Question

Assertion :If $a>0$ and $b^2-ac+c<0$,then domain of the function $f\left(x\right)=\sqrt{a{x}^2+2bx+c}\text{is}R$ Reason: If $b^2-ac<0$,then $a{x}^2+2bx+c=0$ has imaginary roots.

• 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

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

If $a>0,$ then graph of $y=a{x}^2+2bx+c$ is concave upward.

Also if $b^2-ac<0,$ then the graph always lies above $x$ -axis.

Hence, $a{x}^2+2bx+c>0$ for all real values of $x.$ Thus, domain of function $f\left(x\right)=\sqrt{a{x}^2+2bx+c}$ is $R$

If $b^2-ac<0,$ then $a{x}^2+2bx+c=0$ has imaginary roots.

Then the graph of $y=a{x}^2+2bx+c$ never cuts $x$ -axis, or $y$ is either always positive or always negative.

Hence, both the statements are correct but statement 2 is not correct explanation of statement 1