1. Statement A (Assertion): If one zero of the polynomial p(x) = (k² + 4)x² + 9x + 4k is the reciprocal of the other zero then k = 2. Statement B (Reason): If (x - a) is a factor of the polynomial p(x), then a is a zero of p(x). a. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A) b. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A) c. Assertion (A) is true but Reason (R) is false. d. Assertion (A) is false but Reason (R) is true.
Question
- Statement A (Assertion): If one zero of the polynomial p(x) = (k² + 4)x² + 9x + 4k is the reciprocal of the other zero then k = 2. Statement B (Reason): If (x - a) is a factor of the polynomial p(x), then a is a zero of p(x). a. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A) b. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A) c. Assertion (A) is true but Reason (R) is false. d. Assertion (A) is false but Reason (R) is true.
Solution
The answer is (b). Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Here's why:
Assertion (A): If one zero of the polynomial p(x) = (k² + 4)x² + 9x + 4k is the reciprocal of the other zero then k = 2.
This is true. If one zero is the reciprocal of the other, then the product of the zeros (which is c/a, where c is the constant term and a is the coefficient of x²) is 1. In this case, the product of the zeros is 4k/(k² + 4). Setting this equal to 1 and solving for k gives k = 2.
Reason (R): If (x - a) is a factor of the polynomial p(x), then a is a zero of p(x).
This is also true. By the Factor Theorem, a polynomial p(x) has a factor (x - a) if and only if p(a) = 0, i.e., a is a zero of p(x).
However, Reason (R) is not the correct explanation for Assertion (A). The reason Assertion (A) is true has to do with the relationship between the zeros of a quadratic polynomial and its coefficients, not the Factor Theorem.
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