 \cdot 12$
$( k - 1 ) ^ { 2 } < 12$
$- \sqrt { 12 } < k - 1 < \sqrt { 12 }$
$- 3.46 \leq k - 1 \leq 3.46$
$- 2.46 \leq k \leq 4.46$
The largest integral value that k can take is 4.
Now, we need to minimize $\frac { k } { 4 y } + 9 y$ where k takes the largest integral value and y is positive…
$\frac { k } { 4 y } + 9 y = \frac { 4 } { 4 y } + 9 y = \frac { 1 } { y } + 9 y$
$\frac { 1 } { y } \& 9 y$ are both positive real numbers, therefore, their A.M is greater than equal to their G.M.
$A M \left( \frac { 1 } { y } , 9 y \right) \geq G M \left( \frac { 1 } { y } , 9 y \right)$
$\frac { \frac { 1 } { y } + 9 y } { 2 } \geq \sqrt { \frac { 1 } { y } ( 9 y ) }$
$\frac { \frac { 1 } { y } + 9 y } { 2 } \geq 3$
$\frac { 1 } { y } + 9 y \geq 6$
$\therefore \left( \frac { k } { k y } + 9 y \right) = 6$

The question is " Let $k$ be the largest integer such that the equation $(x-1)^2+2 k x+11=0$ has no real roots. If $y$ is a positive real number, then the least possible value of $\frac{k}{4 y}+9 y$ is "

Hence, the answer is '6'

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