im getting d.....[4][4][4].........
y=f(x) is a curve defined as
x = 1-3t2
y = t-3t3
h(x) = |f(x)| + f(x)
g(x) = |f(x)| - f(x)
then a∫b h(x) dx is equal to
(A) a∫b g(x) dx
(B) b∫a g(x) dx
(C) b∫a g(x) dx only if a,b →[0,1]
(D) None of these
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11 Answers
make t da subject from "x" √ [(1-x)/3]
then substitute t in y.
u get f(x), then get h, g then solve everything
y = x√(1-x)/3 =f(x)
h(x) = |f(x)| + f(x) ----------1
g(x) = |f(x)| - f(x) ----------2
subtract 2 from 1.
h(x) - g(x) = 2 f(x)
\int_{a}^{b}{h(x)} - \int_{a}^{b}{g(x)} = 2/\sqrt{3}. \int_{a}^{b}{x\sqrt{1-x} dx}
even if a=0, b=1..
∫h(x) ≠∫g(x) ...
so. shud be D.
good one!
yeh toh soche nahi...
ho sakta hai...
par as it is not mentioned beside the options ...
so [3]
waise, subjective me ayega ,, toh main bagal mein likh dungi : 'priyam says' :P [3]
ya but..........not given in that cases a=b=0........so not possible .....