JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 2)
If $${\Delta _1} = \left| {\matrix{
x & {\sin \theta } & {\cos \theta } \cr
{ - \sin \theta } & { - x} & 1 \cr
{\cos \theta } & 1 & x \cr
} } \right|$$ and
$${\Delta _2} = \left| {\matrix{ x & {\sin 2\theta } & {\cos 2\theta } \cr { - \sin 2\theta } & { - x} & 1 \cr {\cos 2\theta } & 1 & x \cr } } \right|$$, $$x \ne 0$$ ;
then for all $$\theta \in \left( {0,{\pi \over 2}} \right)$$ :
$${\Delta _2} = \left| {\matrix{ x & {\sin 2\theta } & {\cos 2\theta } \cr { - \sin 2\theta } & { - x} & 1 \cr {\cos 2\theta } & 1 & x \cr } } \right|$$, $$x \ne 0$$ ;
then for all $$\theta \in \left( {0,{\pi \over 2}} \right)$$ :
$${\Delta _1} - {\Delta _2}$$ = x (cos 2$$\theta $$ – cos 4$$\theta $$)
$${\Delta _1} + {\Delta _2}$$ = - 2x3
$${\Delta _1} + {\Delta _2}$$ = – 2(x3 + x –1)
$${\Delta _1} - {\Delta _2}$$ = - 2x3
Explanation
$${\Delta _1} = \left| {\matrix{
x & {\sin \theta } & {\cos \theta } \cr
{ - \sin \theta } & { - x} & 1 \cr
{\cos \theta } & 1 & x \cr
} } \right|$$
= x(–x2 –1) – sin$$\theta $$(–xsin$$\theta $$ – cos$$\theta $$) + cos$$\theta $$(–sin$$\theta $$+ xcos$$\theta $$)
= –x3 – x + xsin2$$\theta $$ + sin$$\theta $$cos$$\theta $$ – cos$$\theta $$sin$$\theta $$ + xcos2$$\theta $$
= –x3 – x + x = –x3
Similarly $${\Delta _2} = - {x^3}$$
$${\Delta _1} + {\Delta _2} = - 2{x^3}$$
= x(–x2 –1) – sin$$\theta $$(–xsin$$\theta $$ – cos$$\theta $$) + cos$$\theta $$(–sin$$\theta $$+ xcos$$\theta $$)
= –x3 – x + xsin2$$\theta $$ + sin$$\theta $$cos$$\theta $$ – cos$$\theta $$sin$$\theta $$ + xcos2$$\theta $$
= –x3 – x + x = –x3
Similarly $${\Delta _2} = - {x^3}$$
$${\Delta _1} + {\Delta _2} = - 2{x^3}$$
Comments (0)
