Tag: maths

$\mathop {\lim }\limits _{x \to 0} {{\sin x + \log \left( {1 - x} \right)} \over {{x^2}}}$

${0 \over 0}form$

$ = \mathop {\lim }\limits _{x \to 0} {{\cos x - {1 \over {1 - x}}} \over {2x}}$

(L-hospital)

$ = \mathop {\lim }\limits _{x \to 0} {{\left( {1 - x} \right)\cos x - 1} \over {2x\left( {1 - x} \right)}}$

$ = \mathop {\lim }\limits _{x \to 0} {{\cos x - x\cos x - 1} \over {2x\left( {1 - x} \right)}}$

${0 \over 0}form$

$ = \mathop {\lim }\limits _{x \to 0} {{ - \sin x + x\sin x - \cos x} \over { - 2x + 2\left( {1 - x} \right)}}$

$ = {{0 + 0 - 1} \over {0 + 2}}$

$ =  - {1 \over 2}$

$f\left( x \right) ={ \left( 2011+x \right)  }^{ n }\\ f\left( 0 \right) +f^{ ' }\left( 0 \right) +\dfrac { f^{ '' }\left( 0 \right)  }{ 2! } +\dfrac { f^{ ''' }\left( 0 \right)  }{ 3! } +.........\dfrac { f^{ n-1 }\left( 0 \right)  }{ (n-1)! } \\ =2011+\dfrac { n{ (2011) }^{ n-1 } }{ 2! } +\dfrac { n(n-1){ (2011) }^{ n-2 } }{ 3! } +......\dfrac { (n(n-1)....2!)(2011) }{ (n-1)! } \\ ={ C } _{ 0 }^{ n }(2011)+{ C } _{ 1 }^{ n }{ (2011) }^{ n-1 }+......{ C } _{ n-1 }^{ n }(2011)+{ C } _{ n }^{ n }{ (2011) }^{ 0 }-1$

$ ={ \left( 2011+1 \right)  }^{ n }-1$

$ ={ \left( 2012 \right)  }^{ n }-1$

Hence, option $C$ is correct

The taylor series expansion of $\log x$ is $f(x) = \ln(x)$

$\begin{array}{l} \frac { 1 }{ { \left( { 1-2x } \right) \left( { 1+3x } \right)  } }  \ \Rightarrow { \left( { 1-2x } \right) ^{ -1 } }{ \left( { 1+3x } \right) ^{ -1 } }=\frac { 1 }{ 2 } \cdot \frac { 1 }{ 3 } { \left( { \frac { 1 }{ 2 } -x } \right) ^{ -1 } }{ \left( { \frac { 1 }{ 3 } +x } \right) ^{ -1 } } \ \Rightarrow { { by } }\, \, { { using } }\, \, { { binomial } }\, \, { { of } }\, \, { { rational } }\, \, { { index } } \ \Rightarrow { \left( { 1+x } \right) ^{ n } } \ \left| x \right| <1 \ Hence, \ \Rightarrow \left| x \right| <\frac { 1 }{ 2 }  \ \Rightarrow \left| x \right| <\frac { 1 }{ 3 }  \ \therefore \left| x \right| <\frac { 1 }{ 6 }  \end{array}$

The Taylor series is given by $f(x)=\sum _{k=0}^{n}\dfrac{f^{(k)}(a)}{k!}(x-a)^k$

We have $f(x)=x^4+x^2-2, a=1$

Since we have to find the coefficient of the fourth term, let us take $n=5$.

$\therefore f(x)\approx\sum _{k=0}^{5}\dfrac{f^{(k)}(1)}{k!}(x-1)^k$

$f^{(0)}(x)=x^4+x^2-2, \Rightarrow f^{(0)}(1)=1+1-2=0$

$f^{(1)}(x)=4x^3+2x, \Rightarrow f^{(1)}(1)=4+2=6$

$f^{(2)}(x)=12x^2+2, \Rightarrow f^{(2)}(1)=12+2=14$

$f^{(3)}(x)=24x, \Rightarrow f^{(3)}(1)=24$

$f^{(4)}(x)=24, \Rightarrow f^{(4)}(1)=24$

$f^{(5)}(x)=0, \Rightarrow f^{(5)}(1)=0$

$\therefore f(x) \approx 0+\dfrac{6}{1!}(x-1)+\dfrac{14}{2!}(x-1)^2+\dfrac{24}{3!}(x-1)^3+\dfrac{24}{4!}(x-1)^4+0$

$\Rightarrow f(x)\approx 6(x-1)+7(x-1)^2+4(x-1)^3+(x-1)^4$

Thus the coefficient of fourth term is $1$.

We know that taylor series of $e^x$ is given by

Taylor Series expansion of $\log \ x = (x-1) - \dfrac{1}{2} (x-1)^{2} + \dfrac{1}{3} (x-1)^{3} -   ...$

The taylor series is given by $\sum _{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}(x-a)^k$

The Maclaurin series is given by $f(x)=\sum _{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}x^k$ where $a=0$

We have $f(x)=xe^{-x}$

Since we have to find the third term, let us take $n=5$.

$\therefore f(x)\approx\sum _{k=0}^{5}\dfrac{f^{(k)}(0)}{k!}x^k$

$f^{(0)}(x)=xe^{-x}, \Rightarrow f^{(0)}(0)=0$

$f^{(1)}(x)=(-x+1)e^{-x}, \Rightarrow f^{(1)}(0)=1$

$f^{(2)}(x)=(x-2)e^{-x}, \Rightarrow f^{(2)}(0)=-2$

$f^{(3)}(x)=(-x+3)e^{-x}, \Rightarrow f^{(3)}(0)=3$

$f^{(4)}(x)=(x-4)e^{-x}, \Rightarrow f^{(4)}(0)=-4$

$f^{(5)}(x)=(-x+5)e^{-x}, \Rightarrow f^{(5)}(0)=5$

$\therefore f(x) \approx \dfrac{0}{0!}x^0+\dfrac{1}{1!}x^1+\dfrac{-2}{2!}x^2+\dfrac{3}{3!}x^3+\dfrac{-4}{4!}x^4+\dfrac{5}{5!}x^5$

$\Rightarrow f(x)\approx x-x^2+\dfrac{1}{2}x^3-\dfrac{1}{6}x^4+\dfrac{1}{24}x^5$

Thus the third term is $\dfrac{1}{2}x^3$.