A) \[\left( 1,\frac{35}{24} \right)\]
B) \[\left( 1,-\frac{35}{24} \right)\]
C) \[\left( 2,\frac{35}{12} \right)\]
D) \[\left( 2,-\frac{35}{12} \right)\]
Correct Answer: B
Solution :
\[\frac{{{(1-3x)}^{1/2}}+{{(1-x)}^{5/3}}}{2{{\left[ 1-\frac{x}{4} \right]}^{1/2}}}\] \[=\frac{\left[ 1+\frac{1}{2}(-3x)+\frac{1}{2}\left( -\frac{1}{2} \right)\frac{1}{2}{{(-3x)}^{2}}+.... \right]+\left[ 1+\frac{5}{3}(-x)+\frac{5}{3}\frac{2}{3}\frac{1}{2}{{(-x)}^{2}}+.... \right]}{2\left[ 1+\frac{1}{2}\left( -\frac{x}{4} \right)+\frac{1}{2}\left( -\frac{1}{2} \right)\frac{1}{2}{{\left( -\frac{x}{4} \right)}^{2}}+.... \right]}\] \[=\frac{\left[ 1-\frac{19}{12}x+\frac{53}{144}{{x}^{2}}-.... \right]}{\left[ 1-\frac{x}{2}-\frac{1}{8}{{x}^{2}}-.... \right]}=1-\frac{35}{24}x+....\] Neglecting higher powers of \[x\], then \[a+bx=1-\frac{35}{24}x\Rightarrow a=1,b=-\frac{35}{24}\].
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