Let f(x) is an even function such that f(x)+f(x−λ)=x(x−λ)+λ2,λ>0, then value of ∑λ=112∫0λf(x)x2−λx+λ2dx is

I=∫0λ f(x)x2−λx+λ2dx also ∵f(x) is even function ⇒f(x−λ)=f(λ−x) also I=∫0λ f(λ−x)x2−λx+λ2dx ⇒I=12∫0λ dx⇒I=λ2⇒∑λ=112 λ2=12⋅(12)(13)2=39

Let f(x) is an even function such that f(x)+f(x−λ)=x(x−λ)+λ2,λ>0, then value of ∑λ=112∫0λf(x)x2−λx+λ2dx is

I=∫0λ f(x)x2−λx+λ2dx also ∵f(x) is even function ⇒f(x−λ)=f(λ−x) also I=∫0λ f(λ−x)x2−λx+λ2dx ⇒I=12∫0λ dx⇒I=λ2⇒∑λ=112 λ2=12⋅(12)(13)2=39

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Let f(x) is an even function such that $f (x) + f (x - λ) = x (x - λ) + λ^{2}, λ > 0,$ then value of $\sum_{λ = 1}^{12} (\int_{0}^{λ} \frac{f (x)}{x^{2} - λx + λ^{2}} dx) is$

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answer is D.

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Detailed Solution

$I = \int_{0}^{λ} \frac{f (x)}{x^{2} - λx + λ^{2}} dx also ∵ f (x) is even function \Rightarrow f (x - λ) = f (λ - x) also I = \int_{0}^{λ} \frac{f (λ - x)}{x^{2} - λx + λ^{2}} dx \Rightarrow I = \frac{1}{2} \int_{0}^{λ} dx \Rightarrow I = \frac{λ}{2} \Rightarrow \sum_{λ = 1}^{12} \frac{λ}{2} = \frac{1}{2} \cdot \frac{(12) (13)}{2} = 39$