Engineering Mathematics Multiple Choice Questions 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)
Answer: a
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
Answer: d
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)………)
Answer: a
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 +….
Answer: b
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+…..)
Answer: a
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+….)
Answer: a
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+…)
Answer: b
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
Answer: a
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!}))
Answer: b
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+…)
Answer: b
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 -…..]
Answer: b
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!}+..)
Answer: d
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!}+…)