Question

Assertion(A): $f:R\to R$ defined by $f\left(x\right)=(x-9)(x-10)(x-11)$ is onto but not one one
Reason (R): If $f:A\to B;x_1,x_2\in A$ and $f\left(x_1\right)\ne f\left(x_2\right)\Rightarrow x_1=x_2$, then $f$ is one-one.

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is not correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

Answer: (C)

A is true but R is false

Let us draw a rough sketch of $f\left(x\right)$. We have three points where $f\left(x\right)$ becomes $0$.

These are $x=9,10$ and $11$.

For $x<9$, we have $f\left(x\right)<0$ (since all the three terms are negative here).

For $9<x<10$, we have $f\left(x\right)>0$ (since only the first term is positive here).

For $10<x<11$, we have $f\left(x\right)<0$ (since only the third term is negative here).

For $x>11$, we have $f\left(x\right)>0$ (since all the three terms are positive here).

So, $f\left(x\right)$ should look as shown.

We see that we may have same values of $f\left(x\right)$ for different values of $x$ (e.g. for $x<9$ and $10<x<11$).

So, $f\left(x\right)$ is not one-one.

We also see that $f\left(x\right)$ can vary from $-\infty$ to $+\infty$.

So, $f\left(x\right)$ is onto.

Hence, the assertion is true.

Now, the correct statement is: if $A\to B,x_1,x_2\in A$ and $x_1=x_2⟹f\left(x_1\right)\ne f\left(x_2\right)$, then $f\left(x\right)$ is one-many.

Hence, the reason is false.