Tag: applications of differential calculus

Quizzes (2) Questions (20)

Questions Related to applications of differential calculus

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

If $\sin { x } +\sin ^{ 2 }{ x } =1$, then the value of $\cos ^{ 12 }{ x } +3\cos ^{ 10 }{ x } +3\cos ^{ 8 }{ x } +\cos ^{ 6 }{ x } -2$ is equal to

  1. $0$
  2. $-1$
  3. $-2$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$sinx+sin^2x=1$

or, $sinx=1−sin^2x=cos^2x $

or, $sin^2x=cos^4x$

$cos^{12}x+3cos^{10}x+3cos^8x+cos^6x-2$

$=cos^6x(cos^6x+3cos^4x+3cos^2x+1)-2$

$=cos^6x((cos^2x)^3+3(cos^2x)^2.1+3.cos^2x.1^2+1^3)-2$

$=(cos^2x)^3(cos^2x+1)^3-2$

$=(cos^4x+cos^2x)^3-2$

$=(sin^2x+cos^2x)^3-2$

$=1^3-2$

$=1-2=-1$
Multiple choice taylor's and maclaurin's series applications of differential calculus maths

In the Taylor series expansion of $\exp \left( x \right) + \sin \left( x \right)$ about the point $x = \pi $, the coefficient of ${\left( {x = \pi } \right)^2}$ is

  1. $\exp \left( \pi \right)$
  2. $0.5\exp \left( \pi \right)$
  3. $\exp \left( \pi \right) + 1$
  4. $\exp \left( \pi \right) - 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice taylor's and maclaurin's series applications of differential calculus maths

If the sum of the series $\dfrac{3}{1!}+\dfrac{5}{2!}+\dfrac{7}{3!}+\dfrac{9}{4!}+...\infty=Ae+B$
Find the value of $A+B$

  1. $1$
  2. $7$
  3. $0$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Its general term is $\dfrac{2n+1}{n!}$

So we have to calculate $\sum^{n=\infty} _{n=0}\dfrac{2n+1}{n!}=2\sum^{k=\infty} _{k=0}\dfrac{1}{k!}+\sum^{n=\infty} _{n=0}\dfrac{1}{n!}-2=3e-2$ (using  taylor's expansion for $e^x$)

So A+B=1

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

The value of $\mathop {\lim }\limits _{x \to 0} \frac{{\sin x + \log \left( {1 - x} \right)}}{{{x^2}}}$ equals

  1. $0$
  2. $\frac{1}{2}$
  3. $\frac{{ - 1}}{2}$
  4. $-1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

${0 \over 0}form$

$ = \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}$

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

$\ln{(1+x)}< x-\cfrac{{x}^{2}}{2}+\cfrac{{x}^{3}}{3}$ for $x> 0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Maclaurin series for ln(1+x) is x - x^2/2 + x^3/3 - x^4/4 + ... for -1 < x <= 1. For x > 0, the next term is -x^4/4, which is negative. Therefore, ln(1+x) is strictly less than the sum of the first three terms.

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

If $f(x) = (2011 + x)^{n}$, where $x$ is a real variable and $n$ is a positive integer, then the value of
$f(0) + f'(0) + \dfrac {f"(0)}{2!} + .... + \dfrac {f^{(n - 1)}(0)}{(n - 1)!}$ is.

  1. $(2011)^{n}$
  2. $(2012)^{n}$
  3. $(2012)^{n} - 1$
  4. $n(2011)^{n}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

$ ={ \left( 2012 \right)  }^{ n }-1$

Hence, option $C$ is correct

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

The fourth term in Taylor series of $\log\ x$ centered at $a=1$ is?

  1. $\dfrac{(x-1)^3}{3}$
  2. $\dfrac{(x-1)^2}{2}$
  3. $-\dfrac{(x-1)^4}{4}$
  4. $(x-1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

$ = \left(x-1\right)-\dfrac{1}{2}\left(x-1\right)^2 + \dfrac{1}{3} \left(x-1\right)^3-\dfrac{1}{4} \left(x-1\right)^4 + \cdots$ $ f(x) $
$= \displaystyle\sum\limits _{n=1}^{\infty} \left[\frac{\left(-1\right)^{n+1}}{n}\left(x-1\right) ^n\right] $ 

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

If $\dfrac{1}{(1-2x)(1+3x)}$ is to be expanded as a power series of $x$, then

  1. $|x|<1/2$
  2. $|x|<1/6$
  3. $\dfrac{1}{3}$
  4. $|x|<1/3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

The coefficient of the fourth term in Taylor series of $x^4 + x ^2-2$ centered at $a=1$.

  1. $4$
  2. $1$
  3. $3$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 $1$.

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

The coefficient of the third term in the Taylor series of  $(x-1)e^x$ is?

  1. $\dfrac{1}{3}$
  2. $3$
  3. $2$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

$e^x=1+x^1+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}...$

$(x-1)e^x=(x-1)(1+x^1+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}...)$

$=-1+\dfrac{1}{2}x^2+\dfrac{1}{3}x^3+\dfrac{1}{8}x^4+\dfrac{1}{30}x^5+....$

Hence, the cofficient of the third term in a taylor series of $(x-1)e^x $ is $\dfrac{1}{3}$