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Question

Let $f left parenthesis x right parenthesis equals fraction numerator log invisible function application left parenthesis 1 plus x plus x to the power of 2 end exponent right parenthesis plus log invisible function application left parenthesis 1 minus x plus x to the power of 2 end exponent right parenthesis over denominator sec invisible function application x minus cos invisible function application x end fraction$ , x $not equal to$ 1. The value of f(0) so that f is continuous at x = 0 is

1
2
4
none of these

The correct answer is: 1

f(0) = $stack l i m with x rightwards arrow 0 below f left parenthesis x right parenthesis$
$equals stack l i m with x rightwards arrow 0 below fraction numerator log invisible function application left square bracket left parenthesis 1 plus x plus x to the power of 2 end exponent right parenthesis plus log invisible function application left parenthesis 1 minus x plus x to the power of 2 end exponent right parenthesis over denominator sec invisible function application x minus cos invisible function application x end fraction equals stack l i m with x rightwards arrow 0 below fraction numerator log invisible function application left parenthesis left parenthesis 1 plus x to the power of 2 end exponent right parenthesis to the power of 2 end exponent minus x to the power of 2 end exponent right parenthesis over denominator 1 minus cos to the power of 2 end exponent invisible function application x end fraction. cos invisible function application x$
$equals stack l i m with x rightwards arrow 0 below fraction numerator log invisible function application left parenthesis 1 plus x to the power of 4 end exponent plus x to the power of 2 end exponent right parenthesis over denominator x to the power of 2 end exponent. fraction numerator sin to the power of 2 end exponent invisible function application x over denominator x to the power of 2 end exponent end fraction end fraction stack l i m with x rightwards arrow 0 below cos invisible function application x equals stack l i m with x rightwards arrow 0 below fraction numerator log invisible function application left square bracket 1 plus left parenthesis x to the power of 2 end exponent plus x to the power of 4 end exponent right parenthesis right square bracket over denominator x to the power of 2 end exponent end fraction$
$equals stack l i m with x rightwards arrow 0 below fraction numerator log invisible function application left parenthesis 1 plus left parenthesis x to the power of 2 end exponent plus x to the power of 4 end exponent right parenthesis right square bracket over denominator x to the power of 2 end exponent plus x to the power of 4 end exponent end fraction equals 1$

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