Question

If $f: R \rightarrow R$ be an injective mapping and $p, q, r$ are non-zero distinct real numbers satisfying:

$f(\frac{p}{r}) = f(\frac{p-q}{q-r})$ and $f(\frac{q}{r}) = f(\frac{r}{p})$.

If the graph of $g(x) = px^2 + qx + r$ passes through $M(1, 6)$ then find the value of $q$.

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (D)

8

Answer 2

Here's how to solve the problem:

1. Use Injectivity: Since $f$ is injective, equal outputs imply equal inputs. Therefore:

   $\frac{p}{r} = \frac{p-q}{q-r}$ and $\frac{q}{r} = \frac{r}{p}$

2. Simplify the Equations:

   • From $\frac{p}{r} = \frac{p-q}{q-r}$, we get $p(q-r) = r(p-q)$, which simplifies to $pq - pr = pr - rq$. Rearranging, we have $q(p+r) = 2pr$, so $q = \frac{2pr}{p+r}$.

   • From $\frac{q}{r} = \frac{r}{p}$, we get $qp = r^2$.

3. Express q and r in terms of p:

   Substitute $q = \frac{2pr}{p+r}$ into $r^2 = pq$:

   $r^2 = p \cdot \frac{2pr}{p+r} = \frac{2p^2r}{p+r}$. Since $r \neq 0$, divide by $r$ to get $r = \frac{2p^2}{p+r}$.

   This gives $r(p+r) = 2p^2$, which simplifies to $r^2 + pr - 2p^2 = 0$.

4. Solve for r:

   Factor the quadratic: $(r-p)(r+2p) = 0$. Thus, $r = p$ or $r = -2p$. Since $p, q, r$ are distinct, $r \neq p$, so $r = -2p$.

5. Solve for q:

   Substitute $r = -2p$ into $q = \frac{2pr}{p+r}$:

   $q = \frac{2p(-2p)}{p+(-2p)} = \frac{-4p^2}{-p} = 4p$.

6. Use the point M(1, 6):

   Since $g(x) = px^2 + qx + r$ passes through $(1, 6)$, we have $p(1)^2 + q(1) + r = 6$, so $p + q + r = 6$.

7. Substitute and solve for p:

   Substitute $q = 4p$ and $r = -2p$ into $p + q + r = 6$:

   $p + 4p - 2p = 6$, which gives $3p = 6$, so $p = 2$.

8. Find q:

   Since $q = 4p$, we have $q = 4(2) = 8$.

Therefore, the value of $q$ is 8.