A) \[9si{{n}^{2}}x\cos x\]
B) \[3{{\cos }^{2}}x\]
C) \[\sin x{{\cos }^{2}}x\]
D) None of these
Correct Answer: B
Solution :
[b] Operating \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] and taking out the common factor from \[{{C}_{1}},\] we have \[\Delta =\sin (x+h)(1+2cos\,h)\] \[\left| \begin{matrix} 1 & \sin (x+h) & \sin (x+2h) \\ 1 & \sin x & \sin (x+h) \\ 1 & \sin (x+2h) & \sin x \\ \end{matrix} \right|\] \[\{Since,\,\,sin(x+h)+sin(x)+sin(x+2h)\] \[=\sin (x+h)(1+2cosh)\}\] By operating \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\], We have \[\Delta =\sin (x+h)(1+2\,cos\,h)\] \[\left| \begin{matrix} 1 & \sin (x+h) & \sin (x+2h) \\ 0 & \sin x-\sin (x+h) & \sin (x+h)-\sin (x+2h) \\ 0 & \sin (x+2h)-\sin (x+h) & \sin x-\sin (x+2h) \\ \end{matrix} \right|\] \[=\left\{ \begin{align} & \left( 2\cos \left( x+\frac{h}{2} \right)\sin +\frac{h}{2} \right)(2cos(x+h)sin\,h) \\ & +\left( 2\cos (x+\frac{3h}{2} \right)\operatorname{si}{{\left. n\frac{h}{2} \right)}^{2}} \\ \end{align} \right\}\] Therefore, \[\underset{h\to 0}{\mathop{\lim }}\,\frac{\Delta }{{{h}^{2}}}=3\,\,{{\cos }^{2}}x\].
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