Let's do some calculus! (20)

Calculus Level 4

lim ⁡ n → ∞ ( ∑ k = 1 n a k − 1 ∫ ( k − 1 ) a k a f ( x ) f ( x ) + f ( ( 2 k − 1 ) a − x ) d x ) = 7 5 \lim_{n \to \infty} \left(\sum_{k=1}^n a^{k-1} \int_{(k-1)a}^{ka} {\dfrac{f(x)}{f(x) + f \left( \left(2k-1\right)a-x \right)}} \ dx \right) = \dfrac{7}{5}

Given that function f ( x ) > 0 f(x) > 0 , ∀ x ∈ R \forall ~ x \in \mathbb{R} is bounded and satisfies the condition above. If a < 1 a < 1 , find a a .

For more problems on calculus, click here .
2 7 \dfrac{2}{7} 7 19 \dfrac{7}{19} 1 19 \dfrac{1}{19} 9 14 \dfrac{9}{14} 1 7 \dfrac{1}{7} 14 19 \dfrac{14}{19} 14 9 \dfrac{14}{9} 9 19 \dfrac{9}{19}

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Chew-Seong Cheong
Sep 24, 2016

Let the limit be L L , then we have,

L = lim ⁡ n → ∞ ∑ k = 1 n a k − 1 ∫ ( k − 1 ) a k a f ( x ) f ( x ) + f ( ( 2 k − 1 ) a − x ) d x By ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x = lim ⁡ n → ∞ ∑ k = 1 n a k − 1 1 2 ∫ ( k − 1 ) a k a ( f ( x ) f ( x ) + f ( ( 2 k − 1 ) a − x ) + f ( ( 2 k − 1 ) a − x ) f ( ( 2 k − 1 ) a − x ) + f ( ( 2 k − 1 ) a − ( 2 k − 1 ) a + x ) ) d x = lim ⁡ n → ∞ ∑ k = 1 n a k − 1 1 2 ∫ ( k − 1 ) a k a ( f ( x ) f ( x ) + f ( ( 2 k − 1 ) a − x ) + f ( ( 2 k − 1 ) a − x ) f ( ( 2 k − 1 ) a − x ) + f ( x ) ) d x = lim ⁡ n → ∞ ∑ k = 1 n a k − 1 1 2 ∫ ( k − 1 ) a k a f ( x ) + f ( ( 2 k − 1 ) a − x ) f ( x ) + f ( ( 2 k − 1 ) a − x ) d x = lim ⁡ n → ∞ 1 2 ∑ k = 1 n a k − 1 ∫ ( k − 1 ) a k a 1 d x = lim ⁡ n → ∞ 1 2 ∑ k = 1 n a k − 1 [ k a − ( k − 1 ) a ] = lim ⁡ n → ∞ 1 2 ∑ k = 1 n a k = a 2 ( 1 − a ) \begin{aligned} L & = \lim_{n \to \infty} \sum_{k=1}^n a^{k-1} \int_{(k-1)a}^{ka} \frac {f(x)}{f(x) + f((2k-1)a-x)} dx \quad \quad \small \color{#3D99F6}{\text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \lim_{n \to \infty} \sum_{k=1}^n a^{k-1} \frac 12 \int_{(k-1)a}^{ka} \left( \frac {f(x)}{f(x) + f((2k-1)a-x)} + \frac {f((2k-1)a-x)}{f((2k-1)a-x) + f((2k-1)a-(2k-1)a+x)} \right) dx \\ & = \lim_{n \to \infty} \sum_{k=1}^n a^{k-1} \frac 12 \int_{(k-1)a}^{ka} \left( \frac {f(x)}{f(x) + f((2k-1)a-x)} + \frac {f((2k-1)a-x)}{f((2k-1)a-x) + f(x)} \right) dx \\ & = \lim_{n \to \infty} \sum_{k=1}^n a^{k-1} \frac 12 \int_{(k-1)a}^{ka} \frac {f(x) + f((2k-1)a-x)}{f(x) + f((2k-1)a-x)} dx \\ & = \lim_{n \to \infty} \frac 12 \sum_{k=1}^n a^{k-1} \int_{(k-1)a}^{ka} 1 \ dx \\ & = \lim_{n \to \infty} \frac 12 \sum_{k=1}^n a^{k-1} [ka - (k-1)a] \\ & = \lim_{n \to \infty} \frac 12 \sum_{k=1}^n a^k \\ & = \frac a{2(1-a)} \end{aligned}

Therefore,

a 2 ( 1 − a ) = 7 5 5 a = 14 − 14 a 19 a = 14 ⟹ a = 14 19 \begin{aligned} \frac a{2(1-a)} & = \frac 75 \\ 5a & = 14-14 a \\ 19a & = 14 \\ \implies a & = \boxed{\dfrac {14}{19}} \end{aligned}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Those stacks didn't look pretty :P . Anyways, sir, as always, you got it right on spot!

Tapas Mazumdar - 4 years, 8 months ago

A perfect problem, Which upgraded me to Calculus Level 5. Nice Sums. And Very nice solution.

Only seems a bit boring and lenghty. Please dont mind sir.

Md Zuhair - 4 years, 4 months ago

sir, why did you multiply with a^(k-1) initially, is it a property related to integrals.

senna s - 3 years, 11 months ago

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I think someone has removed it from the question.

Chew-Seong Cheong - 3 years, 11 months ago

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Even i have same thought s

Md Zuhair - 3 years, 11 months ago

Since there is no restriction on $f(x)$, just replacing $f(x)=1$, then one can easily find then answer...

Daewon Chung - 11 months ago

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