(x**2 – 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it’s numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 – (x-1)(x-5)( 2 (x+1) )
= (2 x – 6) (x+1)**2 – (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) – (x-1)(x-5) = 0
(x**2 – 2 x – 3 ) – (x**2 – 6 x + 5) = 0
x**2 – x**2 – 2 x + 6 x – 3 – 5 = 0
4 x – 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 – 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 – 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3
maternal grand mother
( a ) BQDCJCMF
“TERMINAL” split it into the half we get “TERM” “INAL”
Now the first half is decreased by one and the next half is
increased by one, so we get:
“SDQLJOBM” (pls note in the ques we have “SDQIJOBM” where
‘L’ shud have come instead of ‘I’)
so “CREDIBLE” is to “BQDCJCMF”
125^15 would be 5^18
625^20 would be 5^24
25^10 would yield 5^12
so
The answer is b) i.e. 5^125
19, 26, 33, 46, 59, 74, 91
Answer 33
Adding 7,9.13,15,17
Correct is 35
let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9
so X<=3 and any of the digit in the number is <=9
and also given that 3 pairs sum is 11...
so make trial and error..
if X=1...any of the no is 10 which is wrong trial....
if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296
Solution:
As given, we have,
The cost of one pen = 36 Rs.
So, the cost of 15 pens = 36 × 15 = 540 Rs.
The cost of one book = 45 Rs.
So, the cost of 12 books = 45 × 12 = 540 Rs.
The cost of one pencil = 8 Rs.
So, the cost of 10 pencils = 8 × 10 = 80 Rs.
Now,
the cost of each eraser is 40 Rs. less than the combined costs of pen and pencil.
So,
Combined costs of pen and pencil = 36 + 8 = 44 Rs.
Cost of one eraser = 44 – 40 = 4 Rs.
So, the cost of 5 erasers = 4 × 5 = 20 Rs.
Hence,
The total amount spent is
Hence, the total amount spent is 1180 Rs.
16
1500
pages # figures
1000 1000 4
from 999 to 100 (999-99)*3 2700
from 99 to 10 (99-9)*2 180
from 9 to 1 (9-0)*1 9
total sum= 2893
Y R G
B B G
Y G Y