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It is given that at x = 1, the function x4− 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.

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प्रश्न

It is given that at x = 1, the function x4− 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.

योग
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उत्तर

f (x) = x4 - 62x2 + ax + 9, 0 ≤ x ≤ 2

= f' (x) = 4x3 - 124x + a

We have f (x) attains maximum value at x = 1 ∈ [0, 2]

∴ We must have f' (1) = 0

⇒ 4 - 124 + a = 0

= a = 120

Thus, we have f(x) = x4 - 62x2 + 120x + 9

f (0) = 9, f(1) = 1 - 62 + 120 + 9 = 68 and

f (2) = 24 - 62 × 22 + 120 × 2 + 9 = 17

Clearly, f(x) is maximum at x = 1.

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अध्याय 6: Application of Derivatives - Exercise 6.5 [पृष्ठ २३३]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.5 | Q 11 | पृष्ठ २३३

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