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3 years ago

⦁ Evaluate ⦁ 6! ⦁ 8P5 ⦁ 12C4

Mathematics

2 answers:

lisabon 2012 [21]3 years ago

7 0

6! = 6 * 5 *4*3*2*1 =720

8p5 = 8*7*6*5*4 = 6720

12C4 = (12p4)/4! = (12*11*10*9)/4*3*2*1 = 495

Vesnalui [34]3 years ago

6 0

N! = n(n-1)(n-2)(n-3)....
∴6! = 6×5×4×3×2×1 = 720
⁸P₅ = $\frac{8!}{(8-5)!} = \frac{8*7*6*5*4*3*2*1}{3*2*1} = 6720$
¹²C₄ = $\frac{12!}{4!*(12-4)!} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4!*8!}$
= $\frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1*8*7*6*5*4*3*2*1} = 495$

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3 years ago

Please see attachment

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b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

Step(i):-

Given function

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$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

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Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

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The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

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$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

Final answer:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

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