Question

$\Delta \left(x\right)=\left|\begin{array}{ccc}{e}^x& sin2x& tan{x}^2\\ \ln (1+x)& cosx& sinx\\ cos{x}^2& e^x-1& sin{x}^2\end{array}\right|=A+Bx+C{x}^2+\dots \dots$ then $B$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (A)

$0$

$\Delta \left(x\right)=\left|\begin{array}{ccc}{e}^x& \sin 2x& \tan x^2\\ \ln (1+x)& \cos x& \sin x\\ \cos x^2& e^x-1& \sin x^2\end{array}\right|=A+Bx+C{x}^2+$
Put $x=0$
$\Rightarrow \left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& 0& 0\end{array}\right|=A$
$\Rightarrow A=1$
So, $\Delta \left(x\right)=\left|\begin{array}{ccc}{e}^x& \sin 2x& \tan x^2\\ \ln (1+x)& \cos x& \sin x\\ \cos x^2& e^x-1& \sin x^2\end{array}\right|=1+Bx+C{x}^2+$
${\Delta }^{\prime }\left(x\right)=\left|\begin{array}{ccc}{e}^x& 2\cos 2x& 2x\sec x^2\\ \ln (1+x)& \cos x& \sin x\\ \cos x^2& e^x-1& \sin x^2\end{array}\right|+\left|\begin{array}{ccc}{e}^x& \sin 2x& \tan x^2\\ 1/(1+x)& -\sin x& \cos x\\ \cos x^2& e^x-1& \sin x^2\end{array}\right|+\left|\begin{array}{ccc}{e}^x& \sin 2x& \tan x^2\\ \ln (1+x)& \cos x& \sin x\\ -2x\sin x^2& e^x& 2x\cos x^2\end{array}\right|=B+2Cx+...$
Put $x=0$
$\Rightarrow \left|\begin{array}{ccc}1& 2& 0\\ 0& 1& 0\\ 1& 0& 0\end{array}\right|+\left|\begin{array}{ccc}1& 0& 0\\ 1& 0& 1\\ 1& 0& 0\end{array}\right|+\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 1& 0\end{array}\right|=B$
$\Rightarrow B=0$