Quantitative Aptitude – Algebra - Functions – Let f(x) = min{2x^2,52−5x}
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 )?
Answer: 32
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
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Quantitative Aptitude – Algebra - Logarithms – If log(base12) 81=p then 3
Slot -1 – Quantitative Aptitude – Algebra - Logarithms – If log(base12) 81=p then 3
If log(base12) 81=p then 3 { (4-p)/(4+p)} is equal to
a) log (base2) 8
b) log (base4) 16
c) log (base6) 8
d) log (base6) 16
Solution:
Given , log(base12) 81 = p
4 log (base12) 3 = p
So 3 (4-p)/(4+p) = 3 (1- log(base12) 3 )/(1+ log(base12) 3)=3 × log(base12)(12/3)/log(base12)(12×3) = 3×log(base36) 4 = log(base6)

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Quantitative Aptitude – Algebra - Logarithms – If log(base2)(5 + log(base3) a)
Slot -1 - Quantitative Aptitude – Algebra - Logarithms – If log(base2)(5 + log(base3) a)
If log2(5 + log3a) = 3 and log5(4a + 12 + log2 b) = 3, then a + b is equal to?
a) 59
b) 40
c) 67
d) 32
Solution:
Given, log2(5 + log3 a) = 3
5 + log3 a = 2^3 = 8
log3 a = 3
so a = 3^3 = 27
Now log(base5) (4a+12+log(base2) b)=3
Or 4a + 12 + log(base2) b) = 125
log(base2) b = 125-12-4×27=5
So b = 2^5 = 32
Th

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