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Question 62
If $$f(x^2 - 1) = x^4 - 7x^2 + k_1$$ and $$f(x^3 - 2) = x^6 - 9x^3 +k_2$$ then the value of $$(k_2 - k_1)$$ is
Solution
$$f(x^2 - 1) = x^4 - 7x^2 + k_1$$
Put $$x^2 = 1$$ to make it 0
=> $$f(0) = (1)^2 - 7(1) + k_1 = k_1 - 6$$ --------(i)
Also, $$f(x^3 - 2) = x^6 - 9x^3 +k_2$$
Put $$x^3 = 2$$
=> $$f(0) = (2)^2 - 9(2) + k_2 = k_2 - 14$$ -----------(ii)
Equating (i) & (ii), we get :
=> $$k_1 - 6 = k_2 - 14$$
=> $$k_2 - k_1 = 14 - 6 = 8$$
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