Continuity and Differentiability

Question

If the function f(x) =[(x-2)^{3 ​​​​​​}​/a] sin(x-2) + acos(x-2), [.]  denotes greatest integer function, is continuous & differentiable in (4,6) then find ‘a’ range:

(A) a ϵ (-∞,∞)

(B) a ϵ [64, ∞)

(C) a ϵ [128, ∞)

(D) Not defined
 

Comments

is it a?

Answer 1

Answer: Option B

Explaination:

Since [$(x-2)^3$ /a]  is not continuous and differentiable at integral point (because [ ] functions are not continuous)

Given function f(x) is continuous and differentiable in (4,6) so

[$(x-2)^3$ /a] has to be equal to zero,

hence [$(x-2)^3$ /a] = 0

The maximum value x can take, approximately 6, for [$(x-2)^3$ /a] numerator will be $(6-2)^3$ = $4^3$ = 64

Now for [$(x-2)^3$ /a] to be near 0, the denominator has to be greater than 0.

hence, a ≥ 64 [since if we have a value less than 64, we will get 1 as step function, but we need step function to be’0’]

Answer 2

Go through it

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