MCQ
If $f(x) = \left\{ {\begin{array}{*{20}{c}}{\frac{{{x^2} - 9}}{{x - 3}}\,,}&{{\rm{if \,\,}}x \ne 3}\\{2x + k\,,}&{{\rm{otherwise}}}\end{array}} \right.$, is continuous at $x = 3,$ then $k = $
- A$3$
- ✓$0$
- C$-6$
- D$1/6$
and $f(3) = 2(3) + k = 6 + k$
$\because f $ is continuous at $x = 3$;
$\therefore $ $6 + k = 6 \Rightarrow k = 0$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}$
where
$f(\theta)=\frac{1}{2}\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _e\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _e\left(\frac{\pi}{4}\right) & \tan \pi\end{array}\right|$.
Let $p (x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(\theta)$, and $p(2)=2-\sqrt{2}$. Then, which of the following is/are TRUE ?
$(A)$ $p \left(\frac{3+\sqrt{2}}{4}\right)<0$
$(B)$ $p \left(\frac{1+3 \sqrt{2}}{4}\right)>0$
$(C)$ $p \left(\frac{5 \sqrt{2}-1}{4}\right)>0$
$(D)$ $p \left(\frac{5-\sqrt{2}}{4}\right)<0$