Question

Let $\left|\begin{array}{cc}2\cos x& 3\sin x\\ e^x& 5\end{array}\right|=A+Bx+C{x}^2+\cdots (A,B,C\text{are real constants}),$ then which of the following(s) is/are true

• A. $A+B=13$
• B. $A-B=13$
• C. $A=10$
• D. $B=3$

Answer 1

Answer: (B), (C)

$A=10$

Given: $\left|\begin{array}{cc}2\cos x& 3\sin x\\ e^x& 5\end{array}\right|=A+Bx+C{x}^2+\cdots$
Put $x=0$ on both sides,
$\Rightarrow \left|\begin{array}{cc}2& 0\\ 1& 5\end{array}\right|=A$
$\therefore A=10$
Now, differentiating both sides w.r.t. $x$,
$\Rightarrow \left|\begin{array}{cc}-2\sin x& 3\cos x\\ e^x& 5\end{array}\right|+\left|\begin{array}{cc}2\cos x& 3\sin x\\ e^x& 0\end{array}\right|=B+2Cx+\cdots$
Now, putting $x=0$ on both sides,
$\Rightarrow \left|\begin{array}{cc}0& 3\\ 1& 5\end{array}\right|+\left|\begin{array}{cc}2& 0\\ 1& 0\end{array}\right|=B$
$\therefore B=-3$
$\Rightarrow A+B=7$
$\Rightarrow A-B=13$