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Pythagoras Theorem

If f(x)=x-2 ...

Question

If f(x)=\begin{vmatrix}x-2&(x-1)^2&x^3 x-1&x^2&(x+1)^3 x&(x+1)^2&(x+2)^3\end{vmatrix} , then coefficient of x in f(x) is

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

, find the absolute value of coefficient of

x

Δ (x)

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

Then the coefficient of x in

Δ

- k

Consider the function

f (x) = {\begin{matrix} x^{2} - 5, & x \leq 3 \sqrt{x + 13}, & x > 3 \end{matrix}

.

Find the differential coefficient of

f (x)

x = 12.

f (x) = 1 - x + x^{2} - x^{3} + . . . . - x^{15} + x^{16} + x^{17}

, then the coefficient of

x^{2}

f (x - 1)

f (x) = 1 - x + x^{2} - x^{3} + \dots \dots - x^{15} + x^{16} - x^{17}

, then the coefficient of

x^{2}

f (x - 1)

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Pythagoras Theorem

Standard X Mathematics