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Syllabus
give explaination
a.) there is only one integral 'a' for which f(x) is non-negative for all x belongs to R
b.) for a< 0, the number lies between the zeroes of the polynomial
c.) f(x) =0 has two distinct solutions in (0,1) for 'a' belonging to (1/7,4/7)
d.) the minimum value of f(x) for minimum value of 'a' for which f(x) is non-negative for all x in R is zero
(213.768*4.434)/6.321
1. log 9/5+log 15/9-log 3/2=log 2
2. 2 log 3/7+log 49/9=0
3. logb*ca=logba/1+logbc
(1) (2) (3) (4)
1. logb2a2.logc2b2.loga2c2=1
2. Solve 213.768*4.434 whole divided by 6.321 using log table.
(1) 3 (2) 4
(3) 5 (4) 6
If 2n - 2m = 960 , then n - m = ?