JEE Main 2018 — Mathematics Algebra

$${x}^{2}+(2-\lambda )x+(10-\lambda )=0$$x2+(2−λ)x+(10−λ)=0

\Rightarrow \alpha +\beta =\lambda -2&\alpha \beta =10-\lambda ..........(i)

Let roots are $\alpha and \beta$αandβ.

$$\Rightarrow {\alpha }^{3}+{\beta }^{3}={(\alpha +\beta )}^{3}-3\alpha \beta (\alpha +\beta )$$⇒α3+β3=(α+β)3−3αβ(α+β)

$$={(\lambda -2)}^{3}-3(10-\lambda )(\lambda -2)$$=(λ−2)3−3(10−λ)(λ−2)

$$= {\lambda }^{3}-{6\lambda }^{2}+12\lambda -8-3(10\lambda -{\lambda }^{2}-20+2\lambda )$$=λ3−6λ2+12λ−8−3(10λ−λ2−20+2λ)

$$= {\lambda }^{3}-{3\lambda }^{2}-24\lambda +52$$=λ3−3λ2−24λ+52

$\frac{dz}{d\lambda }=3{\lambda }^{2}-6\lambda -24=3({\lambda }^{2}-2\lambda -8)$dλdz​=3λ2−6λ−24=3(λ2−2λ−8) (where, $z={\alpha }^{3}+{\beta }^{3}$z=α3+β3)

For maximum and critical points, derivative must be zero.

$$\Rightarrow {\lambda }^{2}-2\lambda -8=0$$⇒λ2−2λ−8=0

$$\Rightarrow (\lambda -4)(\lambda +2)=0$$⇒(λ−4)(λ+2)=0

$$\Rightarrow \lambda =-2,4$$⇒λ=−2,4

Now, $\frac{{d}^{2}z}{d{\lambda }^{2}}=6\lambda -6$dλ2d2z​=6λ−6

For $(\lambda =-2),\frac{{d}^{2}z}{d{\lambda }^{2}}<0\Rightarrow {\alpha }^{3}+{\beta }^{3}$(λ=−2),dλ2d2z​<0⇒α3+β3 is maximum and

For $(\lambda =4),\frac{{d}^{2}z}{d{\lambda }^{2}}>0\Rightarrow {\alpha }^{3}+{\beta }^{3}$(λ=4),dλ2d2z​>0⇒α3+β3 is minimum.

$\Rightarrow$⇒Equation will be ${x}^{2}-2x+6=0$x2−2x+6=0.

Using quadratic formula, we get

$$x=\frac{2\pm \sqrt{{(-2)}^{2}-4\times 1\times 6}}{2\times 1}=\frac{2\pm \sqrt{-20}}{2}=\frac{2\pm 2\sqrt{5}i}{2}=1\pm \sqrt{5}i$$x=2×12±(−2)2−4×1×6​​=22±−20​​=22±25​i​=1±5​i

Thus, difference of roots is $|\alpha -\beta |=2\sqrt{5}$∣α−β∣=25​