Free Basic Algebra - 02 Practice Test - CAT
Question 1
The roots of the equation a2x2+abx=b2,a≠0 are:
SOLUTION
Solution : B
D=(ab)2+4a2b2=5a2b2
a2b2 has to be greater than or equal to 0 (i.e.,) D>0. So,roots are always real.
Question 2
Find the value of k so that sum of the squares of roots of equation x2−8x+k=0 is 30.
SOLUTION
Solution : D
Let the roots be a,b.
ab=k,a+b=8
a2+b2=30
(a+b)2=64,a2+b2+2ab=64
30+2k=64,k=17
Question 3
Let p, q be the roots of the equation x2−3x+2=0. The quadratic equation having roots as p+2 and q+2 is:
SOLUTION
Solution : B
Substitute x by (x-2)
The equation becomes: (x−2)2−3(x−2)+2=0,x2−7x+12=0.
Question 4
Let p, q be the roots of the equation x2−3x+2=0.
The quadratic equation having roots 2p and 2q is:
SOLUTION
Solution : D
Substitute x by( x2)
The equation becomes: (x2)2−3(x2)+2=0, x2−6x+8=0.
Question 5
Let p, q be the roots of the equation x2−3x+2=0.
The quadratic equation having roots 1p and 1q is:
SOLUTION
Solution : A
Soln:
Interchange ‘a’ and ‘c’ in the equation ax2+bx+c=0
The equation becomes: 2x2−3x+1=0. Hence option (a)
Question 6
If p, q are roots of an equation x2+5x+2=0 , then find out the value of pq+qp.
SOLUTION
Solution : C
Sum of the roots: p+q=−5
Product of the roots: pq=2
pq+qp=(p2+q2)pq=[(p+q)2−2pq]pq=[(p+q)2]pq−2=252−2=212
Question 7
If the roots of the equation 3x2+bx+3=0 are in the ratio 4:3 , then find out the value of b.
SOLUTION
Solution : A
For a quadratic equation ax2+bx+c=0 , if the roots are in ratio m:n then mnb2=(m+n)2ac.
In the equation given:
4×3×b2=(4+3)2(3×3)
12b2=49×9
b=7×3(2√3)=7√32
Question 8
If equations x2−kx−21=0 and x2−3kx+35=0;k>0 have a common root, then k is equal to:
SOLUTION
Solution : C
The equations x2−kx−21=0 and x2−3kx+35=0 have a common root.
So, equating the two equations:
x2−kx−21=x2−3kx+35
2kx=56
k=28x
Putting in the equation:
x2−28−21=0
x2=49
x=+7
or k=+4,As,k>0,k=4.
Question 9
If the roots of the equation: ax2+bx+c=0,a>0 be each greater than unity, then:
SOLUTION
Solution : A
Both the roots are greater than 1 and a > 0. So, the graph of the equation will be:
f(x)=ax2+bx+c
f(1)>a+b+c>0
Question 10
Find out the value of k for which the expression x2+(k+5)x−k−5=0 has two distinct real roots.
SOLUTION
Solution : B
For the equation given
D=(k+5)2+4(1)(k+5)=k2+10k+25+4k+20=k2+14k+45For the equation to have two distinct roots:
D>0k2+14k+45>0 (k+9)(k+5)>0
Value of k will be (−∞,−9)∪(−5,∞)
Question 11
The minimum value of the expression; a+1a;a>0 is:
SOLUTION
Solution : B
Soln:
a+1a=(√a−1√a)2+2
For minimum value of a+1a ,we want minimum value of (√a−1√a)2 that will be zero.
So, minimum value of a+1a will be 2. Hence option (b)
2nd method:- By, A.M ≥ G.M a+1a2≥√(a×1a thus a+1a>2 (i.e.,)(a+1a)min will be 2 . option (b).
Question 12
If p, q are the roots of the equation x2−3x+2=0 , find the equation which has roots as (2p + 1) and (2q + 1).
SOLUTION
Solution : D
Given equation: x2−3x+2=0
with roots p, qWhen the roots become 2p and 2q,
the equation becomes:
(x2)2−3(x2)+2=0,x2−6x+8=0
When the roots become
(2p+1) & (2q+1)
the equation becomes:(x−1)2−6(x−1)+8=0,x2−8x+15=0.
Hence option (D)
Question 13
For the equation x8+6x7−5x4−3x2+2x−5=0, determine the maximum number of real roots possible.
SOLUTION
Solution : A
f(x)=x8+6x7−5x4−3x2+2x−5
Check the number of positive roots= no. of sign changes in f(x) = 3
Check the number of negative roots = no. of sign changes in f(-x)
f(−x)=x8−6x7−5x4−3x2−2x−5 . No. of sign changes = 1
Now check for zero as a root. f(0)=−5≠0
So, maximum number of real roots = 3 + 1 = 4. Hence option (a)
Question 14
Suppose a and b are two roots of the equation x2−(α−4)x+α=0. Find out the maximum possible value of 5ab−a2−b2.
SOLUTION
Solution : B
a and b are the roots of the equation
a+b=(α−4),ab=α
5ab−a2−b2
5ab−(a2+b2)
5ab−[(a+b)2−2ab]
5α−[(α−4)2−2α]
5α−[α2−10α+16]
−α2+15α−16=−(α2+15α−16)=−(α2+16α−α−16)=−[α(α+16)−1(α+16)]=−(α+16)(α−1)
So,the maximum value =D4a=[152−4(−1)(−16)]4×1=1614.
Question 15
A two digit number is such that the product of its digits is 12. When 9 is added to the number, the digits interchange their places. The number is
SOLUTION
Solution :Let the two digit number be ab.
a×b=12
When 9 is added to the number
10a+b+9=10b+a
9a−9b+9=0
a−b+1=0
a−12a+1=0
a2+a−12=0
(a+4)(a−3)=0
a = -4 is not accepted. So, a = 3, b = 4
The number is 34 .