Quadratic Equation and Inequalities — JEE Main practice

Let $𝑡=|𝑥-1|$.

Then the equation becomes

$𝑡^2-5𝑡+6=0$.

Factorizing,

$(𝑡-3)(𝑡-2)=0$.

Hence,

$𝑡=2$ or $𝑡=3$.

So,

$|𝑥-1|=2$ or $|𝑥-1|=3$.

Therefore,

• $𝑥-1=\pm2$ gives $𝑥=3,-1$.
• $𝑥-1=\pm3$ gives $𝑥=4,-2$.

The sum of all the roots is

$3+(-1)+4+(-2)=4$.

Hence, the correct answer is C.

Since $\alpha<1<\beta$, the point $𝑥=1$ lies between the two roots.

For an upward-opening quadratic, this implies that the value of the polynomial at $𝑥=1$ must be negative.

Substituting $𝑥=1$,

$1+2𝑎+(3𝑎+10)<0$.

Simplifying,

$5𝑎+11<0$,

which gives

$𝑎<-\dfrac{11}{5}$.

Hence, the set of all possible values of $𝑎$ is

$\left(-\infty,-\dfrac{11}{5}\right)$.

Therefore, the correct answer is B.

Split the real line into intervals determined by the critical points $𝑥=-4$ and $𝑥=-2$.

On each interval, remove the absolute value signs according to the signs of the expressions inside them.

Solve the resulting equations and retain only those solutions that satisfy the interval conditions.

Exactly one solution satisfies the original equation.

Hence, the correct answer is D.