NIMCET 2025 — Mathematics PYQ

Step 1: Write Down Given Relations

Since $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2+ax+b=0$:

• Sum of roots: $\alpha +\beta =-a$

• Product of roots: $\alpha \beta =b$

Step 2: Find the Product of the New Roots

The new roots are $\frac{1}{{\alpha }^3+\alpha }$ and $\frac{1}{{\beta }^3+\beta }$.

$$\text{Product of new roots}=\frac{1}{({\alpha }^3+\alpha )({\beta }^3+\beta )}$$

$$\text{Product}=\frac{1}{\alpha ({\alpha }^2+1)\cdot \beta ({\beta }^2+1)}$$

$$\text{Product}=\frac{1}{\alpha \beta \cdot ({\alpha }^2{\beta }^2+{\alpha }^2+{\beta }^2+1)}$$

We know that:

• $\alpha \beta =b$

• ${\alpha }^2{\beta }^2=(\alpha \beta {)}^2=b^2$

• ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =(-a{)}^2-2b=a^2-2b$

Substitute these values into the product formula:

$$\text{Product}=\frac{1}{b\cdot (b^2+a^2-2b+1)}$$

$$\text{Product}=\frac{1}{b(b^2+1+a^2-2b)}$$

Step 3: Find the Sum of the New Roots

$$\text{Sum of new roots}=\frac{1}{{\alpha }^3+\alpha }+\frac{1}{{\beta }^3+\beta }$$

$$\text{Sum}=\frac{({\beta }^3+\beta )+({\alpha }^3+\alpha )}{({\alpha }^3+\alpha )({\beta }^3+\beta )}$$

$$\text{Sum}=\frac{({\alpha }^3+{\beta }^3)+(\alpha +\beta )}{b(b^2+1+a^2-2b)}$$

Now, expand the numerator using the identity ${\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )$:

$${\alpha }^3+{\beta }^3=(-a{)}^3-3(b)(-a)=-a^3+3ab$$

Substitute this and $(\alpha +\beta )=-a$ back into the numerator:

$$\text{Numerator}=(-a^3+3ab)+(-a)=-a^3+3ab-a$$

$$\text{Numerator}=-(a^3+a-3ab)$$

So, the Sum of the new roots becomes:

$$\text{Sum}=\frac{-(a^3+a-3ab)}{b(b^2+1+a^2-2b)}$$

Step 4: Form the Required Quadratic Equation

The standard formula to form a quadratic equation is:

$$x^2-(\text{Sum of roots})x+(\text{Product of roots})=0$$

Substitute our values into the formula:

$$x^2-\left[\frac{-(a^3+a-3ab)}{b(b^2+1+a^2-2b)}\right]x+\frac{1}{b(b^2+1+a^2-2b)}=0$$

$$x^2+\left[\frac{a^3+a-3ab}{b(b^2+1+a^2-2b)}\right]x+\frac{1}{b(b^2+1+a^2-2b)}=0$$

Multiply the entire equation by the denominator $b(b^2+1+a^2-2b)$ to eliminate fractions:

$$b(b^2+1+a^2-2b)x^2+(a^3+a-3ab)x+1=0$$

Correct Option Match:

$$b(b^2+1+a^2-2b)x^2+(a^3+a-3ab)x+1=0$$