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!} f^{n} (0)

b) 1 + ^{x}⁄_{1!} f^{‘} (0) + ^{x2}⁄_{2!} f^{”} (0)…….+^{xn}⁄_{n!} f^{n} (0)

c) f(0) – ^{x}⁄_{1!} f^{‘} (0) + ^{x2}⁄_{2!} f^{”} (0)…….+(-1)^n ^{xn}⁄_{n!} f^{n} (0)

d) f(1) + ^{x}⁄_{1!} f^{‘} (1) + ^{x2}⁄_{2!} f^{”} (1)…….+^{xn}⁄_{n!} f^{n} (1)

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Explanation: By Maclaurin’s series, f(0) +

^{x}⁄

_{1!}f

^{‘}(0) +

^{x2}⁄

_{2!}f

^{”}(0)…….+

^{xn}⁄

_{n!}f

^{n}(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

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Explanation: By Maclaurin’s series, f(0) +

^{x}⁄

_{1!}f

^{‘}(0) +

^{x2}⁄

_{2!}f

^{”}(0)…….+

^{xn}⁄

_{n!}f

^{n}(0)

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

3. The expansion of f(a+h) is

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Explanation: By taylor expansion,

f(a+h) = f(a) +

^{h}⁄

_{1!}f’ (a) +

^{h2}⁄

_{2!}f

^{”}(a)…….

4. The expansion of e^{Sin(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} +….

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Explanation: Now f(x) = e

^{Sin(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) = e^{Sin(x)} = 1 + x + ^{x2}⁄_{2} – ^{x4}⁄_{8} +…. (By mclaurin’ sexpansion)

5. Expansion of y = Sin^{-1}(x) is

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Explanation: Given, y = Sin

^{-1}(x), hence at x = 0, y = 0

Now, differentiating it, we get

6. Find the expansion of f(x) = ln(1+e^{x})

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7. Find the expansion of e^{xSin(x)}

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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

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Explanation: Given f(x) = ln(Cos(x)), f(0) = 0

Differentiating it f'(x) = – tan(x), f'(0) = 0

9. Find the expansion of cos(xsin(t)).

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10. Find the expansion of Sin(lSin^{-1} (x)).

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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} -…..].

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12. Find the solution of differential equation, ^{dy}⁄_{dx} = xy + x^{2}, if y = 1 at x = 0.

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Explanation: Given

^{dy}⁄

_{dx}= xy + x

^{2}

hence, ^{dy}⁄_{dy} (x=0) = 0

and, ^{d2y}⁄_{dx2} = xy_{1} + y + 2x

hence, y_{2} = xy_{1} + y + 2x

hence, ^{d2y}⁄_{dx2} (x=0)=1

**Sanfoundry Global Education & Learning Series – Engineering Mathematics.**

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