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If $x + y + z = 1, a x + b y + c z = k$ , $a^{2} x + b^{2} y + c^{2} z = k^{2}$ has unique solution Then $x =$

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(k−b) (c−k)(a−b) (c−a)

(k−c) (a−k)(b−c) (c−a)

(k−a) (b−k)(b−c) (c−a)

(k−a) (k−b) (k−c)

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detailed solution

Correct option is A

We have Δ= 111abca2b2c2=a-bb-cc-a Δ1=111kbck2b2c2=k-bb-cc-k ∴By Crammar's rule x=Δ1Δ=k-bb-cc-ka-bb-cc-a=k-bc-ka-bc-a

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If x+y+z=1, ax+by+cz=k ,a2x+b2y+c2z=k2 has unique solution Then x=

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

If $x + y + z = 1, a x + b y + c z = k$ , $a^{2} x + b^{2} y + c^{2} z = k^{2}$ has unique solution Then $x =$