250+ TOP MCQs on Taylor Mclaurin Series and Answers

Engineering Mathematics Quiz focuses on “Taylor Mclaurin Series – 4”.

1. The expansion of f(x), about x = a 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)…….

2. Find the expansion of e^x in terms of x + m, m > 0.
a) (e^m [1+(x+m)+frac{(x+m)^2}{2!}+frac{(x+m)^3}{3!}+….])
b) (e^{-m} [1+(x-m)+frac{(x-m)^2}{2!}+frac{(x-m)^3}{3!}+….])
c) (e^m [1+(x-m)+frac{(x-m)^2}{2!}+frac{(x-m)^3}{3!}+….])
d) (e^{-m} [1+(x+m)+frac{(x+m)^2}{2!}+frac{(x+m)^3}{3!}+….])
Answer: d
Explanation: Let, h = x + m = > f(x) = f(h-m) = e^(h-m)
By taylor theorem, putting a = -m, we get,
f(a+h) = (f(a)+frac{h}{1!} f’ (a)+frac{h^2}{2!} f” (a)…)
f(h-m) = (f(-m)+h/1! f'(-m)+frac{h^2}{2!} f” (-m)….(1))
now, f(-m) = f’(-m) = f’’(-m)=e^-m
hence,
f(x)=e^x=f(h-m)=(e^{-m} [1+(x+m)+frac{(x+m)^2}{2!}+frac{(x+m)^3}{3!}+….])

3. Expand ln(x) in the power of (x-m).
a) ln⁡(m)+(frac{h}{m}-frac{1}{2!} (h/m)^2+frac{2}{3!} (h/m)^3-……)
b) ln⁡(m)-(frac{h}{m}-frac{1}{2!} (h/m)^2-frac{2}{3!} (h/m)^3-……)
c) ln⁡(m)-(frac{1}{2!} (h/m)^2+frac{2}{3!} (h/m)^4-……)
d) ln⁡(m)+(frac{h}{m}+frac{2}{3!} (h/m)^3-……)
Answer: a
Explanation: where, h = x-m
Let, h = x – m => f(x) = f(h+m) = e^(h+m)
By taylor theorem, putting a = m , we get,
f(a+h) = (f(a)+frac{h}{1!} f’ (a)+frac{h^2}{2! }f” (a)…)
f(h-m) = (f(m)+frac{h}{1!} f’ (m)+frac{h^2}{2!} f” (m))…….(1)
now,f(m) = ln(m), f’(m)=1/m, f” (m)=-1/m², f”’ (m)=2/m³,……
hence,
f(x)=ln(x)=f(h+m)=(ln⁡(m)+frac{h}{m}-frac{h^2}{2!}frac{1}{m^2}+frac{h^3}{3!} frac{2}{m^3}-……)
f(x)=ln(x)=ln⁡(m)+(frac{h}{m}-frac{1}{2!} (h/m)^2+frac{2}{3!} (h/m)^3-……)
where, h = x-m

4. Find the value of √10
a) 3.1633
b) 3.1623
c) 3.1632
d) 3.1645
Answer: b
Explanation: Now f(x)=√x,
Hence, f’ (x)=(frac{1}{2} x^{-1/2})
f” (x)=(-frac{1}{4} x^{-3/2})
f”’ (x)=(frac{3}{8} x^{-5/2})
Hence,
f(x+h)=(sqrt{x+h}=sqrt{x}+frac{h}{2} x^{-1/2}-frac{h^2}{8} x^{-3/2}+frac{h^3}{16} x^{-5/2}+…. )
(By Taylor’s expansion)
Putting,
h=1, and x=9 we get,
f(10)=√10=3+1/6-1/216+1/3888+….=3.1623

5. Expand f(x) = ¹⁄_x about x = 1.
a) 1 – (x-1) + (x-1)² – (x-1)³ + ….
b) 1 + (x-1) + (x-1)² + (x-1)³ + ….
c) 1 + (x-1) – (x-1)² + (x-1)³ + ….
d) 1 – (x+1) + (x+1)² – (x+1)³ + ….
Answer: a
Explanation: Given f(x) = ¹⁄_x
Let, x – 1 = h
Hence, x = 1 + h
Hence, f(x) = f(1 + h) = f(1) + ^h⁄_1! f’ (1) + ^h2⁄_2! f^” (1) +^h3⁄_3! f^”’ (1)+…
Now, f(1) = 1, f'(1) = -1, f”(1) = 2 ,f”'(1) = -6,…….
Hence, f(1 + h) = 1 – h + h² – h³+….
hence, 1 – (x-1) + (x-1)² – (x-1)³ +….

6. Find the expansion of f(x) = ^ex ⁄_1+e^x, given ∫f(x)dx = ln⁡(2), for x = 0
a) (frac{1}{2}-frac{x}{4}-frac{x^3}{48}-…)
b) (frac{1}{2}+frac{x}{4}-frac{x^3}{48}+….)
c) (frac{1}{2}+frac{x}{4}+frac{x^3}{48}+….)
d) (frac{1}{2}+frac{x}{4}-frac{x^3}{48}+….)
Answer: b
Explanation:
Given,f(x)=(frac{e^x}{1+e^x})
Differentiating it we get
(f^1 (x)=ln⁡(1+e^x)+C), now putting x=0 we get,c=0
Hence,
(f^1 (x)=ln⁡(1+e^x))
Now,(f^1 (x)=ln⁡(1+e^x)=ln⁡(2)+frac{x}{2}+frac{x^2}{8}-frac{x^4}{192}+..)(By,Mclaurin’s expansion)
Hence, Differentiating it we get,
f(x)=(frac{1}{2}+frac{x}{4}-frac{x^3}{48}+….).

7. Find the value of e^π⁄_4√2
a) 1.74
b) 1.84
c) 1.94
d) 1.64
Answer: a
Explanation: Let, f(x) = e^xSin(x), f(0) = 1
Now, the expansion of xSin(x) is (x^2-frac{x^3}{3!}+frac{x^6}{5!}+….)
Hence, (e^xSin(x)=e^y=1+y+frac{y^2}{2!}+frac{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!}+frac{x^6}{6!}+…..)^2}{2!})
(+frac{(x^2-frac{x^4}{3!}+frac{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}+……)
Putting, x = π/4,
We get,
f(π/4)=e^{π/4 Sin(π/4)}=e^π/(4√2)=1+(π/4)²+1/3 (π/4)⁴+….=1+.6168+.1268=1.74

8. Find the value of ln(sin(31^o)) if ln(2) = 0.69315
a) -0.653
b) -0.663
c) -0.764
d) -0.662
Answer: b
Explanation: Let, f(x) = ln⁡(sin⁡(x+h))
Then, f(x) = ln⁡(sin⁡(x)), if h=0
f’ (x)=cot⁡(x), f” (x)=-cosec² (x), f”’ (x)=2cosec² (x)cot⁡(x)
Hence, by Taylor’s theorem,
f(x+h)=f(x)+hf'(x)+(frac{h^2}{2!}) f” (x)+(frac{h^3}{3!}) f”’ (x)+⋯..
Hence, ln⁡(sin⁡(x+h))=ln⁡(sin⁡(x))+h cot⁡(x)-(frac{h^2}{2!}) cosec² (x)+(frac{h^3}{3!}) (2cosec² (x) cot⁡(x))+⋯.
Now let, x=30^o, h=1^o;
ln(sin(31^o)) = (ln(sin(30))+frac{π}{180} cot⁡(frac{π}{6})-frac{1}{2!} (frac{π}{180})^2 cosec^2 (frac{π}{6})+⋯)
ln(sin(31^o)) = -0.6935+.030231-.000304+0
ln(sin(31^o)) = -0.663

9. The expansion of f(x,y), is
a) f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}-2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]+….)
b) f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]+…)
c) f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}-2xy frac{∂^2 f}{∂x∂y}-y^2 frac{∂^2 f}{∂y^2}]+…)
d) f(0,0)-([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]-…)
Answer: b
Explanation: By taylor expansion,
f(x,y) = f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]+…)

10. The expansion of f(x, y)=e^{x Sin(y)}, is
a) x + xy + ….
b) y + y² x + ….
c) x + x² y + ….
d) y + xy + ……..
Answer: d
Explanation: Now, f(x, y)=e^{x Sin(y)}, f(0,0) = 0
Therefore,
f_x (x,y) = e^x Sin(y), hence f_x (0,0) = 0

f_y (x,y) = e^x Cos(y), hence f_y (0,0) = 1

f_xx (x,y) = e^x Sin(y), hence f_xx (0,0) = 0

f_yy (x,y) = -e^x Sin(y), hence f_yy (0,0) = 0

f_xy (x,y) = e^x Cos(y), hence f_xy (0,0) = 1
By taylor expansion,
f(x,y) = f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]+…)
f(x,y) = 0 + 0 + y + (frac{1}{2!}) [0 + 2xy + 0] +…..
f(x,y) = y + xy + ……..