If k - b is positive real number- then the roots of the equation x 2- bx -2- k x 2-3 x -2-0 areA- real and distinct-B- imaginary-C- real and equal-D- none of these

The correct option is A real and distinct.x^2+bx 2+k(x^2+3x+2)=0 ⇒ (k+1)x^2+(b+3k)x+2(k 1)=0 D= (b+3k)^2 8(k+1)(k 1) =b^2+9k^2+6bk 8k^2+8 =b^2+k^2+6bk+8 ...

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Standard XII

Mathematics

Strictly Increasing Functions

If k , b is p...

Question

If $k, b$ is positive real number, then the roots of the equation $x^{2} + b x - 2 + k (x^{2} + 3 x + 2) = 0$ are

real and distinct.

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imaginary.

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real and equal.

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none of these

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Solution

The correct option is A real and distinct.
$x^{2} + b x - 2 + k (x^{2} + 3 x + 2) = 0 \Rightarrow (k + 1) x^{2} + (b + 3 k) x + 2 (k - 1) = 0 D = (b + 3 k)^{2} - 8 (k + 1) (k - 1) = b^{2} + 9 k^{2} + 6 b k - 8 k^{2} + 8 = b^{2} + k^{2} + 6 b k + 8 > 0 \Rightarrow D > 0$
$∴$ roots will be real and distinct

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If k,b is positive real number, then the roots of the equation x2+bx−2+k(x2+3x+2)=0 are

The correct option is A real and distinct.x2+bx−2+k(x2+3x+2)=0⇒(k+1)x2+(b+3k)x+2(k−1)=0D=(b+3k)2−8(k+1)(k−1) =b2+9k2+6bk−8k2+8 =b2+k2+6bk+8>0⇒D>0 ∴ roots will be real and distinct

If $k, b$ is positive real number, then the roots of the equation $x^{2} + b x - 2 + k (x^{2} + 3 x + 2) = 0$ are