13Adding another-
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If f(x) is a continuous function which takes +ve values for x≥0 and satisfies-
0∫x f(t).dt = x√f(x), with f(1)=1/2.
Then, the value of 1/f(√2+1) = ?
Ans- 4
1) Find the value of y(log4), if
y2 -7y1 +12y=0,
given that- y(0)=2 & y1(0)=7.
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If a function f(x) satisfies-
i) f(x+y)= f(x).f(y)
ii) f(x)= 1 + x.g(x), where Lim(x→0) g(x)=1.
Find the value of logf(8).
(Ans - 8 )
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7 Answers
13
11.
This is only I could think of..........
a bit of observation
shows tht the function is an exponential function
Assuming d function to be ae^c1x+Be^c2x
got F(x)=e^{3x}+e^{4x}
by using the given data.........
f(log4)=320
1061. not in JEE... but doable if u know the method
For such expressions, substitute yr = Dr, r≥0
then find its roots.. (if real roots exist)
Then its solution is y=Aea1x + Bea2x .... where a1 a2 are the real roots and A,B are arbitrary constants...
There are other methods for finding solution where real roots cannot be found... there is a different sol. for repeated roots.. I dont think tis will ever be asked in JEE
29ya i agree that it's not in jee..but there was one question in which the differential equation given was of second degree...
http://www.targetiit.com/iit-jee-forum/posts/1999-2m-15183.html