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Need solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise 17.1  question 1

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Minimum value is 3 and maxima value does not exist.


f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a,b] and if f(x) ≥ f(c) then f(x) has minimum value.


f(x)=4 x^{2}-4 x+4 \text { on } R

We have,

 \begin{aligned} &f(x)=4 x^{2}-4 x+4 \text { on } R \\ &=4 x^{2}-4 x+1+3 \\ &=(2 x-1)^{2}+3 \\ &\because(2 x-1)^{2} \geq 0 \\ &(2 x-1)^{2}+3 \geq 3 \\ &f(x) \geq f\left(\frac{1}{2}\right) \end{aligned}

Thus, at x = 1/2 , minimum value of f(x) is 3.

Since the value of f(x) is increasing rapidly. That is why it does not attain the max value.


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