NIMCET 2008 — Mathematics PYQ

Since $a$ and $b$ are the roots of $x^2+px+1=0$, we can write:

Similarly, for the second equation:

2. Analyze the Expression $E$

The expression is $E=\underset{Term1}{\underbrace{(a-c)(b-c)}}\cdot \underset{Term2}{\underbrace{(a+d)(b+d)}}$.

For Term 1:

Notice that $(a-c)(b-c)$ is the same as $(c-a)(c-b)$.

From our definition of $f(x)$:

So, Term 1 $=c^2+pc+1$.

For Term 2:

Notice that $(a+d)(b+d)$ can be written as $(-d-a)(-d-b)$.

From our definition of $f(x)$:

So, Term 2 $=d^2-pd+1$.

3. Substitute and Simplify

Now, $E=(c^2+pc+1)(d^2-pd+1)$.

Since $c$ and $d$ are roots of $x^2+qx+1=0$, we know:

• $c^2+1=-qc$

• $d^2+1=-qd$

Substitute these into the expression for $E$:

4. Use Product of Roots

For the equation $x^2+qx+1=0$, the product of roots $cd=\frac{\text{constant term}}{\text{coefficient of }{x}^2}=\frac{1}{1}=1$.

Substitute $cd=1$:

Conclusion

The simplified value of the given expression is $q^2-p^2$.

Correct Option: (b)