JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 19)
Let f(x) be a polynomial of degree 3 such that
$$f(k) = - {2 \over k}$$ for k = 2, 3, 4, 5. Then the value of 52 $$-$$ 10f(10) is equal to :
$$f(k) = - {2 \over k}$$ for k = 2, 3, 4, 5. Then the value of 52 $$-$$ 10f(10) is equal to :
Answer
26
Explanation
$$k\,f(k) + 2 = \lambda (x - 2)(x - 3)(x - 4)(x - 5)$$ .... (1)
put x = 0
we get $$\lambda = {1 \over {60}}$$
Now, put $$\lambda$$ in equation (1)
$$ \Rightarrow kf(k) + 2 = {1 \over {60}}(x - 2)(x - 3)(x - 4)(x - 5)$$
Put x = 10
$$ \Rightarrow 10f(10) + 2 = {1 \over {60}}(8)(7)(6)(5)$$
$$ \Rightarrow 52 - 10f(10) = 52 - 26 = 26$$
put x = 0
we get $$\lambda = {1 \over {60}}$$
Now, put $$\lambda$$ in equation (1)
$$ \Rightarrow kf(k) + 2 = {1 \over {60}}(x - 2)(x - 3)(x - 4)(x - 5)$$
Put x = 10
$$ \Rightarrow 10f(10) + 2 = {1 \over {60}}(8)(7)(6)(5)$$
$$ \Rightarrow 52 - 10f(10) = 52 - 26 = 26$$
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