If for any real x- we have -1-x2-nx-2-x2-3x-4- 2- then n belongs to-

The correct option is C [ 1, √(40) 6] ...

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Integration of Irrational Algebraic Fractions - 2

If for any re...

Question

If for any real x, we have $- 1 \leq \frac{x^{2} + n x - 2}{x^{2} - 3 x + 4} \leq 2$ , then $n$ belongs to.

[- \sqrt{40} + 6, - 1]

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[- \sqrt{40} + 6, \sqrt{40} - 6]

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[- 1, \sqrt{40} - 6]

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None of these

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\sqrt{40 - x^{2}} - 6 = 0

(1 - x - x^{2})^{20} = 40 \sum r = 0 a_{r} x^{r}

a_{2} + 2 a_{4} + 3 a_{6} + \dots \dots \dots \dots + 20 a_{40} =

Factorise:

(i) x² + 6x + 8; (ii) x² + 4x + 3; (iii) a² + 5a + 6; (iv) a² − 5a + 6; (v) a² − 3a − 40; (vi) x² − x − 72.

{sin}^{- 1} (x - 4) + {cos}^{- 1} (x - 6) + {tan}^{- 1} (\frac{40}{15 + x^{2}}) = m

^{'} m^{'}

If x² - 6x - 40 = (x + 4)(x + a), then the value of a is ______.

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