119363
What is an approximate value of \(\sqrt{199}\) corrected to 4 decimal places?
1 14.1608
2 14.0168
3 14.1086
4 14.1068
Explanation:
D Given, \( \sqrt{199}=(199)^{\frac{1}{2}}\) \(\text { By the approximation formula }\) \(\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot \Delta \mathrm{x}+\mathrm{f}(\mathrm{x})\) \(\text { Let } \mathrm{f}(\mathrm{x})=(199)^{1 / 2}=(196+3)^{1 / 2}\) \(\mathrm{x}=196=(14)^2\) \(\Delta \mathrm{x}=3\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{\frac{1}{2}-1}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{-1 / 2}\) \(\mathrm{f}^{\prime}(\mathrm{x}+\Delta \mathrm{x}) =\frac{1}{2} \mathrm{x}^{\frac{-1}{2}} \Delta \mathrm{x}+\mathrm{x}^{1 / 2}\) \(= \frac{1}{2}\left(14^2\right)^{\frac{-1}{2}}+\left(14^2\right)^{\frac{1}{2}}=\frac{3}{28}+14\) \(= 0.1071+14\) \(=14.1071\) \(\approx 14.1068\)
AP EAMCET-22.09.2020
Binomial Theorem and its Simple Application
119364
Given positive integers \(r>1, n>2\) and the coefficient of \((3 \mathrm{r})^{\text {th }}\) and \((r+2)^{\text {nd }}\) the terms in the binomial expansion of \((1+x)^{2 \mathrm{n}}\) are equal, then
1 \(\mathrm{n}=2 \mathrm{r}\)
2 \(n=2 r+1\)
3 \(\mathrm{n}=3 \mathrm{r}\)
4 None of these
Explanation:
A Given, expansion \((1+\mathrm{x})^{2 \mathrm{n}}, \mathrm{r}>1\) and \(\mathrm{n}>2\). Then, \((3 \mathrm{r})^{\text {th }}\) term in the expansion of \((1+\mathrm{x})^{2 \mathrm{n}}-\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1} \mathrm{x}^{3 \mathrm{r}-1}\) And, \((\mathrm{r}+2)^{\mathrm{nd}}\) the term in the expansion of \(\left(1+\mathrm{x}^{2 \mathrm{n}}\right)\) \(\mathrm{T}_{\mathrm{r}+2}=\mathrm{T}_{(\mathrm{r}+1)+1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1} \mathrm{x}^{\mathrm{r}+1}\) Since, given that the binomial coefficients of \((3 \mathrm{r})^{\text {th }}\) and \((\mathrm{r}+2)^{\text {nd }}\) terms are equal. \(\therefore { }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1}\) \(\Rightarrow 3 r-1=r+1\) \(\text { or } 2 n=(3 r-1)+(r+1)\) \({\left[\because{ }^n C_x={ }^n C_y \Rightarrow n=x+y\right]}\) \(\Rightarrow 2 r=2 \text { or } 2 n=4 r\) \(\Rightarrow r=1 \text { or } n=2 r\) \(\text { But, } r>1\) \(\text { So, } n=2 r\) But, \(\quad \mathrm{r}>1\) So, \(\quad \mathrm{n}=2 \mathrm{r}\)
Manipal UGET-2012
Binomial Theorem and its Simple Application
119365
The first negative coefficient in the terms occurring in the expansion of \((1+x)^{\frac{21}{5}}\) is
119363
What is an approximate value of \(\sqrt{199}\) corrected to 4 decimal places?
1 14.1608
2 14.0168
3 14.1086
4 14.1068
Explanation:
D Given, \( \sqrt{199}=(199)^{\frac{1}{2}}\) \(\text { By the approximation formula }\) \(\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot \Delta \mathrm{x}+\mathrm{f}(\mathrm{x})\) \(\text { Let } \mathrm{f}(\mathrm{x})=(199)^{1 / 2}=(196+3)^{1 / 2}\) \(\mathrm{x}=196=(14)^2\) \(\Delta \mathrm{x}=3\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{\frac{1}{2}-1}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{-1 / 2}\) \(\mathrm{f}^{\prime}(\mathrm{x}+\Delta \mathrm{x}) =\frac{1}{2} \mathrm{x}^{\frac{-1}{2}} \Delta \mathrm{x}+\mathrm{x}^{1 / 2}\) \(= \frac{1}{2}\left(14^2\right)^{\frac{-1}{2}}+\left(14^2\right)^{\frac{1}{2}}=\frac{3}{28}+14\) \(= 0.1071+14\) \(=14.1071\) \(\approx 14.1068\)
AP EAMCET-22.09.2020
Binomial Theorem and its Simple Application
119364
Given positive integers \(r>1, n>2\) and the coefficient of \((3 \mathrm{r})^{\text {th }}\) and \((r+2)^{\text {nd }}\) the terms in the binomial expansion of \((1+x)^{2 \mathrm{n}}\) are equal, then
1 \(\mathrm{n}=2 \mathrm{r}\)
2 \(n=2 r+1\)
3 \(\mathrm{n}=3 \mathrm{r}\)
4 None of these
Explanation:
A Given, expansion \((1+\mathrm{x})^{2 \mathrm{n}}, \mathrm{r}>1\) and \(\mathrm{n}>2\). Then, \((3 \mathrm{r})^{\text {th }}\) term in the expansion of \((1+\mathrm{x})^{2 \mathrm{n}}-\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1} \mathrm{x}^{3 \mathrm{r}-1}\) And, \((\mathrm{r}+2)^{\mathrm{nd}}\) the term in the expansion of \(\left(1+\mathrm{x}^{2 \mathrm{n}}\right)\) \(\mathrm{T}_{\mathrm{r}+2}=\mathrm{T}_{(\mathrm{r}+1)+1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1} \mathrm{x}^{\mathrm{r}+1}\) Since, given that the binomial coefficients of \((3 \mathrm{r})^{\text {th }}\) and \((\mathrm{r}+2)^{\text {nd }}\) terms are equal. \(\therefore { }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1}\) \(\Rightarrow 3 r-1=r+1\) \(\text { or } 2 n=(3 r-1)+(r+1)\) \({\left[\because{ }^n C_x={ }^n C_y \Rightarrow n=x+y\right]}\) \(\Rightarrow 2 r=2 \text { or } 2 n=4 r\) \(\Rightarrow r=1 \text { or } n=2 r\) \(\text { But, } r>1\) \(\text { So, } n=2 r\) But, \(\quad \mathrm{r}>1\) So, \(\quad \mathrm{n}=2 \mathrm{r}\)
Manipal UGET-2012
Binomial Theorem and its Simple Application
119365
The first negative coefficient in the terms occurring in the expansion of \((1+x)^{\frac{21}{5}}\) is
119363
What is an approximate value of \(\sqrt{199}\) corrected to 4 decimal places?
1 14.1608
2 14.0168
3 14.1086
4 14.1068
Explanation:
D Given, \( \sqrt{199}=(199)^{\frac{1}{2}}\) \(\text { By the approximation formula }\) \(\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot \Delta \mathrm{x}+\mathrm{f}(\mathrm{x})\) \(\text { Let } \mathrm{f}(\mathrm{x})=(199)^{1 / 2}=(196+3)^{1 / 2}\) \(\mathrm{x}=196=(14)^2\) \(\Delta \mathrm{x}=3\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{\frac{1}{2}-1}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{-1 / 2}\) \(\mathrm{f}^{\prime}(\mathrm{x}+\Delta \mathrm{x}) =\frac{1}{2} \mathrm{x}^{\frac{-1}{2}} \Delta \mathrm{x}+\mathrm{x}^{1 / 2}\) \(= \frac{1}{2}\left(14^2\right)^{\frac{-1}{2}}+\left(14^2\right)^{\frac{1}{2}}=\frac{3}{28}+14\) \(= 0.1071+14\) \(=14.1071\) \(\approx 14.1068\)
AP EAMCET-22.09.2020
Binomial Theorem and its Simple Application
119364
Given positive integers \(r>1, n>2\) and the coefficient of \((3 \mathrm{r})^{\text {th }}\) and \((r+2)^{\text {nd }}\) the terms in the binomial expansion of \((1+x)^{2 \mathrm{n}}\) are equal, then
1 \(\mathrm{n}=2 \mathrm{r}\)
2 \(n=2 r+1\)
3 \(\mathrm{n}=3 \mathrm{r}\)
4 None of these
Explanation:
A Given, expansion \((1+\mathrm{x})^{2 \mathrm{n}}, \mathrm{r}>1\) and \(\mathrm{n}>2\). Then, \((3 \mathrm{r})^{\text {th }}\) term in the expansion of \((1+\mathrm{x})^{2 \mathrm{n}}-\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1} \mathrm{x}^{3 \mathrm{r}-1}\) And, \((\mathrm{r}+2)^{\mathrm{nd}}\) the term in the expansion of \(\left(1+\mathrm{x}^{2 \mathrm{n}}\right)\) \(\mathrm{T}_{\mathrm{r}+2}=\mathrm{T}_{(\mathrm{r}+1)+1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1} \mathrm{x}^{\mathrm{r}+1}\) Since, given that the binomial coefficients of \((3 \mathrm{r})^{\text {th }}\) and \((\mathrm{r}+2)^{\text {nd }}\) terms are equal. \(\therefore { }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1}\) \(\Rightarrow 3 r-1=r+1\) \(\text { or } 2 n=(3 r-1)+(r+1)\) \({\left[\because{ }^n C_x={ }^n C_y \Rightarrow n=x+y\right]}\) \(\Rightarrow 2 r=2 \text { or } 2 n=4 r\) \(\Rightarrow r=1 \text { or } n=2 r\) \(\text { But, } r>1\) \(\text { So, } n=2 r\) But, \(\quad \mathrm{r}>1\) So, \(\quad \mathrm{n}=2 \mathrm{r}\)
Manipal UGET-2012
Binomial Theorem and its Simple Application
119365
The first negative coefficient in the terms occurring in the expansion of \((1+x)^{\frac{21}{5}}\) is
119363
What is an approximate value of \(\sqrt{199}\) corrected to 4 decimal places?
1 14.1608
2 14.0168
3 14.1086
4 14.1068
Explanation:
D Given, \( \sqrt{199}=(199)^{\frac{1}{2}}\) \(\text { By the approximation formula }\) \(\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot \Delta \mathrm{x}+\mathrm{f}(\mathrm{x})\) \(\text { Let } \mathrm{f}(\mathrm{x})=(199)^{1 / 2}=(196+3)^{1 / 2}\) \(\mathrm{x}=196=(14)^2\) \(\Delta \mathrm{x}=3\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{\frac{1}{2}-1}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{-1 / 2}\) \(\mathrm{f}^{\prime}(\mathrm{x}+\Delta \mathrm{x}) =\frac{1}{2} \mathrm{x}^{\frac{-1}{2}} \Delta \mathrm{x}+\mathrm{x}^{1 / 2}\) \(= \frac{1}{2}\left(14^2\right)^{\frac{-1}{2}}+\left(14^2\right)^{\frac{1}{2}}=\frac{3}{28}+14\) \(= 0.1071+14\) \(=14.1071\) \(\approx 14.1068\)
AP EAMCET-22.09.2020
Binomial Theorem and its Simple Application
119364
Given positive integers \(r>1, n>2\) and the coefficient of \((3 \mathrm{r})^{\text {th }}\) and \((r+2)^{\text {nd }}\) the terms in the binomial expansion of \((1+x)^{2 \mathrm{n}}\) are equal, then
1 \(\mathrm{n}=2 \mathrm{r}\)
2 \(n=2 r+1\)
3 \(\mathrm{n}=3 \mathrm{r}\)
4 None of these
Explanation:
A Given, expansion \((1+\mathrm{x})^{2 \mathrm{n}}, \mathrm{r}>1\) and \(\mathrm{n}>2\). Then, \((3 \mathrm{r})^{\text {th }}\) term in the expansion of \((1+\mathrm{x})^{2 \mathrm{n}}-\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1} \mathrm{x}^{3 \mathrm{r}-1}\) And, \((\mathrm{r}+2)^{\mathrm{nd}}\) the term in the expansion of \(\left(1+\mathrm{x}^{2 \mathrm{n}}\right)\) \(\mathrm{T}_{\mathrm{r}+2}=\mathrm{T}_{(\mathrm{r}+1)+1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1} \mathrm{x}^{\mathrm{r}+1}\) Since, given that the binomial coefficients of \((3 \mathrm{r})^{\text {th }}\) and \((\mathrm{r}+2)^{\text {nd }}\) terms are equal. \(\therefore { }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1}\) \(\Rightarrow 3 r-1=r+1\) \(\text { or } 2 n=(3 r-1)+(r+1)\) \({\left[\because{ }^n C_x={ }^n C_y \Rightarrow n=x+y\right]}\) \(\Rightarrow 2 r=2 \text { or } 2 n=4 r\) \(\Rightarrow r=1 \text { or } n=2 r\) \(\text { But, } r>1\) \(\text { So, } n=2 r\) But, \(\quad \mathrm{r}>1\) So, \(\quad \mathrm{n}=2 \mathrm{r}\)
Manipal UGET-2012
Binomial Theorem and its Simple Application
119365
The first negative coefficient in the terms occurring in the expansion of \((1+x)^{\frac{21}{5}}\) is
119363
What is an approximate value of \(\sqrt{199}\) corrected to 4 decimal places?
1 14.1608
2 14.0168
3 14.1086
4 14.1068
Explanation:
D Given, \( \sqrt{199}=(199)^{\frac{1}{2}}\) \(\text { By the approximation formula }\) \(\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot \Delta \mathrm{x}+\mathrm{f}(\mathrm{x})\) \(\text { Let } \mathrm{f}(\mathrm{x})=(199)^{1 / 2}=(196+3)^{1 / 2}\) \(\mathrm{x}=196=(14)^2\) \(\Delta \mathrm{x}=3\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{\frac{1}{2}-1}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{-1 / 2}\) \(\mathrm{f}^{\prime}(\mathrm{x}+\Delta \mathrm{x}) =\frac{1}{2} \mathrm{x}^{\frac{-1}{2}} \Delta \mathrm{x}+\mathrm{x}^{1 / 2}\) \(= \frac{1}{2}\left(14^2\right)^{\frac{-1}{2}}+\left(14^2\right)^{\frac{1}{2}}=\frac{3}{28}+14\) \(= 0.1071+14\) \(=14.1071\) \(\approx 14.1068\)
AP EAMCET-22.09.2020
Binomial Theorem and its Simple Application
119364
Given positive integers \(r>1, n>2\) and the coefficient of \((3 \mathrm{r})^{\text {th }}\) and \((r+2)^{\text {nd }}\) the terms in the binomial expansion of \((1+x)^{2 \mathrm{n}}\) are equal, then
1 \(\mathrm{n}=2 \mathrm{r}\)
2 \(n=2 r+1\)
3 \(\mathrm{n}=3 \mathrm{r}\)
4 None of these
Explanation:
A Given, expansion \((1+\mathrm{x})^{2 \mathrm{n}}, \mathrm{r}>1\) and \(\mathrm{n}>2\). Then, \((3 \mathrm{r})^{\text {th }}\) term in the expansion of \((1+\mathrm{x})^{2 \mathrm{n}}-\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1} \mathrm{x}^{3 \mathrm{r}-1}\) And, \((\mathrm{r}+2)^{\mathrm{nd}}\) the term in the expansion of \(\left(1+\mathrm{x}^{2 \mathrm{n}}\right)\) \(\mathrm{T}_{\mathrm{r}+2}=\mathrm{T}_{(\mathrm{r}+1)+1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1} \mathrm{x}^{\mathrm{r}+1}\) Since, given that the binomial coefficients of \((3 \mathrm{r})^{\text {th }}\) and \((\mathrm{r}+2)^{\text {nd }}\) terms are equal. \(\therefore { }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1}\) \(\Rightarrow 3 r-1=r+1\) \(\text { or } 2 n=(3 r-1)+(r+1)\) \({\left[\because{ }^n C_x={ }^n C_y \Rightarrow n=x+y\right]}\) \(\Rightarrow 2 r=2 \text { or } 2 n=4 r\) \(\Rightarrow r=1 \text { or } n=2 r\) \(\text { But, } r>1\) \(\text { So, } n=2 r\) But, \(\quad \mathrm{r}>1\) So, \(\quad \mathrm{n}=2 \mathrm{r}\)
Manipal UGET-2012
Binomial Theorem and its Simple Application
119365
The first negative coefficient in the terms occurring in the expansion of \((1+x)^{\frac{21}{5}}\) is
