A positive function f is such that f2-h-4- f2-h-3 as h-0- has a minimum at x-2- Then the value of f2 can be

For minimum at x=2 : f(2 h) gt;f(2) lt;f(2+h) as h→0^+ ⇒ 4 gt;f(2) and f(2) lt;3 As, f(x) gt;0, nbsp; So, possible values of f(2) is (0,3) Now, from ...

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Standard XII

Mathematics

Monotonicity in an Interval

A positive fu...

Question

A positive function $f$ is such that $f (2 - h) = 4, f (2 + h) = 3$ as $h \to 0^{+}$ has a minimum at $x = 2.$ Then the value of $f (2)$ can be

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Solution

The correct option is A $2$
For minimum at $x = 2 :$
$f (2 - h) > f (2) < f (2 + h)$ as $h \to 0^{+}$
$\Rightarrow 4 > f (2)$ and $f (2) < 3$
As, $f (x) > 0,$
So, possible values of $f (2)$ is $(0, 3)$
Now, from the given options only $2 \in (0, 3) .$
$∴$ The value of $f (2)$ can be $2.$

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Monotonicity in an Interval

Standard XII Mathematics

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A positive function f is such that f(2−h)=4,f(2+h)=3 as h→0+ has a minimum at x=2. Then the value of f(2) can be

The correct option is A 2For minimum at x=2: f(2−h)>f(2)<f(2+h) as h→0+ ⇒4>f(2) and f(2)<3 As, f(x)>0, So, possible values of f(2) is (0,3) Now, from the given options only 2∈(0,3). ∴ The value of f(2) can be 2.

A positive function $f$ is such that $f (2 - h) = 4, f (2 + h) = 3$ as $h \to 0^{+}$ has a minimum at $x = 2.$ Then the value of $f (2)$ can be

The correct option is A $2$
For minimum at $x = 2 :$
$f (2 - h) > f (2) < f (2 + h)$ as $h \to 0^{+}$
$\Rightarrow 4 > f (2)$ and $f (2) < 3$
As, $f (x) > 0,$
So, possible values of $f (2)$ is $(0, 3)$
Now, from the given options only $2 \in (0, 3) .$
$∴$ The value of $f (2)$ can be $2.$