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1.Relation and Function
hard
Let $\mathrm{f}(\mathrm{x})$ 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 :
A
$26$
B
$36$
C
$52$
D
$87$
(JEE MAIN-2021)
Solution
$\mathrm{k}\, \mathrm{f}(\mathrm{k})+2=\lambda(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)(\mathrm{x}-5) \ldots(1)$
put $\mathrm{x}=0$
we get $\lambda=\frac{1}{60}$
Now put $\lambda$ in equation $(1)$
$\Rightarrow \mathrm{k\,f}(\mathrm{k})+2=\frac{1}{60}(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)(\mathrm{x}-5)$
Put $\mathrm{x}=10$
$\Rightarrow 10 \,\mathrm{f}(10)+2=\frac{1}{60}(8)(7)(6)(5)$
$\Rightarrow 52-10 \,\mathrm{f}(10)=52-26=26$
Standard 12
Mathematics
