1is the ans of 2nd prob 1?
1)Find the value of "t" for which-
e2x= t √x , has only one solution.
2)Let "g" be the inverse function of a differenciable function "f" such that,
G(x) = 1/g(x).
If f(4)=2 & f'(4)= 1/16, then (G'(2))2 =?
3) Value of "a" for which -
ax + sec-1√2x2-x4 + cosec-1√2x2-x4 =0 is ?
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8 Answers
13Thnx akari [1], But how does it explain tht there exists no other solution ?
1second one :
g(f(x))=x
g(f(4))=4
g(2)=4
now g'(x)=1/f'(g(x)) ....
see here dude : http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation
now g'(2)=1/f'(g(2))
g'(2)=1/f'(4)
g'(2)=16
now G'(x) = -g'(x)/(g(x))2
G'(2)= -1
G'(2)2=1
1First answer is t = 2e12
Second answer is 1. (Both solved by Akari)
For third answer, apply identity directly.
cosec-1 A + sec-1 A = ∩2
Hence for tht qsn, the cosec and sec part becomes pi/2.
Now, we get:
ax + ∩2 = 0.
Hence, a = -∩2x where x belongs to (-∞,1]U[1,∞)
13Thnx akari n ankur...
Some More :-
An extension to Qn 1) (maybe..)
4) Find the integral values of "t" for which-
e2x= t √x , has exactly 2 distinct solutions.
5) If all the roots of x4-14x2+24x-k=0 are real & distinct, find the possible integral values of "k".