Question

If $\frac{42-19x}{(x^{2}+1)(x-4)}=\frac{Ax+B}{x^{2}+1}+\frac{C}{x-4}$ then $B =$

• A. -11
• B. 11
• C. -2
• D. 2

Hint:

To find constants in partial fractions, substitute values of $x$ that make terms zero or compare coefficients of like powers.

Answer 1

The Correct Option is D

Step 1: Concept
Use the method of partial fractions. Multiply through by the denominator $(x^{2}+1)(x-4)$.

Step 2: Meaning
$42 - 19x = (Ax + B)(x - 4) + C(x^{2} + 1)$. We need to find the constant $B$.

Step 3: Analysis
Put $x=4$: $42 - 19(4) = C(4^{2} + 1) \implies 42 - 76 = 17C \implies -34 = 17C \implies C = -2$. Now compare the constant terms on both sides: $42 = -4B + C$.

Step 4: Conclusion
$42 = -4B - 2 \implies 44 = -4B \implies B = -11$. Final Answer: (D)