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Question

Let $f left parenthesis x right parenthesis$ and $g left parenthesis x right parenthesis$ be differentiable for $0 less or equal than x less or equal than 1 comma$ such that $f open parentheses 0 close parentheses equals 2 comma g open parentheses 0 close parentheses equals 0 comma f open parentheses 1 close parentheses equals 6.$ Let there exist a real number $c$ in $open square brackets 0 comma blank 1 close square brackets$ such that $f to the power of ´ end exponent open parentheses c close parentheses equals 2 g to the power of ´ end exponent open parentheses c close parentheses comma$ then the value of $g left parenthesis 1 right parenthesis$ must be

1
2
$- 2$
$- 1$

The correct answer is: 2

Consider the function $ϕ open parentheses x close parentheses equals f open parentheses x close parentheses minus 2 blank g left parenthesis x right parenthesis$ defined on $open square brackets 0 comma blank 1 close square brackets$
As $f left parenthesis x right parenthesis$ and $g left parenthesis x right parenthesis$ are differentiable for $0 less or equal than x less or equal than 1.$ Therefore, $ϕ left parenthesis x right parenthesis$ is differentiable on $open parentheses 0 comma blank 1 close parentheses$ and continuous on $open square brackets 0 comma blank 1 close square brackets$

We have,

$ϕ open parentheses 0 close parentheses equals f open parentheses 0 close parentheses minus 2 blank g open parentheses 0 close parentheses equals 2 minus 0 equals 2$

$ϕ open parentheses 1 close parentheses equals f open parentheses 1 close parentheses minus 2 blank g open parentheses 1 close parentheses equals 6 minus 2 blank g left parenthesis 1 right parenthesis$

Now, $ϕ to the power of ´ end exponent open parentheses x close parentheses equals f to the power of ´ end exponent open parentheses x close parentheses minus 2 blank g ´ left parenthesis x right parenthesis$

$rightwards double arrow ϕ to the power of ´ end exponent open parentheses c close parentheses equals f to the power of ´ end exponent open parentheses c close parentheses minus 2 blank g to the power of ´ end exponent open parentheses c close parentheses equals 0$ [Given]

Thus, $ϕ left parenthesis x right parenthesis$ satisfies Rolle’s theorem on $open square brackets 0 comma blank 1 close square brackets$

$therefore ϕ open parentheses 0 close parentheses equals ϕ left parenthesis 1 right parenthesis$

$rightwards double arrow 2 equals 6 minus 2 blank g open parentheses 1 close parentheses rightwards double arrow g open parentheses 1 close parentheses equals 2$

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