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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