Question

maths question.plz answer

Answer 1

Hey !!!

8sec²¢ - 6sec¢ + 1 = 0

let us consider sec¢ = x

then it will be formed a quadratic equations

8x² - 6x + 1 = 0

=> 8x² - 4x - 2x + 1 = 0 [Using spliting term method ]

=> 4x ( 2x - 1 ) - 1 ( 2x - 1 ) = 0

=> (4x - 1 ) ( 2x - 1 ) = 0

=> 4x - 1 = 0

x = 1/4

similarly

(2x - 1 ) = 0

x = 1/2 ..

means , sec¢ = 1/4 , 1/2

but range of sec¢ is (-∞,0)to ( ∞ , 1)

since , it has no real roots

hence option ( 4) is correct option
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Hope it helps you !!!

@Rajukumar111

Answer 2

Answer: Option(4):

Given Equation is 8sec^2A - 6secA + 1 = 0.

On comparing with ax^2 + bx + c = 0, we get a = 8, b = -6, c = 1.

$= > x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$

$= > \frac{-(-6) + \sqrt{(-6)^2 - 4(8)(1)} }{2(8)}$

$= > \frac{6 + \sqrt{36 - 32}}{16}$

$= > \frac{6 + 2}{16}$

$= > \frac{1}{2}$

(ii)

$= > \frac{- (-6) - \sqrt{(6)^2 - 4(8)(1) }}{2(8)}$

$= > \frac{6 - \sqrt{36 - 32}}{16}$

$= > \frac{6 - 2}{16}$

$= > \frac{4}{16}$

$= > \frac{1}{4}$

Now,

SecA = (1/2) (or) SecA = 1/4

As we know, that range of sectheta = (-infinity,-1) U (1,infinity).

Therefore, the given equations has 0 roots.

Hope this helps!