## Slot -2 – Quantitative Aptitude – Algebra – Functions – Let f(x)=max{5x, 52-2x^2}

Let f(x)=max{5x, 52-2x^2}, where x is any positive real number. Then the minimum possible value of f(x) is?

Solution:
For f(x) to be minimum , 5𝑥 = 52 − 2𝑥^2
2𝑥^2+5𝑥 −52 =0
𝑥−42𝑥+13=0
𝑆𝑜 𝑥 = 4
Thus minimum value of f(x) = 5x = 5*4 = 20

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## Slot -1 – Quantitative Aptitude – Algebra – Functions – If f(x + 2) = f(x) + f(x + 1)

If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals?

Solution:

Given , f(x + 2) = f(x) + f(x + 1)

f(15) = f(13) + f(14)

f(13) + f(14) = 617 —————1)

f(12) + f(13) = f(14) ————-2)

f(11) + f(12) = f(13)————-3)

from eq 1) , 2) & 3)

2f(11) + 3f(12) = 617

Putting f(11) = 91

f(12) = (617 – 2*91)/3 = 145

f(12) = f(11) + f(10)

so f(10) = f(12) – f(11) = 145 – 91 =54

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## Slot -1 – Quantitative Aptitude – Algebra – Functions – Let f(x) = min{2x2,52−5x}

Let f(x) = min{2x2,52−5x}, where x is any positive real number. Then the maximum possible value of f(x) is ( TITA )?

Solution: for maximum possible value , 2x2= 52−5x

2x2+ 5x – 52 = 0

(x -4)*(x+6.5) = 0

So x = 4 ( as x is positive real number )

Maximum possible value of f(x) = 2x2= 52−5x = 32

## Online Coaching Course for CAT Exam Preparation

a) 750+ Videos covering entire CAT syllabus
b) 2 Live Classes (online) every week for doubt clarification
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e) Previous Year Questions solved on video