Engineering Mathematics Quiz

This set of 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) \(hf(a)+\frac{h^2}{1!} f’ (a)+\frac{h^3}{2!} f” (a)…..\)
View Answer

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) \(e^{-m} [1+(x+m)+\frac{(x+m)^2}{2!}+\frac{(x+m)^3}{3!}+….]\)
View Answer

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)+\(\frac{h}{m}+\frac{2}{3!} (h/m)^3-……\)
View Answer

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
View Answer

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 + ….
View Answer

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 +….

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6. Find the expansion of f(x) = ^{e^x} ⁄_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) \(\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+….\)
View Answer

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
View Answer

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
View Answer

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)-\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]-…\)
View Answer

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 + ……..
View Answer

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 + ……..

11. The expansion of f(x, y) = e^x ln(1 + y), is
a) f(x,y) = y + xy – ^y²⁄₂ +…….
b) f(x,y) = y – xy + ^y²⁄₂ -…….
c) f(x,y) = y + x – ^y²⁄₂ +……..
d) f(x,y) = x + y – ^x²⁄₂ +……..
View Answer

Answer: a
Explanation: Now, f(x, y) = e^x ln(1 + y) , f(0,0) = 0
Therefore,
\(f_x (x,y)=e^x ln(1+y)\), hence f_x (0,0) = 0
\(f_y (x,y)=\frac{e^x}{(1+y)}\), hence f_y (0,0) = 1
\(f_{xx} (x,y)=e^x ln(1+y)\), hence f_xx (0,0) = 0
\(f_{yy} (x,y)=-\frac{e^x}{(1+y)^2}\), hence f_yy (0,0) = -1
\(f_{xy} (x,y)=\frac{e^x}{(1+y)}\), hence f_xy (0,0) = 1
By taylor expansion,
f(x,y) = f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+1/2! [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…\)
f(x,y) = y + xy – ^y²⁄₂ +…….

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