AP EAMCET · Maths · Quadratic Equation
Let \(Q(x)\) be a polynomial of degree \(n\). If \(Q(\mathrm{l})=1\) and \(\frac{Q(2 x)}{Q(x+1)}+\frac{56}{x+7}-8=0\), then the value of \({ }^n C_0+{ }^n C_1+\ldots+{ }^n C_n\) is equal to
- A 32
- B 64
- C 8
- D 16
Answer & Solution
Correct Answer
(C) 8
Step-by-step Solution
Detailed explanation
\(Q(x)\) be a polynomial of degree \(n\). \[ Q(1)=1 \text { and } \frac{Q(2 x)}{Q(x+1)}+\frac{56}{x+7}-8=0...(i) \] Put \(x=0\) in Eq. (i), we get \[ \begin{aligned} \frac{Q(0)}{Q(1)}+\frac{56}{7}-8 & =0 \\ \Rightarrow \quad \frac{Q(0)}{1}+8-8 & =0 \\ Q(0) & =0 \end{aligned} \]…
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