Question

integrate the function:
$\frac{1}{ \sqrt{ {(2 - x)}^{2} + 1} }$

Answer 1

Step-by-step explanation:

Step-by-step explanation:

$\huge\mathfrak\green{\bold{\underline{☘{ ℘ɧεŋσɱεŋศɭ}☘}}}$

$\red{\bold{\underline{\underline{❥Question᎓}}}}$integrate the function:

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

$\huge\tt\underline\blue{❯Answer❮</p><p> }$

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Let 2-x=t

differentiating both sides w.r.t.x

$0 - 1 = \frac{dt}{dx}$

$dx = - dt$

Integrating the function w.r.t.x

$∫ \frac{1}{ \sqrt{ {(2 - x)}^{2} + t} } dx$

Put the value of 2-x=t and dx=-dt

$= ∫ \frac{ - dt}{ \sqrt{ {t}^{2} + 1} }$

$= - ∫ \frac{dt}{ \sqrt{ {t}^{2} + {(1)}^{2} } }$

It is the form of :

$∫ \frac{1}{ \sqrt{ {x}^{2} + {a}^{2} } } dx = log |x + \sqrt{ {x}^{2} + {a}^{2} } | + c$

∴ Replace a by 1 and x by t we get

$= - log |t + \sqrt{ {t}^{2} + 1 } | + c$

$= log { |t + \sqrt{ {t}^{2} + 1} | }^{ - 1} + c$

$= log \frac{1}{ |t + \sqrt{ {t}^{2} + 1} | } + c$

$= log \frac{1}{ |2 - x + \sqrt{ {(2 - x)}^{2} + 1 } | } + c$

$= log \frac{1}{ |2 - x + \sqrt{4 + {x}^{2} - 4x + 1} | } + c$

$= log \frac{1}{ |2 - x + \sqrt{ {x}^{2} - 4x + 5} | } + c$

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Answer 2

Step-by-step explanation:

f(x) = kx³ – 8x² + 5

Roots are α – β , α & α +β

Sum of roots = – (-8)/k

Sum of roots = α – β + α + α +β = 3α

= 3α = 8/k

= k = 8/3α

or we can solve as below

f(x) = (x – (α – β)(x – α)(x – (α +β))

= (x – α)(x² – x(α+β + α – β) + (α² – β²))

= (x – α)(x² – 2xα + (α² – β²))

= x³ – 2x²α + x(α² – β²) – αx² +2α²x – α³ + αβ²

= x³ – 3αx² + x(3α² – β²) + αβ² – α³

= kx³ – 3αkx² + xk(3α² – β²) + k(αβ² – α³)

comparing with

kx³ – 8x² + 5

k(3α² – β²) = 0 => 3α² = β²

k(αβ² – α³) = 5

=k(3α³ – α³) = 5

= k2α³ = 5

3αk = 8 => k = 8/3α

(8/3α)2α³ = 5

=> α² = 15/16

=> α = √15 / 4