Question

Let $\left|\begin{array}{cc}2\sec x& 3\tan x\\ e^x& {\tan }^{-1}x\end{array}\right|=A+Bx+C{x}^2+\cdots (A,B,C\text{are real constants}),$ then which of the following(s) is/are true?

• A. $A=0$
• B. $B=1$
• C. $A+B=1$
• D. $A-B=1$

Answer 1

Answer: (A), (D)

$A-B=1$

Given : $\left|\begin{array}{cc}2\sec x& 3\tan x\\ e^x& {\tan }^{-1}x\end{array}\right|=A+Bx+C{x}^2+\cdots$
Put $x=0$ on both sides,
$\Rightarrow A=\left|\begin{array}{cc}2& 0\\ 1& 0\end{array}\right|=0$
Differentiating both sides w.r.t. $x$,
$\Rightarrow \left|\begin{array}{cc}2\sec x\tan x& 3{\sec }^{2}x\\ e^x& {\tan }^{-1}x\end{array}\right|+\left|\begin{array}{cc}2\sec x& 3\tan x\\ e^x& \frac{1}{1+x^2}\end{array}\right|=0+B+2Cx+\cdots$
Now putting $x=0$ on both sides,
$B=\left|\begin{array}{cc}0& 3\\ 1& 0\end{array}\right|+\left|\begin{array}{cc}2& 0\\ 1& 1\end{array}\right|=-1$
$\Rightarrow A+B=-1$
$\Rightarrow A-B=1$