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) + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)
b) 1 + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)
c) f(0) – x⁄1! f‘ (0) + x2⁄2! f” (0)…….+(-1)^n xn⁄n! fn (0)
d) f(1) + x⁄1! f‘ (1) + x2⁄2! f” (1)…….+xn⁄n! fn (1)
View Answer
Explanation: By Maclaurin’s series, f(0) + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! 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
View Answer
Explanation: By Maclaurin’s series, f(0) + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! 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) \(hf(a)+\frac{h^2}{1!} f'(a)+\frac{h^3}{2!} f”(a)………\)
View Answer
Explanation: By taylor expansion,
f(a+h) = f(a) + h⁄1! f'(a) + h2⁄2! f” (a)….
4. The expansion of eSin(x) is?
a) 1 + x + x2⁄2 + x4⁄8 +….
b) 1 + x + x2⁄2 – x4⁄8 +….
c) 1 + x – x2⁄2 + x4⁄8 +….
d) 1 + x + x3⁄6 – x5⁄10 +….
View Answer
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 + x2⁄2 – x4⁄8 +…. (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) \(x+\frac{x^2}{6}+\frac{3}{40} x^2+\frac{5}{112} x^2+…..\)
View Answer
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) \(ln(2)+x/2+x^3/8+x^5/192+….\)
View Answer
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) \(e^{xSin(x)}=x+x^3/3-x^5/120+…\)
View Answer
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(π⁄2)).
a) -1.741
b) 1.741
c) 1.563
d) -1.563
View Answer
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) \(\sum_{n=0}^∞ (\frac{x^n [Sin(nt)]}{n!})\)
View Answer
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) \(1+lx^2+\frac{l(1-l^2)}{3!} x^4+\frac{l(1-l^2)(9-l^2)}{5!} x^6+…\)
View Answer
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)1⁄x, gives ___________
a) e[1 + x⁄2 + 11x2⁄24 -…..]
b) e[1 – x⁄2 + 11x2⁄24 -…..]
c) e[x⁄2 – 11x2⁄24 -…..]
d) e[x⁄2 + 11x2⁄24 -…..]
View Answer
Explanation: Given, y = (1 + x)1⁄x
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 – x⁄2 + 11x2⁄24 -…..].
12. Find the solution of differential equation, dy⁄dx = 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) \(1+\frac{x^2}{2!}+\frac{3x^4}{4!}+\frac{15x^6}{6!}+..\)
View Answer
Explanation: Given dy⁄dx = xy + x2
hence, dy⁄dy (x=0) = 0
and, d2y⁄dx2= xy1 + y + 2x
hence, y2 = xy1 + y + 2x
hence, d2y⁄dx2(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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