1using L hospital.........
f'(a+h)(0+1)-0+f'(a-h)(0-1)/2h
=f'(a+h)-f'(a-h)/2h
=f''(a+h)(0+1)-f''(a-h)(0-1)/2
=f''(a+h)+f''(a-h)/2
=2f''(a)/2.................as h is zero
=f''(a)
ans B
Q1) f(x).f(y) = f(x) +f(y) +f(xy) -- 2 and that f(2)= 5.then f(3) =
a.10
b.24
c.15
d.N.O.T.
(f is a differential function of x)
Q2) lim (h→0) [ f(a+h)-2f(a)+f(a-h) ] / h2
a. 2 f'(a)
b. f''(a)
c. f'(a)
d. f'(a)+f''(a)
(f is twice differential function of x)
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UP 0 DOWN 0 0 11
11 Answers
11f(x).f(y) = f(x) +f(y) +f(xy) -- 2
is that -2???? i mean r u subtracting 2 also??
1f(x).f(y) = f(x) +f(y) +f(xy) - 2
let x=y=0
f(0)2 =3f(0)-2
let f(0)=k
k2-3k+2=0
(k-1)(k-2)=0
k=1 or 2
case 1...let f(0)=1
so take x as 3 and y as 2
so
f(3).f(2) = f(3) +f(2) +f(6) - 2
5f(3)=f(3)+f(6)+3
11answer to Q1 is A
f(1)f(2)=f(1)+f(2)+3
f(-1)f(-2)=f(-1)+f(-2)+3
f(1)=f(-1)
so the function is even
and u can even check that f(1)=2 [take x=1 and y=0]
u get the function as x2+1
hence f(3)=10
1f(x).f(y) = f(x) +f(y) +f(xy) - 2
f(0) is 1 or 2................previous post.
let x=1 and y=0
f(1)f(0)=f(1)+f(0)+f(0)-2
for k=1
it satisfies
for k=2
2f(1)=f(1)+2+2-2
f(1)=2.
thts all i can see....sorry cudnt solve fully
1how did you get that the function is even ?? I am sorry if my question appears to be silly
1i can see the function is even......cox f(x)=f(-x)
but hw d heck is f(1)=f(-1)?????
11mere bhai akand
maine kaha tha
f(1)f(2)=f(1)+f(2)+3
f(-1)f(-2)=f(-1)+f(-2)+3
ab aggar f(-1)=f(1) and f(-2)=f(2) hoi
tabhie to both the equations same hoongi na