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Maths-

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Question

In [0, 1] Lagranges Mean value theorem is NOT applicable to

$f (x) = \{\begin{matrix} \frac{1}{2} - x, x < \frac{1}{2} \\ {(\frac{1}{2} - x)}^{2}, x \geq \frac{1}{2} \end{matrix}$
$f (x) = \{\begin{matrix} \frac{\sin x}{x}, x \neq 0 \\ 1, x = 0 \end{matrix}$
$f (x) = x |x|$
$f (x) = | x |$

The correct answer is: $f open parentheses x close parentheses equals open curly brackets table row cell fraction numerator 1 over denominator 2 end fraction minus x comma blank x less than fraction numerator 1 over denominator 2 end fraction end cell row cell open parentheses fraction numerator 1 over denominator 2 end fraction minus x close parentheses to the power of 2 end exponent comma blank x greater or equal than fraction numerator 1 over denominator 2 end fraction end cell end table close$

For Lagrange’s Mean value theorem we know, $f left parenthesis x right parenthesis$ should be continuous in [a, b] and differentiable in] a, b [.
$open parentheses a close parentheses G i v e n comma f open parentheses x close parentheses equals open curly brackets table row cell fraction numerator 1 over denominator 2 end fraction minus x comma blank x less than fraction numerator 1 over denominator 2 end fraction end cell row cell open parentheses fraction numerator 1 over denominator 2 end fraction minus x close parentheses to the power of 2 end exponent comma blank x greater or equal than fraction numerator 1 over denominator 2 end fraction end cell end table close$

Which is clearly not differentiable at $x equals fraction numerator 1 over denominator 2 end fraction semicolon a s blank R H D$ at $open parentheses x equals 1 divided by 2 close parentheses equals negative 1 blank a n d blank$ LHD at $open parentheses x equals 1 divided by 2 close parentheses equals 0 rightwards double arrow$ Lagrange’s Mean Value is not applicable.

Where option, (b), (c), (d) are continuous and differentiable.

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