CAT 2025 Question Paper | Quant Slot 1

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 8 : A value of \(c\) for which the minimum value of \(f(x)=x^2-4 c x+8 c\) is greater than the maximum value of \(g(x)=-x^2+3 c x-2 c\), is

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Video Explanation

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Explanatory Answer

The function is:
\(f(x)=x^2-4cx+8c\)

Since the coefficient of \(x^2\) is positive (1 > 0), \(f(x)\) opens upwards and achieves its minimum at the vertex:
\(x=-\frac{b}{2a}\)
\(x=\frac{4c}{2}=2c\)

Substituting \(x=2c\) into \(f(x)\):
\(f_{\min}=(2c)^2-4c(2c)+8c\)
\(=4c^2-8c^2+8c\)
\(=-4c^2+8c\)

The function is:
\(g(x)=-x^2+3cx-2c\)

Since the coefficient of \(x^2\) is negative (-1 < 0), \(g(x)\) opens downwards and achieves its maximum at its vertex:
\(x=-\frac{b}{2a}\)
\(x=\frac{-3c}{2(-1)}=\frac{3c}{2}\)

Substituting \(x=\frac{3c}{2}\) into \(g(x)\):
\(g_{\max}=-\left(\frac{3c}{2}\right)^2+3c\left(\frac{3c}{2}\right)-2c\)
\(=-\frac{9c^2}{4}+\frac{9c^2}{2}-2c\)
\(=\frac{9c^2}{4}-2c\)

We set up the condition:
\(f_{\min}>g_{\max}\)

\(-4c^2+8c>\frac{9c^2}{4}-2c\)
\(\frac{25c^2}{4}-10c<0\)
\(5c\left(\frac{5c}{4}-2\right)<0\)
\(0<c<\frac{8}{5}\)

The only option which is in this range is \(\frac{1}{2}\).

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The question is " A value of \(c\) for which the minimum value of \(f(x)=x^2-4 c x+8 c\) is greater than the maximum value of \(g(x)=-x^2+3 c x-2 c\), is "

Hence, the answer is '\(\frac{1}{2}\)'

Choice 4 is the correct answer.

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