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Maths-

General

Easy

Question

The value of $c$ in (0, 2) satisfying the Mean value theorem for the function $f open parentheses x close parentheses equals x open parentheses x minus 1 close parentheses to the power of 2 end exponent comma blank x element of left square bracket 0 comma blank 2 right square bracket$ is equal to

$\frac{3}{4}$
$\frac{4}{3}$
$\frac{1}{3}$
$\frac{2}{3}$

The correct answer is: $fraction numerator 4 over denominator 3 end fraction$

$f open parentheses x close parentheses equals x open parentheses x minus 1 close parentheses to the power of 2 end exponent$
$f to the power of ´ end exponent open parentheses x close parentheses equals 2 x open parentheses x minus 1 close parentheses plus open parentheses x minus 1 close parentheses to the power of 2 end exponent$

$equals open parentheses x minus 1 close parentheses open parentheses 2 x plus x minus 1 close parentheses equals open parentheses x minus 1 close parentheses open parentheses 3 x minus 1 close parentheses$

$therefore blank f to the power of ´ end exponent open parentheses c close parentheses equals fraction numerator f open parentheses 2 close parentheses minus f left parenthesis 0 right parenthesis over denominator 2 minus 0 end fraction$

$rightwards double arrow blank open parentheses c minus 1 close parentheses open parentheses 3 c minus 1 close parentheses equals fraction numerator 2 minus 0 over denominator 2 end fraction equals 1$

$rightwards double arrow blank 3 c to the power of 2 end exponent minus 4 c equals 0$

$rightwards double arrow blank c open parentheses 3 c minus 4 close parentheses equals 0$

$rightwards double arrow blank c equals 0$ or $c equals fraction numerator 4 over denominator 3 end fraction$

$therefore$ The value of $c$ in $left parenthesis 0 comma blank 2 right parenthesis$ is $fraction numerator 4 over denominator 3 end fraction$

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