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Maths-

General

Easy

Question

If f(x) is a real valued function discontinuous at all integral points lying in [0,n] and if $left parenthesis f left parenthesis x right parenthesis right parenthesis to the power of 2 end exponent equals 1 for all x element of left square bracket 0 comma n right square bracket$ then number of functions f(x) are

$2^{n + 1}$
$6 \times 3^{n}$
$2 \times 3^{n - 1}$
$3^{n + 1}$

The correct answer is: $2 cross times 3 to the power of n minus 1 end exponent$

There are four possible functions defined in of them 2 are continues and two are discontinuous now for each of the points (1,2,, n-1), keep functions fixed from left of the point so there are 4 possible functions defined in between next two consecutive integral points of them only is continuous and at last for x=n there is only one possibility of discontinuity of the function. So total number of functions= $2 cross times 3 to the power of n minus 1 end exponent cross times 1$

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