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Question: If \(2B + C = - 1\)then the number of values of x which are integral multiples of \(A = 1\)is....

If 2B+C=−12B + C = - 1then the number of values of x which are integral multiples of A=1A = 1is.

A

4

B

12

C

3

D

None of these

Answer

4

Explanation

Solution

x=−1x = - 1

2(−1)+3=a(−1−3)⇒1=−4a⇒a=−142( - 1) + 3 = a( - 1 - 3) \Rightarrow 1 = - 4a \Rightarrow a = \frac{- 1}{4}, x=32(3)+3=b(3+1)9=4bx = 32(3) + 3 = b(3 + 1)9 = 4b

b=94b = \frac{9}{4} Integral multiple of a+b=−14+94=2a + b = \frac{- 1}{4} + \frac{9}{4} = 2 will be 3x+4(x+1)2(x−1)=A(x−1)+B(x+1)+C(x+1)2\frac{3x + 4}{(x + 1)^{2}(x - 1)} = \frac{A}{(x - 1)} + \frac{B}{(x + 1)} + \frac{C}{(x + 1)^{2}}

Number of required values = 4.