1.Algebra Booster

7.76 Algebra Booster
34. We have
fi
fi
3
x sin x cos x
f() x = 6 -1 0
p
2
p
3
p
2
3x cos x -sin
x
d
[ f ( x )] = 6 -1 0
dx
2 3
p p p
6 x –sin x - cos x
= 6 -1 0
2 3
p p p
6 -cos x -sin
x
3
d
[ f( x)] = 6 -1 0
3
dx
2 3
p p p
Now,
6 -1 0
Ê
3
d ˆ
Á [ f( x)] = 6 -1 0
Ë 3 ˜
dx ¯
p p p
x=
0 2 3
= 0
which is independent of p.
35. We have,
6i
-3i
1 26i
0 0
4 3i
- 1 = 4 3i -1 ( R1Æ R1+
iR3)
20 3 i 20 3 i
= 26i(–3 + 3)
= 0
Thus, x + iy = 0
fi x = 0, y = 0
36. We have,
1 x
x + 1
f() x = 2 x x( x - 1) x( x + 1)
3 xx ( -1) xx ( -1)( x- 2) xx ( + 1)( x-1)
1 x x + 1
2
= x ( x -1) 2 ( x - 1) ( x + 1)
3 ( x - 2) ( x + 1)
1 x x + 1
2
ÊR2 Æ R2 - R1ˆ
= x ( x -1) 1 -1 0 Á
ËR3 Æ R3 - R ˜
1¯
2 -2 0
2 2
1 -1
= x ( x -1) 2 - 2
= 0
37. We have,
sin q cos q sin q
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sin Áq + cos q + sin q +
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sin Áq - cos q - sin q -
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
sin q cos q sin q
Ê2pˆ Ê2pˆ Ê4pˆ
= 2 sin q cosÁ 2 cos q cos 2 sin 2q
cos
Ë
˜
3 ¯ Ë
Á
3 ¯
˜ Á
Ë
˜
3 ¯
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sin Áq - cos q - sin q -
Ë
˜ Á ˜ Á
3 ¯ Ë 3 ¯ Ë
˜
3 ¯
sin q cos q sin q
= -sin q -cos q -sin 2q
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sinÁq - cos q - sin q -
Ë
˜
3 ¯
Á
Ë
˜ Á ˜
3 ¯ Ë 3 ¯
sin q cos q sin q
= 0 0 0
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sinÁq - cos q - sin q -
Ë
˜
3 ¯
Á
Ë
˜ Á ˜
3 ¯ Ë 3 ¯
(R 2
Æ R 2
+ R 1
)
= 0
38. Since the given system of equation has a non zero solution,
then
1 -k
-1
k -1 - 1 = 0
1 1 -1
fi 1(2) + k(–k + 1) – (k + 1) = 0
fi 2 – k 2 + k – k – 1 = 0
fi 2 – k 2 – 1 = 0
fi k 2 = 1
fi k = ±1
Hence, the values of k are 1 and –1.
39. The given equation is
fi
fi
fi
sin x cos x cos x
cos x sin x cos x = 0
cos x cos x sin x
sin x+
2 cos x cos x cos x
sin x+ 2 cos x sin x cos x = 0
sin x+
2 cos x cos x sin x
1 cos x cos x
(sin x+ 2 cos x) 1 sin x cos x = 0
1 cos x sin x
1 cos x cos x
(sin x+ 2cos x) 0 sin x- cos x 0 = 0
0 0 sinx-cosx