The quadratic equation whose roots are 'l' and 'm' where l = Ltθ→03sinθ-4sin3θθ and m=Ltθ→02tanθθ1-tan2θ is

l=Ltα→03sinθ−4sin3θθl=Ltθ→0sin3θθl=3M=Ltθ→02Tanθθ1−Tan−2θ=Ltθ→0Tan2θθm=2l=3,m=2 are roots of equationx2−5x+6=0

The quadratic equation whose roots are 'l' and 'm' where l = Ltθ→03sinθ-4sin3θθ and m=Ltθ→02tanθθ1-tan2θ is

l=Ltα→03sinθ−4sin3θθl=Ltθ→0sin3θθl=3M=Ltθ→02Tanθθ1−Tan−2θ=Ltθ→0Tan2θθm=2l=3,m=2 are roots of equationx2−5x+6=0

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The quadratic equation whose roots are 'l' and 'm' where l = $\underset{θ \to 0}{L t} \frac{3 \sin θ - 4 \sin^{3} θ}{θ} a n d m = \underset{θ \to 0}{L t} \frac{2 \tan θ}{θ (1 - \tan^{2} θ)}$ is

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$x^{2} - 5 x - 6 = 0$

$x^{2} + 5 x - 6 = 0$

$x^{2} + 5 x + 6 = 0$

$x^{2} - 5 x + 6 = 0$

answer is B.

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Detailed Solution

$\begin{array}{l} l = \underset{α \to 0}{L t} \frac{3 \sin θ - 4 \sin^{3} θ}{θ} \\ l = \underset{θ \to 0}{L t} \frac{\sin 3 θ}{θ} \\ l = 3 \\ M = \underset{θ \to 0}{L t} \frac{2 Tan θ}{θ (1 - {Tan}^{- 2} θ)} \\ = \underset{θ \to 0}{L t} \frac{Tan 2 θ}{θ} \\ m = 2 \\ l = 3, m = 2 a r e r o o t s o f e q u a t i o n \\ x^{2} - 5 x + 6 = 0 \end{array}$