Constant term in f x - begin vmatrix x -1-sin x- 3 -cos x1 -ln 1-x - 2 x- 2 -1-x- 2 - 0lend vmatrix when f x is expressed polynomial in x- isA- 1B- 2C- 0D- -1

The correct option is A 1f(0) = 1 ...

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Implicit Differentiation

Constant term...

Question

Constant term in f(x) = $∣ ∣ ∣ ∣ ∣ \begin{matrix} x & (1 + s i n x)^{3} & c o s x 1 & l n (1 + x) & 2 x^{2} & (1 + x)^{2} & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$ when f(x) is expressed polynomial in x, is

-1

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f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + x)^{a} & (2 + x)^{b} & 1 1 & (1 + x)^{a} & (2 + x)^{b} (2 + x)^{b} & 1 & (1 + x)^{a} \end{matrix} ∣ ∣ ∣ ∣ ∣,

f (x)

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x + 1 2 x & x (x - 1) & x (x + 1) 3 x (x - 1) & x (x - 1) (x - 2) & x (x^{2} - 1) \end{matrix} ∣ ∣ ∣ ∣ ∣

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{5} & x^{2} & 1 1 & x^{3} - 1 & 1 + x + x^{2} 2 & x^{2} - 2 x + 1 & (1 - x)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

x^{n}

If f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + x^{n} & (1 - x)^{n} & 2 + x^{n} (2 + x)^{n} & (2 + x)^{n} & 1 (3 - x)^{n} & 1 & 3 + x \end{matrix} ∣ ∣ ∣ ∣ ∣

then the constant term in the expansion is

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} cos x & x & 1 2 sin x & x^{2} & 2 x tan x & x & 1 \end{matrix} ∣ ∣ ∣ ∣ .

lim x \to 0 \frac{f (x)}{x}

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