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1.Relation and Function
hard
If $f\left( x \right) = {\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} - 1$ , $x \in R$ , then the equation $f(x) = 0$ has
A
no solution
B
one solution
C
two solution
D
more than two solutions
(JEE MAIN-2014)
Solution
$f\left( x \right) = {\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} – 1$
Put $f\left( x \right) = 0$
$ \Rightarrow 0 = {\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} – 1$
$ \Rightarrow {\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} – 1$
$ \Rightarrow {3^x} + {4^x} = {5^x}$
For $x=1$
${3^1} + {4^1} > {5^1}$
for $x=3$
${3^3} + {4^3} = 91 < {5^3}$
Only for $x=2$, equation $(1)$ Satisfy
So, only one solution $(x=2)$
Standard 12
Mathematics
