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  • Provide solution for rd sharma class 12 chapter 17 Maxima and Minima excercise 17.1 question 8

Provide solution for rd sharma class 12 chapter 17 Maxima and Minima  excercise 17.1 question 8

Answers (1)

best_answer

Answer:

Minimum value is 24 and maximum value does not exist.

Hint:

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.

Given:

f(x)=16 x^{2}-16 x+28 \text { on } R

Explanation:

We have,

\begin{aligned} &f(x)=16 x^{2}-16 x+28 \text { on } R\\ &=16 x^{2}-16 x+4+24\\ &=(4 x-2)^{2}+24\\ &\text { Now, }(4 x-2)^{2} \geq 0 \text { for all } \mathrm{x} \in R\\ &(4 x-2)^{2}+24 \geq 24 \text { for all } \mathrm{x} \in R\\ &f(x) \geq f\left(\frac{1}{2}\right) \end{aligned}

So, minimum value of f(x) is 24 at x=\frac{1}{2}

Since, value of f(x) increases rapidly. So it does not have maximum value.

 

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