Unit-2:Algebra
(d)Quadratic equations
(e)Arithmetic progressions
Q1.
x = 1, is a solution of equation kx2 + 2x - 3 = 0, then the value of k is.
(a) 1 (b) 2 (c) 8 (d) 3
Solution.
Put x = 1 in the equation and equate with 0.
k(1)2 + 2(1) - 3 = 0
k + 2 - 3 = 0
k - 1 = 0
k = 1
Solution
Q2.
Which of the following is not a quadratic equation.
(a) x + 1/x = 1
(b) x2 - 8x - 10 = 0
(c) √5x2 - 2x + 1/2 = c
(d) x2 - 2√x + √5 = 0
Solution.
(d) x2 - 2√x + √5 = 0
Solution
Q3.
At the time of solving equation 9x2 - 15x + 6 = 0 by the method of completing the square which number is to be added both sides.
(a) (1/6)2 (b) 1/5 (c) (5/6)2 (d) (6/5)2
Solution.
(c) (5/6)2
Solution
Q4.
Write the discriminant of the quadratic equation x2 - 4x + 2 = 0
(a) 2 (b) 8 (c) 4 (d) 6
Solution.
Discriminant (D) = b2 - 4ac
= (-4)2 - 4 x 1 x 2 = 8
Solution
Q5.
x2 - 5x + k = 0 has real and equal roots if k is _________.
(a) 5 (b) 25 (c) 4 (d) 25/4
Solution.
Discriminant (D) = 0
b2 - 4ac = 0
25 - 4 x 1 x k = 0
25 = 4k
k = 25/4
Solution
Q6.
The sum of two numbers is 15. If the sum of their recipocals is 3/10, find the numbers .
(a) 7,8 (b) 9,6 (c) 10,5 (d)
12,3
Solution.
Let the numbers be x and 15 - x.
then 1/x + 1/15 - x = 3/10
after solving we get x = 10 or 5
Solution
Q7.
If Tn =
3n + 2/5
, find T6 .
(a) 5 (b) 2 (c) 3 (d) 4
Solution.
T
6 =
3 x 6 + 2/5
=
20 /5
= 4
Solution
Q8.
Find the 9th term from the end of the given A.P 17, 14, 11, ........ -40
(a) -16 (b) -8 (c) 24 (d) 8
Solution.
nth term from the end = l - (n - 1) d
so 9th term from the end
= -40 - (9 - 1)(-3)
= - 40 - 8(-3)
= -40 + 24 = - 16
Solution
Q9.
If the 8th term of an A.P. is 31 and the 15th term is 16 more than the 11th term, find a and d.
(a) 4,7 (b) 3,4 (c) 3,2 (d) 3,1
Solution.
T8 = 31 and T15 = 16 + T11
a + 14d = 16 + a + 10d
d = 4
a + 7d = 31
a + 7 x 4 = 31
a = 31 - 28 = 3
Solution
Q10.
If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is.
(a) 5n (b) 2n2
(c) 4n + 3 (d) 4n + 6
Solution.
Tn = Sn - Sn-1
Tn = 2n2 + 5n - {2(n - 1)2 + 5(n-1)}
T n = 2n2 + 5n - {2n2 + 2 - 4n + 5n - 5}
Tn = 2n2 + 5n - 2n2
+ 3 - n
Tn = 4n + 3
Solution
