JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 4)
The number of distinct real roots of the equation,
$$\left| {\matrix{ {\cos x} & {\sin x} & {\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0$$ in the interval $$\left[ { - {\pi \over 4},{\pi \over 4}} \right]$$ is :
$$\left| {\matrix{ {\cos x} & {\sin x} & {\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0$$ in the interval $$\left[ { - {\pi \over 4},{\pi \over 4}} \right]$$ is :
4
3
2
1
Explanation
Given,
$$\left| {\matrix{ {\cos x} & {\sin x} & {\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0$$
R1 $$ \to $$ R1 $$-$$ R3
R1 $$ \to $$ R2 $$-$$ R3
$$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & {\sin x - \cos x} \cr {\sin x} & {\sin x} & { \cos x} \cr } } \right| = 0$$
C3 $$ \to $$ C3 + C2
$$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & 0 \cr {\sin x} & {\sin x} & {\sin x + \cos x} \cr } } \right| = 0$$
Expanding using first column,
(cosx $$-$$ sinx)(cos $$-$$ sinx) (sinx + cos x)
+ sinx (cosx $$-$$ sinx) (sinx $$-$$ cosx) = 0
$$ \Rightarrow $$ (cosx $$-$$ sinx)2 (sinx + cosx)
$$+$$ sinx (cosx $$-$$ sinx)2 = 0
$$ \Rightarrow $$ (cosx $$-$$ sinx)2 (sinx + cosx $$+$$ sinx) = 0
$$ \Rightarrow $$ (2sinx + cosx )(cosx $$-$$ sinx)2 = 0
$$ \therefore $$ cosx = -2sinx or cosx = sinx
$$ \Rightarrow $$ tanx = $$ - {1 \over 2}$$ or tanx = 1
$$ \therefore $$ x = $$ - {\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$, $${\pi \over 4}$$
$$ \therefore $$ Number of solutions = 2
$$\left| {\matrix{ {\cos x} & {\sin x} & {\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0$$
R1 $$ \to $$ R1 $$-$$ R3
R1 $$ \to $$ R2 $$-$$ R3
$$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & {\sin x - \cos x} \cr {\sin x} & {\sin x} & { \cos x} \cr } } \right| = 0$$
C3 $$ \to $$ C3 + C2
$$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & 0 \cr {\sin x} & {\sin x} & {\sin x + \cos x} \cr } } \right| = 0$$
Expanding using first column,
(cosx $$-$$ sinx)(cos $$-$$ sinx) (sinx + cos x)
+ sinx (cosx $$-$$ sinx) (sinx $$-$$ cosx) = 0
$$ \Rightarrow $$ (cosx $$-$$ sinx)2 (sinx + cosx)
$$+$$ sinx (cosx $$-$$ sinx)2 = 0
$$ \Rightarrow $$ (cosx $$-$$ sinx)2 (sinx + cosx $$+$$ sinx) = 0
$$ \Rightarrow $$ (2sinx + cosx )(cosx $$-$$ sinx)2 = 0
$$ \therefore $$ cosx = -2sinx or cosx = sinx
$$ \Rightarrow $$ tanx = $$ - {1 \over 2}$$ or tanx = 1
$$ \therefore $$ x = $$ - {\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$, $${\pi \over 4}$$
$$ \therefore $$ Number of solutions = 2
Comments (0)
