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 Other posts related to Quantitati
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)
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