# Engineering Mathematics Questions and Answers – Taylor Mclaurin Series – 3

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Taylor Mclaurin Series – 3”.

1. Expansion of function f(x) is?
a) f(0) + x1! f (0) + x22! f (0)…….+xnn! fn (0)
b) 1 + x1! f (0) + x22! f (0)…….+xnn! fn (0)
c) f(0) – x1! f (0) + x22! f (0)…….+(-1)^n xnn! fn (0)
d) f(1) + x1! f (1) + x22! f (1)…….+xnn! fn (1)

Explanation: By Maclaurin’s series, f(0) + x1! f (0) + x22! f (0)…….+xnn! fn (0)

2. The necessary condition for the maclaurin expansion to be true for function f(x) is __________
a) f(x) should be continuous
b) f(x) should be differentiable
c) f(x) should exists at every point
d) f(x) should be continuous and differentiable

Explanation: By Maclaurin’s series, f(0) + x1! f (0) + x22! f (0)…….+xnn! fn (0)

Where, f(x) should be continuous and differentiable upto nth derivative.

3. The expansion of f(a+h) is ______
a) $$f(a)+\frac{h}{1!} f'(a)+\frac{h^2}{2!} f”(a)…….+\frac{h^n}{n!} f^n (a)$$
b) $$f(a)+\frac{h}{1!} f'(a)+\frac{h^2}{2!} f”(a)…….$$
c) $$hf(a)+\frac{h^2}{1!} f'(a)+\frac{h^3}{2!} f”(a)…….+\frac{h^n}{n!} f^n (a)$$
d) $$hf(a)+\frac{h^2}{1!} f'(a)+\frac{h^3}{2!} f”(a)………$$

Explanation: By taylor expansion,
f(a+h) = f(a) + h1! f'(a) + h22! f (a)….

4. The expansion of eSin(x) is?
a) 1 + x + x22 + x48 +….
b) 1 + x + x22x48 +….
c) 1 + x – x22 + x48 +….
d) 1 + x + x36x510 +….

Explanation: Now f(x) = eSin(x), f(0) = 1
Hence, f (x)=f(x)Cos(x), f (0) = 1
f (x)=f (x)Cos(x) – f(x)Sin(x),f (0)=1
f”’ (x)=f (x)Cos(x) – 2f (x)Sin(x) – f(x) cos⁡(x),f”’ (x) = 0
f”” (x)=f”’ (x)Cos(x) – 3f (x)Sin(x) – 3f (x) cos⁡(x) + f(x) sin⁡(x), f”” (x) = -3
Hence,
f(x) = eSin(x) = 1 + x + x22x48 +…. (By mclaurin’ sexpansion)

5. Expansion of y = Sin-1(x) is?
a) $$x+\frac{x^3}{6}+\frac{3}{40} x^5+\frac{5}{112} x^7+…..$$
b) $$x-\frac{x^3}{6}+\frac{3}{40} x^5-\frac{5}{112} x^7+…..$$
c) $$\frac{x^3}{6}-\frac{3}{40} x^5+\frac{5}{112} x^7…..$$
d) $$x+\frac{x^2}{6}+\frac{3}{40} x^2+\frac{5}{112} x^2+…..$$

Explanation: Given, y = Sin-1(x), hence at x = 0, y = 0
Now, differentiating it, we get
$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}=(1-x^2)^{-1/2}$$
On expanding the R.H.S. by Binomial Theorem we get,
$$\frac{dy}{dx}=1+\frac{x^2}{2}+\frac{3}{8} x^4+\frac{5}{16} x^6+…$$
On integrating we get,
$$y=x+\frac{x^3}{6}+\frac{3}{40} x^5+\frac{5}{112} x^7+….+C$$
By putting x=0 hence we get,
y=c=0
Hence,
$$y=x+\frac{x^3}{6}+\frac{3}{40} x^5+\frac{5}{112} x^7+….$$

6. Find the expansion of f(x) = ln⁡(1+ex)?
a) $$ln(2)+x/2+x^2/8-x^4/192+….$$
b) $$ln⁡(2)+x/2+x^2/8+x^4/192+….$$
c) $$ln⁡(2)+x/2+x^3/8-x^5/192+….$$
d) $$ln⁡(2)+x/2+x^3/8+x^5/192+….$$

Explanation: Given, f(x) = ln⁡(1+ex), f(0) = ln⁡(2)
Differentiating it we get
$$f{‘}(x)=\frac{e^{x}}{1+e^x}=1-1/(1+e^x), f{‘}(0)=1/2$$
Again differentiating we get
$$f”(x)=e^x/(1+e^x)^2 =(f'(x))/(1+e^x),f”(0)=1/4$$
$$f”'(0)=((1+e^x) f”(x)-f'(x) e^x)/(1+e^x)^2 =(f”(x))/(1+e^x)-f'(x) f”(x)$$, hence f”'(0)=0
$$f””(0)=(1(1+e^x) f”'(x)-f”(x) e^x)/(1+e^x)^2 -(f”(x))^2-f'(x) f”'(x)$$, hence f””(0)=1/8
Hence, by mclaurin’s series,
$$f(x)=ln⁡(1+e^x)=ln⁡(2)+x/2+x^2/8-x^4/192+…$$

7. Find the expansion of exSin(x)?
a) $$e^{xSin(x)}=1+x^2-x^4/3+x^6/120-…$$
b) $$e^{xSin(x)}=1+x^2+x^4/3+x^6/120+…$$
c) $$e^{xSin(x)}=x+x^3/3+x^5/120+..$$
d) $$e^{xSin(x)}=x+x^3/3-x^5/120+…$$

Explanation: Given, f(x) = exSin(x), f(0) = 1
Now, the expansion of xSin(x) is $$x^2-x^3/3!+x^6/5!+…$$
Hence, $$e^xSin(x)=e^y=1+y+y^2/2!+y^3/3!+…$$
Hence,
$$e^{xSin(x)}=1+(x^2-\frac{x^4}{3!}+\frac{x^6}{6!}+..)+\frac{(x^2-\frac{x^4}{3!}+x^6/6!+..)^2}{2!}+\frac{(x^2-\frac{x^4}{3!}+x^6/6!+…)^3}{6}+..$$
$$e^{(xSin(x))}=1+x^2-\frac{x^4}{3!}+\frac{x^6}{5!}+\frac{x^4}{2}-\frac{x^6}{6}+\frac{x^6}{6}+….$$ (we neglect all other other terms by considering the options given)
Hence, $$e^{xSin(x)}=1+x^2+\frac{x^4}{3}+\frac{x^6}{120}+…$$

8. Given f(x)= ln⁡(cos⁡(x)),calculate the value of ln⁡(cos⁡(π2)).
a) -1.741
b) 1.741
c) 1.563
d) -1.563

Explanation: Given f(x) = ln⁡(Cos(x)), f(0) = 0
Differentiating it f'(x) = – tan⁡(x), f'(0) = 0
Again $$f^{”}(x)=-sec^2⁡(x),f^{”}(0)=-1$$
And $$f^{”’}(x)=-2 sec⁡(x) \,sec⁡(x) \,tan⁡(x)=-2f^{”}(x)f'(x)$$, hence f”(0)=0
f””(x)=-2(f”'(x) f'(x)+(f”(x))2), hence f””(0)=-2
Now, by mclaurins’s series
f(x)=ln⁡(Cos(x))=$$0+0-x^2/2!+0-x^4/12+..$$
Therefore, f(x)=$$-x^2/2!-x^4/12+…$$
Hence,
ln⁡(cos⁡(π/2))=-1.741

9. Find the expansion of cos(xsin(t)).
a) $$\sum_{n=1}^∞ (\frac{x^n [Cos(nt)]}{n!})$$
b) $$\sum_{n=0}^∞ (\frac{x^n [Cos(nt)]}{n!})$$
c) $$\sum_{n=1}^∞ (\frac{x^n [Sin(nt)]}{n!})$$
d) $$\sum_{n=0}^∞ (\frac{x^n [Sin(nt)]}{n!})$$

Explanation:
Given, f(x)=Cos(xSin(t))=real part of (eixSin(t))
=real part of(exCos(t) eixSin(t))
=real part of(ex[Cos(t)+iSin(t)])
=Real part of $$\sum_{n=0}^∞ \frac{x^n [Cos(nt)+iSin(nt)]}{n!}$$
=$$\sum_{n=0}^∞ \frac{x^n [Cos(nt)]}{n!}$$

10. Find the expansion of Sin(lSin-1 (x)).
a) $$lx-\frac{l(1-l^2)}{3!}x^3+\frac{l(1-l^2)(9-l^2)}{5!} x^5-…$$
b) $$lx+\frac{l(1-l^2)}{3!} x^3+\frac{l(1-l^2)(9-l^2)}{5!} x^5+…$$
c) $$1-lx^2+\frac{l(1-l^2)}{3!} x^4-\frac{l(1-l^2)(9-l^2)}{5!} x^6+…$$
d) $$1+lx^2+\frac{l(1-l^2)}{3!} x^4+\frac{l(1-l^2)(9-l^2)}{5!} x^6+…$$

Explanation: Given, y = f(x) = Sin(lSin-1(x))
Now, differentiating,
$$\frac{dy}{dx}=Cos(lSin^{-1} (x))(\frac{l}{\sqrt{1-x^2}})$$
Hence,
$$(1-x^2)(\frac{dy}{dx})^2=lCos(lSin^{-1}(x))^2$$
$$(1-x^2)(y_1)^2=l^2 Cos(lSin^{-1}(x))^2=l^2 [1-y^2]$$
Hence, differentiating again we get,
$$(1-x^2)2y_1 y_2-2xy_1^2=-2l^2 yy_1$$
$$(1-x^2) y_2-xy_1+l^2 y=0$$
Hence by Leibniz theorem,
$$(1-x^2) y_{(n+2)}-(2n+1)xy_{(n+1)}-(n^2-m^2) y_n=0$$
Therefore by putting x=0, we get,
$$y_{(n+2)}(0)=(n^2-l^2)y_n (0)$$
putting ,n=1,2,3,4,…..
$$y_3 (0)=(1-l^2) y_1 (0)=l(1-l^2)$$
$$y_4 (0)=(4-l^2) y_2 (0)=0$$
$$y_5 (0)=(9-l^2)3(0)=l(1-l^2)(9-l^2)$$
Hence,$$y=Sin(lSin^{-1}(x))=lx+\frac{l(1-l^2)}{3!}x^3+\frac{l(1-l^2)(9-l^2)}{5!} x^5+…$$.

11. Expand (1 + x)1x, gives ___________
a) e[1 + x2 + 11x224 -…..]
b) e[1 – x2 + 11x224 -…..]
c) e[x211x224 -…..]
d) e[x2 + 11x224 -…..]

Explanation: Given, y = (1 + x)1x
Hence,
ln⁡(y)=$$\frac{ln⁡(1+x)}{x}$$
Hence, ln⁡(y)=$$\frac{1}{x} [x-\frac{x^2}{2}+\frac{x^3}{3}-……]=1-\frac{x}{2}+\frac{x^2}{3}$$-……
Hence, $$y=e^{1-\frac{x}{2}+\frac{x^2}{3}-……}=e^{1+z}$$, where, $$z=\frac{-x}{2}+\frac{x^2}{3}-……$$
Hence, $$y=e.e^z=e(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+…)$$
y=$$e[1+(\frac{-x}{2}+\frac{x^2}{3}-……)+\frac{1}{2!}(\frac{-x}{2}+\frac{x^2}{3}-…)^2+…]$$
y=$$e[1-x/2+x^2/e+x^2/8+…]$$
y = e[1 – x2 + 11x224 -…..].

12. Find the solution of differential equation, dydx = xy + x2, if y = 1 at x = 0.
a) $$1-\frac{x^2}{2!}+\frac{3x^4}{4!}-\frac{15x^6}{6!}+…$$
b) $$\frac{x}{1!}+\frac{3x^3}{4!}+\frac{15x^5}{6!}+…$$
c) $$\frac{x}{1!}-\frac{3x^3}{4!}+\frac{15x^5}{6!}-…$$
d) $$1+\frac{x^2}{2!}+\frac{3x^4}{4!}+\frac{15x^6}{6!}+..$$

Explanation: Given dydx = xy + x2
hence, dydy (x=0) = 0
and, d2ydx2= xy1 + y + 2x
hence, y2 = xy1 + y + 2x
hence, d2ydx2(x=0)=1
Differentiating it n times we get,
$$y_{n+2}=xy_{n+1}+ny_n+y_n=xy_{n+1}+(n+1)y_n$$
Putting x=0 we get,
$$y_{n+2} (0)=(n+1)y_n (0)$$
Now putting the values of n as 1, 2, 3, 4, 5 we get,
$$y_3 (0)=0, \,y_4 (0)=3, \,y_5 (0)=0, \,y_6(0)=15……$$ and so on
By mclaurin’s series,
$$y=1+\frac{x^2}{2!}+\frac{3x^4}{4!}+\frac{15x^6}{6!}+…$$

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