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TS EAMCET 19-July-2022 Shift 1 Solved Paper

Section: Mathematics

Question : 15 of 160

Marks: +1, -0

x^{10}-3 x^{8}+5 x^{6}-5 x^{4}+3 x^{2}-1=0

Solution: 👈: Video Solution

P(x): x^{10}-3 x^8+5 x^6-5 x^4+3 x^2-1=0

\because P(1)

and

P(-1)

are zero

\Rightarrow (x-1)

and

(x+1)

are factors.

P(x)= (x^2-1) (x^8-2 x^6+3 x^4-2 x^2+1)

Let

g(x)=x^8-2 x^6+3 x^4-2 x^2+1

\; = x^2 (x^6-2 x^4+3 x^2-2) +1

\; = x^2 (x^2-1) (x^4-x^2+2) +1

\; = x^2 (x^2-1) \left\{ \left(x^2-\frac{1}{2}\right)^2+\frac{7}{4} \right\} +1

g(x) \neq 0

for any real value.

\therefore

Total number of non-real roots

=10-2=8

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