NIMCET 2020 — Mathematics PYQ

Step 1: Use the property of roots satisfying the equation

Since $n$ and $m$ are the roots of the quadratic equation $a{x}^2-2bx+c=0$, they must satisfy the equation.

Substituting $x=n$:

$$a{n}^2-2bn+c=0⟹a{n}^2+c=2bn$$

Substituting $x=m$:

$$a{m}^2-2bm+c=0⟹a{m}^2+c=2bm$$

Step 2: Substitute these values into the given expression

We need to find the value of:

$$\frac{b}{a{n}^2+c}+\frac{b}{a{m}^2+c}$$

Replace $a{n}^2+c$ with $2bn$ and $a{m}^2+c$ with $2bm$:

$$=\frac{b}{2bn}+\frac{b}{2bm}$$

Cancel out $b$ from the numerator and denominator of both terms:

$$=\frac{1}{2n}+\frac{1}{2m}$$

$$=\frac{1}{2}\left(\frac{1}{n}+\frac{1}{m}\right)$$

$$=\frac{1}{2}\left(\frac{m+n}{nm}\right)$$

Step 3: Apply sum and product of roots relations

For the standard quadratic equation $a{x}^2-2bx+c=0$:

• Sum of roots ($n+m$): $-\frac{(-2b)}{a}=\frac{2b}{a}$

• Product of roots ($nm$): $\frac{c}{a}$

Substitute these relationships back into our simplified expression:

$$=\frac{1}{2}\left(\frac{\frac{2b}{a}}{\frac{c}{a}}\right)$$

The denominator $a$ cancels out:

$$=\frac{1}{2}\left(\frac{2b}{c}\right)$$

$$=\frac{b}{c}$$

Conclusion

The final simplified value of the given expression is $\frac{b}{c}$.

Correct Answer: D