If α,β are two distinct real roots of the equation ax3+x−1−a=0,(a≠−1,0) none of which is equal to unity, then the value of limx→(1/α)(1+a)x3−x2−a(e1−ax−1)(x−1) is al(kα-β)α. Find the value of kl.

ax3+x−1−a=0⇒ ax3−1+(x−1)=0⇒(x−1)ax2+ax+(a+1)=0 It is given that α.,β are two distinct roots of ax3+x−1−a=0 or Therefore, α,β are roots of ax2+ax+(a+1)=0 and hence, 1α,1β are roots of (a+1)x2+ax+a=0. ∴ (a+1)x−1αx−1β=(a+1)x2+ax+alimx→1/α (1+a)x3−x2−ae1−αx−1(x−1)=limx→1/α ax3−1+x2(x−1)e1−αx−1(x−1)=limx→1/α ax2+x+1+x2e1−αx−1=limx→1/α (a+1)x2+ax+a1−αx= =limx→1/α (a+1)x−1α(x-1β)1−αx =−a+1−1α(1α-1β)∴alα(kα−β)=(a+1)α2β(α−β)⇒ alα(kα−β)=(a+1)aα(a+1)(α−β) ∵αβ=a+1a⇒ l(kα−β)=(α−β)⇒l=k=1⇒kl=1

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If $α, β$ are two distinct real roots of the equation $a x^{3} + x - 1 - a = 0, (a \neq - 1, 0)$ none of which is equal to unity, then the value of $\lim_{x \to (1 / α)}$ $\frac{(1 + a) x^{3} - x^{2} - a}{(e^{1 - a x} - 1) (x - 1)}$ is $\frac{a l (k α - β)}{α}$ . Find the value of kl.

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Detailed Solution

$\begin{array}{l} a x^{3} + x - 1 - a = 0 \\ \Rightarrow a (x^{3} - 1) + (x - 1) = 0 \Rightarrow (x - 1) \{a x^{2} + a x + (a + 1)\} = 0 \end{array}$

$It is given that α.,β are two distinct roots of a x^{3} + x - 1 - a = 0 or$

$Therefore, α, β are roots of a x^{2} + a x + (a + 1) = 0 and hence, \frac{1}{α}, \frac{1}{β} are roots of (a + 1) x^{2} + a x + a = 0 . ∴ (a + 1) (x - \frac{1}{α}) (x - \frac{1}{β}) = (a + 1) x^{2} + a x + a$

$\begin{array}{l} lim_{x \to 1 / α} \frac{(1 + a) x^{3} - x^{2} - a}{(e^{1 - α x} - 1) (x - 1)} \\ = lim_{x \to 1 / α} \frac{a (x^{3} - 1) + x^{2} (x - 1)}{(e^{1 - α x} - 1) (x - 1)} \\ = lim_{x \to 1 / α} \frac{a (x^{2} + x + 1) + x^{2}}{(e^{1 - α x} - 1)} \\ = lim_{x \to 1 / α} \frac{(a + 1) x^{2} + a x + a}{1 - α x} = \\ = lim_{x \to 1 / α} \frac{(a + 1) (x - \frac{1}{α}) (x - \frac{1}{β})}{1 - α x}) \\ = - (a + 1) (- \frac{1}{α}) (\frac{1}{α} - \frac{1}{β}) \\ ∴ \frac{a l}{α} (k α - β) = \frac{(a + 1)}{α^{2} β} (α - β) \end{array}$

$\begin{aligned} \Rightarrow \frac{a l}{α} (k α - β) = \frac{(a + 1) a}{α (a + 1)} (α - β) [∵ α β = \frac{a + 1}{a}] \\ \Rightarrow l (k α - β) = (α - β) \Rightarrow l = k = 1 \Rightarrow k l = 1 \end{aligned}$