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 :
Answer (integer)
26
Solution
$k\,f(k) + 2 = \lambda (x - 2)(x - 3)(x - 4)(x - 5)$ .... (1)<br><br>put x = 0<br><br>we get $\lambda = {1 \over {60}}$<br><br>Now, put $\lambda$ in equation (1)<br><br>$\Rightarrow kf(k) + 2 = {1 \over {60}}(x - 2)(x - 3)(x - 4)(x - 5)$<br><br>Put x = 10<br><br>$\Rightarrow 10f(10) + 2 = {1 \over {60}}(8)(7)(6)(5)$<br><br>$\Rightarrow 52 - 10f(10) = 52 - 26 = 26$
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Quadratic Equations
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