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Best Answer:
f((x−y)²) = f²(x)−2xf(y)+y² … (i)

Reverse x and y in (i) to get f((y−x)² = f²(y)−2yf(x)+x²

Hence f²(x)−2xf(y)+y² = f²(y)−2yf(x)+x² … (ii)

y=0 in (ii) ⟹ f²(x)−2xf(0) = f²(0)+x² … (iii)

x=y=0 in (i) ⟹ f(0) = f²(0) ⟹ f(0)=0 or 1

Now use these values in (iii) to determine f(x) explicitly …

f(0)=0 ⟹ f(x)=±x and f(x)=−x is inconsistent with (i) so f(x)=x

f(0)=1 ⟹ f(x)=±(x+1) and f(x)=−(x+1) is inconsistent with (i) so f(x)=x+1

If f(x)=x then g(x) = ∫ exp(5x)cos4x dx = (1/41)exp(5x)(4sin4x+5cos4x) + C (see **)

g(0)=5/41 so 5/41 = (1/41)(5)+C ⟹ C=0

∴ g(π/2) = (1/41)exp(5π/2)(0+5) = (5/41)exp(5π/2)

If f(x)=x+1 then g(x) = ∫ ( exp(5x)cos4x + 1 )dx = (1/41)exp(5x)(4sin4x+5cos4x) + x + C

g(0)=5/41 so 5/41 = (1/41)(5)+0+C ⟹ C=0

∴ g(π/2) = (1/41)exp(5π/2)(0+5)+π/2 = (5/41)exp(5π/2)+π/2

**

∫ exp(5x)cos4x dx = ℝ{ ∫ exp((5+4i)x)dx } = ℝ{ (1/(5+4i) exp((5+4i)x) }

= (1/41)ℝ{ (5−4i)exp(5x)(cos4x+isin4x) } = (1/41)exp(5x)(5cos4x+4sin4x)

Reverse x and y in (i) to get f((y−x)² = f²(y)−2yf(x)+x²

Hence f²(x)−2xf(y)+y² = f²(y)−2yf(x)+x² … (ii)

y=0 in (ii) ⟹ f²(x)−2xf(0) = f²(0)+x² … (iii)

x=y=0 in (i) ⟹ f(0) = f²(0) ⟹ f(0)=0 or 1

Now use these values in (iii) to determine f(x) explicitly …

f(0)=0 ⟹ f(x)=±x and f(x)=−x is inconsistent with (i) so f(x)=x

f(0)=1 ⟹ f(x)=±(x+1) and f(x)=−(x+1) is inconsistent with (i) so f(x)=x+1

If f(x)=x then g(x) = ∫ exp(5x)cos4x dx = (1/41)exp(5x)(4sin4x+5cos4x) + C (see **)

g(0)=5/41 so 5/41 = (1/41)(5)+C ⟹ C=0

∴ g(π/2) = (1/41)exp(5π/2)(0+5) = (5/41)exp(5π/2)

If f(x)=x+1 then g(x) = ∫ ( exp(5x)cos4x + 1 )dx = (1/41)exp(5x)(4sin4x+5cos4x) + x + C

g(0)=5/41 so 5/41 = (1/41)(5)+0+C ⟹ C=0

∴ g(π/2) = (1/41)exp(5π/2)(0+5)+π/2 = (5/41)exp(5π/2)+π/2

**

∫ exp(5x)cos4x dx = ℝ{ ∫ exp((5+4i)x)dx } = ℝ{ (1/(5+4i) exp((5+4i)x) }

= (1/41)ℝ{ (5−4i)exp(5x)(cos4x+isin4x) } = (1/41)exp(5x)(5cos4x+4sin4x)

Indica
· 1 month ago

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