Maths Mania # 138
DIRECTIONS: For the following questions, four options are given. Choose the correct option.
1.
If 5x 3 + 5x 2 – 6x + 9 is divided by (x + 3 ), the remainder is
(1) 135
(2) – 1 3 5
(3) 6 3
(4) –63
2.
Let f(x) = a0 x n + a1 x n–1 + a2 x n–2 + .... + an–1x + an, where a0 , a1 , a 2 , ...., an are constants. If f(x) is
divided by (ax – b), then the remainder is
FG b IJ
H aK
F bI
f G− J
H aK
F aI
f G J
H bK
F aI
f G− J
H bK
(1) f
(2)
(3)
(4)
3.
If f(x) is divided by (2x + 3), the remainder is
FG 2 IJ
H 3K
F 2I
f G− J
H 3K
F 3I
f G J
H 2K
F 3I
f G− J
H 2K
(1) f
(2)
(3)
(4)
4.
If (x 11 + 1) is divided by (x + 1), the remainder is
(1) 0
(2) 2
(3) 1 1
(4) 1 2
5.
The value of expression (16x 2 + 24x + 9) for x = –
3
is
4
(1) 2
(2) 1
(3) 0
(4) – 1
6.
If 2x 3 + 5x 2 – 4x – 6 is divided by 2x + 1, the remainder is
(1) −
13
2
(2) 3
(3) – 3
(4) 6
7.
If x3 + 5x 2 + 10k leaves remainder – 2x when divided by x2 + 2, then the value of k is
(1) – 2
(2) – 1
(3) 1
(4) 2
8.
When (x 3 – 2x 2 + px – q) is divided by (x 2 – 2x – 3), the remainder is (x – 6). The values of p and q
are
(1) p = – 2, q = – 6
(2) p = 2, q = –6
(3) p =– 2, q = 6
(4) p = 2, q = 6
9.
The sum of (x 2 + 1) and the reciprocal of (x 2 – 1) is
(1) 2x 2
(2)
(3)
(4)
x4
2
x −1
x2 − 2x + 2
x2 − 1
2x2
x4 − 1
10. For making (x 4 – 11x 2 y 2 + y4 ) a perfect square, the expression to be added is
(1) 5x 2y2
(2) 9x 2y2
(3) –5x 2 y2
(4) 7x 2y2
Detailed Solutions
1.
Remainder is f(–3) = 5 × (–3) 3 + 5 × (–3) 2 – 6 × (–3) + 9 = –63. Ans.(4)
2.
ax – b = 0 ⇒ x =
3.
2x + 3 = 0 ⇒ x = –
4.
Remainder is f(–1) = (–1) 11 + 1 = 0. Ans.(1)
5.
16x 2 + 24x + 9 = (4x + 3) 2
FG IJ
H K
3
F 3I
. So, remainder is f G − J. Ans.(4)
2
H 2K
b
b
. So, remainder is f
. Ans.(1)
a
a
3
will make it zero. Ans.(3)
4
∴ 4x + 3 is a factor, i.e. x = –
6.
2x + 1 = 0 ⇒ x = –
1
.
2
∴ Remainder is
FG 1 IJ = 2 × FG − 1 IJ
H 2 K H 2K
f −
7.
3
FG 1 IJ
H 2K
FG 1 IJ − 6 = −3. Ans.(3)
H 2K
2
+5× −
−4× −
On actually dividing x3 + 5x 2 + 10k by (x 2 + 2), the remainder obtained is –2x + 10k –10.
∴ –2x + 10k – 10 = –2x or k = 1. Ans.(3)
8.
On actual division, remainder is (p + 3) x – q.
∴ (p + 3) x – q = x – 6 ⇔ p + 3 = 1 & q = 6 ⇔ p = –2, q = 6. Ans.(3)
9.
Required sum = (x 2 + 1) +
ex
1
2
j
−1
=
x4 − 1 + 1
x4
=
. Ans.(2)
x2 − 1
x2 − 1
10. x 2 – 11x 2 y 2 + y 4 = [(x 2 ) 2 + (y 2 ) 2 – 2x 2 y 2 ] –9x 2 y 2 .
So, to make it a perfect square, we will have to add 9x 2 y 2 . Ans.(2)