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F(X) = 16x2 \[-\] 16x + 28 on R ? - CBSE (Arts) Class 12 - Mathematics

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Question

f(x) = 16x2 \[-\] 16x + 28 on R ?

Solution

Given: f(x) = 16x2 − 16x + 28

\[\Rightarrow\] f(x) = 4(4x2 - 4x + 1) + 24

\[\Rightarrow\] f(x) = 4(2x − 1)2 + 24

Now,
4(2x − 1)\[\geq\] 0 for all x \[\in\] R

\[\Rightarrow\] f(x) = 4(2x − 1)2 + 24 \[\geq\] 24 for all x \[\in\] R

\[\Rightarrow\] f(x)\[\geq\] 24 for all x \[\in\] R.

The minimum value of f is attained when (2x − 1) = 0.
(2x − 1) = 0

⇒ x = \[\frac{1}{2}\]

Therefore, the minimum value of f  at x =\[\frac{1}{2}\] is 24.

Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.

Hence, the function f does not have a maximum value.

  Is there an error in this question or solution?

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Solution F(X) = 16x2 \[-\] 16x + 28 on R ? Concept: Graph of Maxima and Minima.
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