Question

Assertion: If $p{x^2} + qx + r = 0$ is a quadratic equation $\left( {p,\,q,\,r \in R} \right)$ such that its roots are $\alpha ,\,\beta$ and $p + q + r < 0,$ $p - q + r < 0$ and $r > 0$ , then $\left[ \alpha \right] + \left[ \beta \right] = - 1$ where $\left[ . \right]$ denotes greatest integer function.
Reason: If for any two real numbers $a$ and $b,$ function $f(x)$ is such that $f\left( a \right)f\left( b \right) < 0 \Rightarrow f\left( x \right)$ has at least one real root lying in $\left( {a,b} \right)$ .
(A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
(B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
(C) Assertion is correct but Reason is incorrect.
(D) Both Assertion and Reason are incorrect.

Answer 1

In this question the values of the function at $1,$ $- 1$ and $0$ are given and the value of the function at this point are greater than and less than $0$ but not equal to zero. Therefore, the roots of the function lies between $1,$ $- 1$ and $0$ . Now, we will check whether the reason is correct or incorrect.

Complete step by step solution:
The given equation is $p{x^2} + qx + r$ and its roots are $\alpha ,\,\beta$ . Also given, $p + q + r < 0,$ $p - q + r < 0$ and $r > 0$ .
If we put $x = 1$ in the equation $f\left( x \right) = p{x^2} + qx + r$ then we will get
$\Rightarrow f\left( 1 \right) = p{\left( 1 \right)^2} + q\left( 1 \right) + r \\\ \Rightarrow f\left( 1 \right) = p + q + r \\\$
It is given in the question that $p + q + r$ is less than $1$ . Therefore, we got $f(1) < 0$ .
Now, put $x = 0$ in the equation $f\left( x \right)$ then we will get
$\Rightarrow f\left( 0 \right) = p{\left( 0 \right)^2} + q\left( 0 \right) + r \\\ \Rightarrow f\left( 0 \right) = r \\\$
It is given in the question that $r$ is greater than $1$ . Therefore, we got $f(0) > 0$ .
Now, put $x = - 1$ in the equation $f\left( x \right)$ then we will get
$\Rightarrow f\left( { - 1} \right) = p{\left( { - 1} \right)^2} + q\left( { - 1} \right) + r \\\ \Rightarrow f\left( { - 1} \right) = p - q + r \\\$
It is given in the question that $p - q + r$ is less than $1$ . Therefore, we got $f( - 1) < 0$ .
From the above observations we can say that If $\alpha ,\,\beta$ are the roots of the equation $p{x^2} + qx + r = 0$ then one of the root lies between $\left( { - 1,0} \right)$ and the other root lies between $\left( {0,1} \right)$ .
Therefore, we can write $\left[ \alpha \right] + \left[ \beta \right] = - 1$ where $\left[ . \right]$ is the greatest integer function and this function gives the greatest integer less than the given value. Example: If we have $\left[ {1.45} \right]$ then it will give value $1$ .
And also from the observation we can write $f\left( 0 \right).f\left( { - 1} \right) < 0$ . Therefore, we can say that at least one root lies between$\left( { - 1,0} \right)$ .
Therefore, both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Hence, option (A) is the correct option.

Note:
In this question we should get the idea by looking at the question that the value of the equation at some points is less than zero and at some points it is greater than zero and between them we will get the roots of the equation. And we should have the knowledge of the greatest integer function.