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Let $f(x)$ be a continuous function defined for $1\leq x\leq 3$. If $f(x)$ takes rational value for all $x$ and $f(2) = 10$ then $f(1.5)$ =

$(a)\;10\qquad(b)\;11\qquad(c)\;12\qquad(d)\;13$

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Since $f(x)$ is given continuous on the closed bounded interval [1,3] $f(x)$ is bounded and assumes all the values lying in the interval [m,M] where
m=min f(x) and M=max f(x)
$1\leq x\leq 3\Rightarrow f(1)\leq f(x)\leq (3)$
If $m\geq M$ then f(x) must assume all the irrational values,we must have m=M i.e f(x) must be constant function.
As f(2)=10 we get
$f(x)=10\forall x\in [1,3]$
$f(1.5)=10$
Hence (a) is the correct answer.
answered Dec 31, 2013 by sreemathi.v
 

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