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Question: If α and β are the roots of the quadratic equation ax<sup>2</sup> + bx + c = 0, then \(\lbrack d + a...

If α and β are the roots of the quadratic equation ax2 + bx + c = 0, then [d+a,d+2a][ad,a+d]\lbrack d + a,d + 2a\rbrack\lbrack a - d,a + d\rbrack =

A

[d+a,ad]\lbrack d + a,a - d\rbrack

B

[da,d+a]\lbrack d - a,d + a\rbrack

C

f(x)=log(x+x2+1)f(x) = \log(x + \sqrt{x^{2} + 1})

D

None of these

Answer

[d+a,ad]\lbrack d + a,a - d\rbrack

Explanation

Solution

ax2 + bx + c = 0 roots α, β 

So, cx2 + bx + a = c (x – 1/α) (x – 1/β)

Limx1/α\operatorname { Lim } _ { x \rightarrow 1 / \alpha }

Limx1/α\operatorname { Lim } _ { x \rightarrow 1 / \alpha } 2sin2c2(x1/α)(x1/β)2α2(x1/α)2\sqrt { \frac { 2 \sin ^ { 2 } \frac { \mathrm { c } } { 2 } ( x - 1 / \alpha ) ( x - 1 / \beta ) } { 2 \alpha ^ { 2 } ( x - 1 / \alpha ) ^ { 2 } } }

Limx1/α\operatorname { Lim } _ { x \rightarrow 1 / \alpha }

Limx1/α\operatorname { Lim } _ { x \rightarrow 1 / \alpha }=c2α(1α1β)\left| \frac { \mathrm { c } } { 2 \alpha } \left( \frac { 1 } { \alpha } - \frac { 1 } { \beta } \right) \right|