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Find the no of solutions of the equation2^x = x^2 + 1Caution : answer is not 2 ;)
2^x = x^2 + 1
Take log both sides with base ''e''
xlog2 = log(x^2 + 1)
x = log(x^2 + 1)/log2
x = log[(x^2 + 1)/(2)]
(x^2 + 1)/(2) = e^x
x^2/2 + 1/2 = e^x
From this graph we conclude that both graph cuts each other at only one point.
Hence there is onle and only one solution.
Plz Approve!
No no no seeu will surely get two values 1 and 0Now if u take f(x)=x^2 + 1 - 2^xfind f''(x) now apply rolles theoremu will get that f(x)=0 has 3 rootsanother root will be b/w 4 and 5
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