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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