(D) For Statement-$I$: We examine the behavior of $f(x)$ near $x = 0$.For $x < 0$,$f(x) = -\frac{x}{2}$. As $x \to 0^-$,$f(x) \to 0$.For $x \geq 0$,$f

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Statement-$I$ is false,Statement-$II$ is true.

(D) For Statement-$I$: We examine the behavior of $f(x)$ near $x = 0$.For $x For $x \geq 0$,$f(x) = 7x + 8$. As $x \to 0^+$,$f(x) \to 8$.At $x = 0$,$f(0) = 8$.Since $f(x)$ is negative for $x 0$,$f(-h) = \frac{h}{2} > 0$,but $f(0) = 8$. However,we need $f(0) Statement-$I$ is false.For Statement-$II$: This is the standard definition of a local minimum at $x = a$. Thus,Statement-$II$ is true.

Class 12 Mathematics Applications of Derivatives Maxima and Minima

Medium

Statement-$I$: Let the function $f(x) = \begin{cases} -\frac{x}{2} & x < 0 \\ 7x + 8 & x \geq 0 \end{cases}$. Then $f(x)$ has a local minimum at $x = 0$.
Statement-$II$: If $f(a) < f(a - h)$ and $f(a) < f(a + h)$ for a sufficiently small $h > 0$,then $f(x)$ has a local minimum at $x = a$.

A
Statement-$I$ is true,Statement-$II$ is true; Statement-$II$ is a correct explanation for Statement-$I$.
B
Statement-$I$ is true,Statement-$II$ is true; Statement-$II$ is not a correct explanation for Statement-$I$.
C
Statement-$I$ is true,Statement-$II$ is false.
D
Statement-$I$ is false,Statement-$II$ is true.

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Applications of Derivatives

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Maxima and Minima

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Statement-$I$: Let the function $f(x) = \begin{cases} -\frac{x}{2} & x < 0 \\ 7x + 8 & x \geq 0 \end{cases}$. Then $f(x)$ has a local minimum at $x = 0$.Statement-$II$: If $f(a) < f(a - h)$ and $f(a) < f(a + h)$ for a sufficiently small $h > 0$,then $f(x)$ has a local minimum at $x = a$.

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Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p(1)=6$ and $p(3)=2$,then $p^{\prime}(0)$ is

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Let $f(x) = |(x-1)(x^2-2x-3)| + x - 3$,$x \in R$. If $m$ and $M$ are respectively the number of points of local minimum and local maximum of $f$ in the interval $(0,4)$,then $m + M$ is equal to

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Statement-$I$: Let the function $f(x) = \begin{cases} -\frac{x}{2} & x < 0 \\ 7x + 8 & x \geq 0 \end{cases}$. Then $f(x)$ has a local minimum at $x = 0$.
Statement-$II$: If $f(a) < f(a - h)$ and $f(a) < f(a + h)$ for a sufficiently small $h > 0$,then $f(x)$ has a local minimum at $x = a$.