119331
If \(x\) is numerically so small so that \(x^2\) and higher powers of \(x\) can be neglected, then \(\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32+5 x)^{-1 / 5}\) is approximately equal to
1 \(\frac{32+31 x}{64}\)
2 \(\frac{31+32 x}{64}\)
3 \(\frac{31-32 \mathrm{x}}{64}\)
4 \(\frac{1-2 \mathrm{x}}{64}\)
Explanation:
A Given, \(\left(1+\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5}\) \(=\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32)^{-1 / 5}\left(1+\frac{5 x}{32}\right)^{-1 / 5}\) We know that, \((1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3+\ldots .\) Neglecting high power of \(x\) Then, \((1+x)^n=1+n x\) So, \(=\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right] \cdot\left(2^5\right)^{-1 / 5}\left[1+\left(\frac{-1}{5}\right)\left(\frac{5 x}{32}\right)\right]\) \(=[1+x] \cdot \frac{1}{2}\left[1-\frac{x}{32}\right]\) \(=\frac{1}{2}\left[1-\frac{x}{32}+x\right] \quad\left\{\because x^2 \text { neglected }\right\}\) \(=\frac{1}{2}\left[1+\frac{31 \mathrm{x}}{32}\right]\) \(=\frac{32+31 \mathrm{x}}{64}\)
VITEEE-2009
Binomial Theorem and its Simple Application
119332
In the expansion of \(\frac{a+b x}{e^x}\) the coefficient of \(x^r\) is
A Given, \((1+\mathrm{x})^{2 \mathrm{n}}\) We know that, The greatest binomial coefficient for the following expansion will be the middle term. So, which is \(\mathrm{T}_{\mathrm{r}+1}\) term \(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\) Hence, the greatest coefficient in the expansion of \((1+\) \(\mathrm{x})^{2 \mathrm{n}}\) is \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}\).
UPSEE-2005
Binomial Theorem and its Simple Application
119334
The coefficient of \(x^5\) in the expansion of \((1+\) \(\left.\mathbf{x}^2\right)^5(1+x)^4\) is
119331
If \(x\) is numerically so small so that \(x^2\) and higher powers of \(x\) can be neglected, then \(\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32+5 x)^{-1 / 5}\) is approximately equal to
1 \(\frac{32+31 x}{64}\)
2 \(\frac{31+32 x}{64}\)
3 \(\frac{31-32 \mathrm{x}}{64}\)
4 \(\frac{1-2 \mathrm{x}}{64}\)
Explanation:
A Given, \(\left(1+\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5}\) \(=\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32)^{-1 / 5}\left(1+\frac{5 x}{32}\right)^{-1 / 5}\) We know that, \((1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3+\ldots .\) Neglecting high power of \(x\) Then, \((1+x)^n=1+n x\) So, \(=\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right] \cdot\left(2^5\right)^{-1 / 5}\left[1+\left(\frac{-1}{5}\right)\left(\frac{5 x}{32}\right)\right]\) \(=[1+x] \cdot \frac{1}{2}\left[1-\frac{x}{32}\right]\) \(=\frac{1}{2}\left[1-\frac{x}{32}+x\right] \quad\left\{\because x^2 \text { neglected }\right\}\) \(=\frac{1}{2}\left[1+\frac{31 \mathrm{x}}{32}\right]\) \(=\frac{32+31 \mathrm{x}}{64}\)
VITEEE-2009
Binomial Theorem and its Simple Application
119332
In the expansion of \(\frac{a+b x}{e^x}\) the coefficient of \(x^r\) is
A Given, \((1+\mathrm{x})^{2 \mathrm{n}}\) We know that, The greatest binomial coefficient for the following expansion will be the middle term. So, which is \(\mathrm{T}_{\mathrm{r}+1}\) term \(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\) Hence, the greatest coefficient in the expansion of \((1+\) \(\mathrm{x})^{2 \mathrm{n}}\) is \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}\).
UPSEE-2005
Binomial Theorem and its Simple Application
119334
The coefficient of \(x^5\) in the expansion of \((1+\) \(\left.\mathbf{x}^2\right)^5(1+x)^4\) is
119331
If \(x\) is numerically so small so that \(x^2\) and higher powers of \(x\) can be neglected, then \(\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32+5 x)^{-1 / 5}\) is approximately equal to
1 \(\frac{32+31 x}{64}\)
2 \(\frac{31+32 x}{64}\)
3 \(\frac{31-32 \mathrm{x}}{64}\)
4 \(\frac{1-2 \mathrm{x}}{64}\)
Explanation:
A Given, \(\left(1+\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5}\) \(=\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32)^{-1 / 5}\left(1+\frac{5 x}{32}\right)^{-1 / 5}\) We know that, \((1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3+\ldots .\) Neglecting high power of \(x\) Then, \((1+x)^n=1+n x\) So, \(=\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right] \cdot\left(2^5\right)^{-1 / 5}\left[1+\left(\frac{-1}{5}\right)\left(\frac{5 x}{32}\right)\right]\) \(=[1+x] \cdot \frac{1}{2}\left[1-\frac{x}{32}\right]\) \(=\frac{1}{2}\left[1-\frac{x}{32}+x\right] \quad\left\{\because x^2 \text { neglected }\right\}\) \(=\frac{1}{2}\left[1+\frac{31 \mathrm{x}}{32}\right]\) \(=\frac{32+31 \mathrm{x}}{64}\)
VITEEE-2009
Binomial Theorem and its Simple Application
119332
In the expansion of \(\frac{a+b x}{e^x}\) the coefficient of \(x^r\) is
A Given, \((1+\mathrm{x})^{2 \mathrm{n}}\) We know that, The greatest binomial coefficient for the following expansion will be the middle term. So, which is \(\mathrm{T}_{\mathrm{r}+1}\) term \(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\) Hence, the greatest coefficient in the expansion of \((1+\) \(\mathrm{x})^{2 \mathrm{n}}\) is \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}\).
UPSEE-2005
Binomial Theorem and its Simple Application
119334
The coefficient of \(x^5\) in the expansion of \((1+\) \(\left.\mathbf{x}^2\right)^5(1+x)^4\) is
119331
If \(x\) is numerically so small so that \(x^2\) and higher powers of \(x\) can be neglected, then \(\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32+5 x)^{-1 / 5}\) is approximately equal to
1 \(\frac{32+31 x}{64}\)
2 \(\frac{31+32 x}{64}\)
3 \(\frac{31-32 \mathrm{x}}{64}\)
4 \(\frac{1-2 \mathrm{x}}{64}\)
Explanation:
A Given, \(\left(1+\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5}\) \(=\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32)^{-1 / 5}\left(1+\frac{5 x}{32}\right)^{-1 / 5}\) We know that, \((1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3+\ldots .\) Neglecting high power of \(x\) Then, \((1+x)^n=1+n x\) So, \(=\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right] \cdot\left(2^5\right)^{-1 / 5}\left[1+\left(\frac{-1}{5}\right)\left(\frac{5 x}{32}\right)\right]\) \(=[1+x] \cdot \frac{1}{2}\left[1-\frac{x}{32}\right]\) \(=\frac{1}{2}\left[1-\frac{x}{32}+x\right] \quad\left\{\because x^2 \text { neglected }\right\}\) \(=\frac{1}{2}\left[1+\frac{31 \mathrm{x}}{32}\right]\) \(=\frac{32+31 \mathrm{x}}{64}\)
VITEEE-2009
Binomial Theorem and its Simple Application
119332
In the expansion of \(\frac{a+b x}{e^x}\) the coefficient of \(x^r\) is
A Given, \((1+\mathrm{x})^{2 \mathrm{n}}\) We know that, The greatest binomial coefficient for the following expansion will be the middle term. So, which is \(\mathrm{T}_{\mathrm{r}+1}\) term \(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\) Hence, the greatest coefficient in the expansion of \((1+\) \(\mathrm{x})^{2 \mathrm{n}}\) is \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}\).
UPSEE-2005
Binomial Theorem and its Simple Application
119334
The coefficient of \(x^5\) in the expansion of \((1+\) \(\left.\mathbf{x}^2\right)^5(1+x)^4\) is
119331
If \(x\) is numerically so small so that \(x^2\) and higher powers of \(x\) can be neglected, then \(\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32+5 x)^{-1 / 5}\) is approximately equal to
1 \(\frac{32+31 x}{64}\)
2 \(\frac{31+32 x}{64}\)
3 \(\frac{31-32 \mathrm{x}}{64}\)
4 \(\frac{1-2 \mathrm{x}}{64}\)
Explanation:
A Given, \(\left(1+\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5}\) \(=\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32)^{-1 / 5}\left(1+\frac{5 x}{32}\right)^{-1 / 5}\) We know that, \((1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3+\ldots .\) Neglecting high power of \(x\) Then, \((1+x)^n=1+n x\) So, \(=\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right] \cdot\left(2^5\right)^{-1 / 5}\left[1+\left(\frac{-1}{5}\right)\left(\frac{5 x}{32}\right)\right]\) \(=[1+x] \cdot \frac{1}{2}\left[1-\frac{x}{32}\right]\) \(=\frac{1}{2}\left[1-\frac{x}{32}+x\right] \quad\left\{\because x^2 \text { neglected }\right\}\) \(=\frac{1}{2}\left[1+\frac{31 \mathrm{x}}{32}\right]\) \(=\frac{32+31 \mathrm{x}}{64}\)
VITEEE-2009
Binomial Theorem and its Simple Application
119332
In the expansion of \(\frac{a+b x}{e^x}\) the coefficient of \(x^r\) is
A Given, \((1+\mathrm{x})^{2 \mathrm{n}}\) We know that, The greatest binomial coefficient for the following expansion will be the middle term. So, which is \(\mathrm{T}_{\mathrm{r}+1}\) term \(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\) Hence, the greatest coefficient in the expansion of \((1+\) \(\mathrm{x})^{2 \mathrm{n}}\) is \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}\).
UPSEE-2005
Binomial Theorem and its Simple Application
119334
The coefficient of \(x^5\) in the expansion of \((1+\) \(\left.\mathbf{x}^2\right)^5(1+x)^4\) is
