Let f(x) be a polynomial of degree

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Question:

Let $f(x)$ be a polynomial of degree 3 such that $\mathrm{f}(\mathrm{k})=-\frac{2}{\mathrm{k}}$ for $\mathrm{k}=2,3,4,5 .$ Then the value of $52-10 \mathrm{f}(10)$ is equal to :

Solution:

$\mathrm{k} \mathrm{f}(\mathrm{k})+2=\lambda(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)(\mathrm{x}-5) \ldots(1)$

put $x=0$

we get $\lambda=\frac{1}{60}$

Now put $\lambda$ in equation (1)

$\Rightarrow \mathrm{kf}(\mathrm{k})+2=\frac{1}{60}(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)(\mathrm{x}-5)$

Put $x=10$

$\Rightarrow 10 \mathrm{f}(10)+2=\frac{1}{60}(8)(7)(6)(5)$

$\Rightarrow 52-10 \mathrm{f}(10)=52-26=26$

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