Question

Assertion :Let f : R $\to$ R be defined as $f\left(x\right)=a{x}^2+bx+c$, where a, b, c $\epsilon$ R and a $\ne$ 0.

If $f\left(x\right)=0$ is having non-real roots, then $\int \frac{dx}{f\left(x\right)}=\lambda ta{n}^{-1}\left(g\right(x\left)\right)+k$, where $\lambda$, $k$ are constants and $g\left(x\right)$ is linear function of $x$.

Reason: $tan\left(ta{n}^{-1}g\right(x\left)\right)=g\left(x\right)\forall x\in R$

• 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. Assertion is incorrect but Reason is correct

Answer 1

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

Given that $f\left(x\right)=a{x}^2+bx+c$ is a quadratic function, and $f\left(x\right)=0$ has non-real roots. So, $f\left(x\right)$ can be written as
$a[{(x+\frac{b}{2a})}^2+\frac{c}{a}-\frac{b^2}{4a^2}]$
$=a[{(x+\frac{b}{2a})}^2+\frac{4ac-b^2}{4a^2}]$
Since $f\left(x\right)$ has non-real roots,
$b^2<4ac$
$\Rightarrow \frac{4ac-b^2}{4a^2}>0$
Suppose $\frac{b}{2a}=c_1$ and $\frac{4ac-b^2}{4a^2}=c_2$
Hence $f\left(x\right)=a[{(x+c_1)}^2+{\left(\sqrt{c_2}\right)}^2]$
So, $\int \frac{dx}{f\left(x\right)}=\frac{1}{a}.\frac{1}{\sqrt{c_2}}{\tan }^{-1}\left(\frac{x+c_1}{\sqrt{c_2}}\right)+k$,
which is in the form $\lambda {\tan }^{-1}\left(g\right(x\left)\right)+k$, where $g\left(x\right)$ is a linear function. Hence Assertion is true.
$\tan \left({\tan }^{-1}g\right(x\left)\right)=g\left(x\right)\forall x\in R$. This statement is always true. But it does not give an explanation for assertion.
Hence, option B is correct.