Mathematics-6.Unit-4.01-Quadratic Equation With Solution
QUADRATIC EQUATIONS
1. INTRODUCTION
An equation of the form anxn + an – 1xn – 1 + an – 2xn – 2 + …. + a1x + a0 = 0 (a0, a1…., an are real coefficients and an 0) is called a polynomial equation of degree n. This equation is called a linear equation if n=1, quadratic equation if n=2, cubic equation if n=3, bi-quadratic equation if n=4…and so on.
1.1 DEFINITION
An equation of the form ax2 + bx + c = 0, where a 0 and a, b, c are real numbers, is called a quadratic equation. The numbers a, b, c are called the coefficients of the quadratic equation.
Root of a quadratic equation:
A root of the quadratic equation is a number (real or complex) such that a2 + b + c = 0. The roots of the quadratic equation are given by x = . Let and be two roots of the given quadratic equation then + = - and = .
Discriminant of a quadratic equation:
The quantity D (D= b2 - 4ac) is known as the discriminant of a quadratic equation.
The quadratic equation has real and equal roots if and only if D = 0
i.e. b2 - 4ac = 0.
The quadratic equation has real and distinct roots if and only if D > 0
i.e. b2 - 4ac > 0
The quadratic equation has complex roots with non-zero imaginary parts if and only if D < 0 i.e. b2 - 4ac < 0. If p + iq (p and q being real) is a root of the quadratic equation where i = , then p - iq is also a root of the quadratic equation.
Illustration 1: If the roots of the quadratic equation x2 – ax + b = 0 are real and differ by a quantity less than 1, then prove that .
Key concept: Since the roots of the given equation differ by less then 1 hence roots are real and distinct, hence the D>0. Also the difference between the roots should be less then one. Apply the identity ( – )2 = ( + )2 – 4
Solution: Clearly D=a2 – 4b > 0 b < ……….…….(1)
Now ( – )2 = ( + )2 – 4= a2 – 4b , where and are roots of the given equation.
| – |2 = a2 – 4b, but given that | – | < 1 |a2 – 4b| < 1
a2 – 4b < 1 < b ……(2)
From (1) and (2), .
Illustration 2: If c ¹ 0 and the equation has the equal roots then find the value of p.
Key concept: First simplify the given equation, then since roots are real and equal apply the condition that discriminant should be equal to zero.
Solution: (2a + 2b p) x2 2c (a b) x + pc2 = 0
for equal roots, c2 (a b)2 pc2 (2a +2b p) = 0
(a b)2 2p (a +b) + p2 = 0
[ p (a + b) ]2 = (a + b)2 (a b)2 = 4ab
p (a+b) = 2 p =
1.2 BASIC RESULTS RELATED TO A QUADRATIC EQUATION
If p + is an irrational root of the quadratic equation, then p - is also a root of the quadratic equation provided that all the coefficients are rational.
Illustration 3: If , be the roots of the equation x2 + 2ax + b = 0, form a quadratic with rational with rational coefficients one of whose roots is + + .
Key concept: since the coefficients of the given equation are rationals and one given root is irrational, hence its second root will also be irrational and conjugate of the given root.
Solution: Given + = – 2a and = b
Clearly roots of the equired equation will be + + and + – .
Hence, the sum of the roots of the required equation is
= 2( + ) = – 4a, and the product of the roots of the required equation is= ( + )2 – (2 + 2) = 2 = 2b
Hence, the required equation is
x2 + 4ax + 2b = 0 .
The quadratic equation has rational roots if D is a perfect square and a, b, c are rational.
If a, b, c are odd integers, then the roots of quadratic equation can’t be rational.
If a = 1 and b, c are integers and the roots of the quadratic equation are rational, then the roots must be integers.
Illustration 4: Find the total number of values of a so that x2 – x – a = 0 has integral roots , where a N, 6 a 100.Key concept: Since all the coefficients of the given equation are integers, hence for roots to be integer discriminant must be a perfect square.
Solution: Now D = 1 + 4a = which is an odd integer.
Hence it will be in the form of D = (2 + 1)2
1 + 4a = 1 + 44 + 4
a = ( + 1), hence a should be in the form of product of two consecutive integers. Since a [6, 100], therefore
a = 6, 12, 20, 30, 42, 56, 72, 90
Thus 'a' can attain 8 different values.
A quadratic equation, whose roots are and can be written as (x - ) (x - ) = 0. i.e., ax2 + bx + c a(x - ) (x - ).
Illustration 5: Let , be the roots of the equation (x - a) (x - b) = c , c 0. Then find the roots of the equation (x - )(x - ) + c = 0 in terms of a and b.
Solution: since the roots of the equation(x - a) (x - b) – c=0 are , hence this equation can be written as (x-a) (x-b) – c= (x-)(x-)
(x-)(x-) +c (x-a) (x-b).
Clearly roots of ( x- ) ( x-) + c = 0 are a and b.
Illustration 6: Find the equation whose roots are negative of the roots of the equation x3 –5x2 –7x –3=0.
Key concept: To obtain an equation whose roots are negative of the roots of a given equation replace x by –x.
Solution: (-x)3 –5 (-x)2 –7(-x) –3 = 0
or -x3 –5x2 + 7x –3 = 0 or x3 + 5x2 –7x + 3 = 0
If a + b + c = 0, then 1 is a root of ax2 + bx + c = 0.
If a + b + c = 0 and a, b, c are rational, then roots of equation ax2 + bx + c = 0 are rational.
Illustration 7: If the roots of (a - b)x2 + (b - c)x + (c - a) = 0 are real and equal, then show that 2a = b + c.
Key concept: Since the roots of the given equation are real and equal hence discriminant must be equal to zero.
Solution 1: Since roots of given equation are real and equal. Hence D = 0
(b – c)2 – 4(a – b)(c – a) = 0.
(2a – b – c)2 = 0 2a = b + c.
Key concept: Since sum of all the coefficients in the given equation is zero hence one will be one root of the given equation. Now given that both the roots are equal hence both the roots will be one and their product will also be equal to one.
Solution 2: Product of the roots 1 . 1 =
2a = b + c
Illustration 8: If a, b, c, d are four non–zero real numbers such that (d + a – b)2 + (d + b – c)2 = 0 and roots of the equation a(b – c)x2 +b(c – a)x + c(a – b) = 0 are real and equal, then show that a=b=c.
Key concept: In the first given equation we will use the concept that if . And in the second equation since sum of all the coefficients is zero and given both the roots are equal product their roots will be equal to one.
Solution: From the first equation(d + a – b)2 + (d + b – c)2 = 0
d + a – b=0 and d + b – c 2b = a + c…………..(1)
And from the second given equation product of the roots is equal to one.b = …(2)
From (1) and (2) = a + c (a – c)2 = 0 a = c
Thus a = b = c
Quadratic expression ax2 + bx + c is a perfect square (i.e. square of a polynomial of degree one) if b2 – 4ac = 0.
Illustration 9: Show that expression x2 + 2 (a + b +c) x + 3(bc + ca + ab) will be a perfect square if a = b = c .
Key concept: The given expression f(x)= x2 + 2 (a + b +c) x + 3(bc + ca + ab)will be a perfect square if both the roots of the equation f(x)=0 are equal. Hence discriminant of f(x)=0 must be equal.
Solution: Discriminant of f(x)=0, D=4(a + b + c)2 – 12(ab + bc + ca) = 0
a2 + b2 + c2 – ab – bc – ca = 0
(a – b)2 + (b – c)2 + (c – a)2 = 0
a = b = c
1.3 IDENTITY
If the quadratic equation is satisfied by more than two distinct numbers (real or complex), then it becomes an identity i.e. a = b = c = 0. For example
is satisfied by three value of x which are a, b and c. Hence this is an identity in x.
Illustration 10: Let p(x) = .c2 + .a2 + .b2,
a b c. Prove that p(x) has the property that p(y) = y2 for all y R
Solution: Note that P(a) = a2, P(b) = b2 and P(c) = c2.
Consider the polynomial Q(x) = P(x) – x2, Q(x) has degree atmost 2.
Also Q(a) = Q(b) = Q(c) = 0 Q(x) has 3 distinct roots.
It follows that Q(x) is identically zero
i.e. Q(y) = 0 y R
P(y) – y2 = 0 y R P(y) = y2 y R
1.4 CONDITION FOR COMMON ROOT(S)
Let ax2 + bx + c = 0 and dx2 + ex + f = 0 have a common root (say). Then
a2 + b + c = 0 and d2 + e + f = 0 Solving for 2 and , we get
i.e. 2 = and =
(dc - af)2 = (bf - ce) (ae - bd)
which is the required condition for the two equations to have a common root.
Important results:
To find the common root of two equations, make the coefficient of second degree terms in both the equations equal and then subtract the two equations. The value of x so obtained is the required common root.
Condition for both the roots to be common is .
Illustration 11: If the equation of x2 + 2xsiny + 1 = 0, where y (0, /2) and ax2 + x + 1 = 0 have a common root, then find the value of a and y.
Key concept: If two quadratic equations with real coefficients have an imaginary root common, then both roots will be common and the two equations will be identical. .
Solution: Since discriminate of x2 + 2xsiny + 1 = 0 is 4sin2y – 4 < 0. Hence roots of this equation are imaginary. Now this equation and ax2 + x + 1 = 0 have a common root, hence roots of ax2 + x + 1 = 0 are also imaginary. This implies given two equations have both the roots common. Hance both the equation are identical
a = 1, y = /6.
Illustration 12: If equations ax3 + 2bx2 + 3cx + 4d = 0 and ax2 + bx + c = 0 have a non-zero common root, then prove that (c2 – 2bd) (b2 – 2ac) 0
Solution: Let be the non-zero common root
so, a3 + 2b2 + 3c + 4d = 0 ....(1)
a2 + b + c = 0 ....(2)
(1) - (2) gives
b2 + 2c + 4d = 0 ....(3)
by (2) and (3)
.
(4ad-bc)2 = 2(b2 – 2ac) (c2 – 2bd) (b2 – 2ac) (c2 – 2bd) 0
1.5 TO FIND THE RANGE OF A RATIONAL EXPRESSTION IN X, WHERE X IS REAL
STEP 1: Put the given rational expression equal to y and form the quadratic equation in x.
STEP 2: Find the discriminant D of the quadratic equation obtained in step 1.
STEP 3: Since x is real, therefore, put D 0. We get an inequation in y.
STEP 4: Solve the above inequation for y. The range of y so obtained determines the range attained by the given rational expression.
Illustration 13: Find the maximum and minimum values of for real values of x.
Solution: Let = y y(x2 + x + 1) = x2 – x + 1
(y – 1)x2 + (y + 1)x + (y – 1) = 0. On solving for x, we get
x= =
Now for real values of x we will get real values of y, hence x to be real
.
1.6 QUADRATIC EXPRESSION
The expression ax2 + bx + c is said to be a real quadratic expression in x where a, b, c are real and a 0. Let f(x) = ax2 + bx + c where a, b, c, R (a 0). Now f(x) can be rewritten as f(x) = = a ……….(1), where D = b2 – 4ac is the discriminant of the quadratic expression. From (1) it is clear that f(x) = ax2 + bx + c will represent a parabola whose axis is parallel to the y-axis, and vertex is at A .
It is also clear that if a > 0, the parabola will be open upward and if a < 0 the parabola will be open downward and it depends on the sign of b2 –4ac that the parabola cuts the x-axis at two points ( b2-4ac > 0), touches the x-axis (b2- 4ac = 0) or never intersects with the x-axis(b2-4ac < 0).
Case I: If a > 0
Sub case A: a > 0 and b2 - 4ac < 0 f(x) > 0 x R.
In this case the parabola always remains open upward and above the x-axis.
Sub case B: a > 0 and b2 – 4ac = 0 f(x) 0 x R.
In this case the parabola touches the x-axis at one point and remains open upward.
Sub case C: a > 0 and b2 - 4ac > 0. Let f(x) = 0 has two real roots and ( < ). Then f(x) > 0 x (-, )(, )and f(x) < 0 x (, )
In this case the parabola cuts the x- axis at two points and and remains open upward.
Greatest and least value of a quadratic expression ax2 + bx + c when a>0:
In this case ax2 + bx + c has no greatest value and it has least value at x = – .
Case II: If a > 0
Sub case A: a < 0 and b2 - 4ac < 0 f(x) < 0 x R.
In this case the parabola remains open downward and always below the x-axis.
Sub case B: a < 0 and b2 – 4ac = 0 f(x) 0 x R.
In this case the parabola touches the x - axis and remains open downward.
Sub case C: a < 0 and b2 - 4ac > 0.Let f(x) = 0 have two real roots and ( < ).Then f(x) < 0 x (-, )(, ) and f(x) > 0 x (, ).
Greatest and least value of a quadratic expression ax2 + bx + c when a<0:
If a < 0, then ax2 + bx + c has no least value and it has greatest value at x = – .
Illustration 14: If min {x2 + (a – b)x + (1 – a – b)} > max (-x2 + (a +b)x – (1 + a + b)) Prove that a2 + b2 < 4
Solution: Given min{x2 + (a – b)x + (1 – a – b)} > max{-x2 + (a +b)x – (1 + a + b)}
min
1 – a – b – > – (1 + a + b)
a2 + b2 < 4
Illustration 15: If ax2- bx + 5 = 0 does not have 2 distinct real roots, then find the minimum value of 5a +b.
Key concept: Since the given equation doesn’t have two real distinct roots, hence its roots will either be imaginary or real and equal. Now the parabola of f(x)= ax2- bx + 5 will either be above the x-axis or will always be below the x-axis. But f(0)=5>0, hence graph of f(x) will always be above the x-axis. Hence f(x)0 for all x.
Solution: Now f(x) 0 x R
In particular f(-5) 0 25a +5 b +5 0 5a +b - 1
Hence the least value of 5a +b is - 1.
1.7 ROOTS LIE IN AN INTERVAL
Here basically we will discuss different necessary and sufficient conditions we should impose on a quadratic equation ax2 + bx + c = 0 such that roots of the given equation lies in a particular interval. Since a 0 , we can take f(x) = x2 +
Case I: Both the roots are positive i.e. they lie in (0, ), then the sum of the roots as well as the product of the roots must be positive.
+ = - and = with b2 – 4ac 0.
Case II: Both the roots are negative i.e. they lie in (- , 0), then the sum of the roots must be negative and the product of the roots must be positive.
i.e. + = - < 0 and = with b2 – 4ac 0.
Case III: One root is positive and other is negative i.e. origin is lying between the roots. Clearly f(0)<0 is the necessary and sufficient condition.
Case IV: Both the roots are greater then a real number k.
D 0 … (1)
f(k) > 0 …(2)
> k …(3)
These are the necessary & sufficient conditions.
Case V: If both the roots are less than a real number k.
D 0 … (1)
f(k) > 0 …(2)
< k …(3)
These are the necessary & sufficient conditions.
Case VI: A real number k is lying between the roots i.e. one root is less then k and other is greater then k.
D > 0 … (1)
f(k) < 0 … (2)
These are the necessary & sufficient conditions.
Case VII: exactly are root is lying between k1 and k2
f(k1) < 0 and f(k2) > 0 f(k1) > 0 and f(k2) < 0
Hence the required condition is f(k1). f(k2) < 0
Illustration 16: For the quadratic equation x2 – (m – 3)x + m = 0, find the value of m for which
(i) one root is smaller than 2 and the other is greater than 2
(ii) both roots are grater than 2
(iii) both roots lie in (1, 2)
(iv) exactly one root lie in (1, 2)
Solution: Let f(x) = x2 – (m – 3)x + m and D is = (m – 1)(m – 9)
(i) (a) D> 0 and (b) f (2) < 0
i.e., m < 1 or m > 9 and, m > 10
m (10, )
(ii) The required necessary and sufficient conditions are
D 0Þ m 1 or m 9………(1)
f(2) > 0 Þ m < 10………..(2)
> 2Þ m > 7……………(3)
From (1),(2) and (3) m [9, 10)
(iii) D 0Þ m1 or m9 ……(1)
af(1) > 0Þ 4>0 mR ……(2)
af(2) > 0Þ m < 10 ……(3)
1 < < 2Þ m>5 and m<7 ……(4)
Taking intersection of these four conditions , we get m .
(iv) D > 0 Þ m < 1 or m > 9 ……(1)
f(1) . f(2) < 0Þ m > 10 ……(2)
Taking intersection of these two conditions, we get m (10, )
Illustration 17: Find ‘a’ for which exactly one root of the equation 2ax2 – 4a. x + 2a – 1 = 0 lies between 1 and 2.
Solution: F(1). F(2) < 0 (2a – 4a + 2a – 1) (4. 2a – 2.4a + 2a – 1) < 0
< 2a < 1 log2(1/2) < a < log21
- 1 < a < 0a (-1, 0)
Illustration 18: Find the values of k for which the expression 12x2 – 10xy + 2y2 + 11x – 5y + k is the product of two linear factors.
Solution: Rearrange as the corresponding quadratic equation in x.
12x2 + x(11 – 10y) + (2y2 – 5y + k) = 0
Solving, x =
The factors will be linear only if
(10y – 11)2 – 48(2y2 – 5y + k) is a perfect square
if 4y2 + 20y + (121 – 48k) is a perfect square
if its discriminant 400 – 16 (121 – 48k) = 0
if k = 2
1.8 RELATION BETWEEN THE ROOTS OF A POLYNOMIAL EQUATION OF DEGREE N
Consider the equation
anxn + an – 1xn – 1 + an – 2xn – 2 + …. + a1x + a0 = 0 . . . . (1)
( where a0, a1…., an are real coefficients and an 0)
Let 1, 2,….,n be the roots of equation (1). Then
anxn + an – 1xn – 1 + an – 2xn – 2 + ….. + a1x + a¬0 an(x - 1) (x - 2) ….. (x - n)
Comparing the coefficients of like powers of x, we get
1 + 2 + 3 + …. + n = -
1¬2 + 13 + 14 + …. + 23 + … + n - 1n =
………………………………
1¬2 . . . . .r + …. + n-r+1n-r+2 … n = ( -1)r
…….…………………………
12 … n = (-1)n
e.g. If , , and are the roots of ax4 + bx3 + cx2 + dx + e = 0 then
+ + + = -b/a
+ + + + + = c/a
+ + + = -d/a
= e/a
Some important results:
A polynomial equation of degree n has n roots (real or imaginary).
If all the coefficients are real then the imaginary roots occur in pairs i.e. number of complex roots is always even.
If the degree of a polynomial equation is odd then the number of real roots will also be odd. It follows that at least one of the roots will be real.
Factor theorem: If is a root of the equation f(x)=0, then f(x) is exactly divisible by (x–) and conversely, if f(x) is exactly divisible by (x–) then is a root of the equation f(x)=0.
Let f(x)=0 be a polynomial equation and p and q are two real numbers, then f(x)=0 will have at least one real root or an odd number of roots between p and q if f(p) and f(q) are of opposite sign. But if f(p) and f(q) are of same signs, then either f(x)=0 has no real roots or an even number of roots between p and q.
If is repeated root repeating r times of a polynomial equation f(x) = 0 of degree n i.e. f(x) = (x - )r g(x) , where g(x) is a polynomial of degree n - r and g()0,then f() = f() = f() = . . . . = f (r-1)() = 0 and f r () 0.
The cubic function f(x) = ax3 + bx2 + cx + d, where x R
Take a > 0, the graph of f has the following properties:
As x , y because the x3 term is positive and will dominate the remaining terms when x is large.
As x , y –
A consideration of (i) and (ii) implies that the graph of f must cross the x–axis at least once, taking this point x = (say), we have
ax3 + bx2 + cx + d = (x – )Q where Q is a quadratic expression in x.
The equation Q = 0 may have two real distinct roots, two real coincident roots or no real roots.
f(x) = 3ax2 + 2bx + c and the equation f(x) = 0 may have two real distinct roots, two real coincident roots or no real roots.
If f(x) = 0 has two real distinct roots p and q (say) where p > q, f(p) is a minimum value of f(x) and f(q) is maximum value of f(x).
If f(x) = 0 has two real coincident roots, r (say) then the stationary value of f(x) at (r, f(r)) is a point of inflexion.
If f(x) = 0 has no real roots, the graph of f has no stationary points.
f(x) = 6ax + 2b and f(x) = 0, f(x) 0 when x = –
This implies that all cubic curves have a point of inflexion.
Illustration 19: If b2 < 2ac, then prove that ax3 + bx2 + cx + d =0 has exactly one real root.
Solution 1: Let , , be the roots of ax3 +bx2 +cx +d =0
Then + + = , + + = and =
Also 2 +2 +2 = ( + + )2 – 2( + +) = .
2 +2 + 2 < 0 , which is not possible if all , , are real.
So atleast one root is non-real, but complex roots occurs in pair. Hence given cubic equation has two non-real and one real roots.
Solution 2: Let f(x) = ax3 + bx2 + cx + d and f(x) = 0 has all the roots real
f (x) = 3ax2 +2bx +c = 0 has two real roots
But its discriminant = (2b)2 – 4. 3.ac = (b2 – 2ac) – ac < 0 ( as b2 < 2ac )
which is a contradiction f(x) =0 will not have all the roots real.
2. OBJECTIVE PROBLEMS
1: If a, b, c be the sides of ABC and equations ax2 + bx + c = 0 and 5x2 + 12x + 13 = 0 have a common root, then C is
(A) 60° (B) 90°
(C) 120° (D) 45°
Solution: (B) Since 5x2 + 12x + 13 = 0 has imaginary roots as D = 144 – 4 5 13 < 0
So, both roots of ax2 + bx + c = 0 and 5x2 + 12x + 13 = 0 will be common
a2 + b2 = c2 C = 90°
2: If f(x) = (x – a1)2 +(x – a2)2 + ……….(x – an)2 find x where f(x) is minimum.
(A) (B)
(C) (D) –
Solution: (B) minimum value of f(x) = nx2 – 2[a1 + a2 + …….an] x + a12 + a22 … an2
exist at x = .
3. The values of a for which both the roots of the equation (1 – a2)x2 + 2ax – 1 = 0 lie between 0 and 1 are given by
(A) a > 2 (B) 1 < a < 2
(C) – < a < (D) none of these
3. (A)
Given equation is (1 – a2)x2 + 2ax – 1 = 0
Its discriminant D = 4 and roots are
Given, 0 < < 1, 0 < < 1
Now, > 0 a > 1,
< 1 – 1 < 0
< 0 a < 1 or a > 2
a > 2 … (1)
< 1 a > – 1
and < 1 – < 0
a < –1 or a > 0
a > 0 … (2)
From (1) and (2) , a > 2
4. The least value of | a | for which sin and cosec are the roots of the equation x2 + ax + b = 0 is
(A) 2 (B) 1
(C) 1/2 (D) 0
4. (A)
sin + cosec = –a
| a | = | sin + cosec |
= | sin | + 2
Hence least value of | a | = 2
5. The sum of all the value of m for which the roots x1 and x2 of the quadratic equation x2 – 2mx + m = 0 satisfy the condition , is :
(A) (B) 1
(C) (D)
5. Given x1 x2 2m x1x2 m
According to given condition
(x1 x2)
(x1 x2) ((x1 x2)2 – 3x1x2) (x1 x2)2 – 2x1x2
2m(4m2 – 3m) 4m2 – 2m
Clearly since is
6. If x2 – x + a – 3 < 0 for atleast one negative value of x, then complete set of values of 'a' is
(A) (–, 4) (B) (–, 2)
(C) (–, 3) (D) (–, 1)
6. (C)
The equation x2 – x + a –3 = 0 must have must have atleast on negative root.
For real roots, D 0
1 – 4(a – 3) 0
a
Both root will be non–negative if
D 0, a – 3 0, 1 0
a , a 3
a
Thus equation will atleast one negative root if a
a (–, 3)
7. If the quadratic equations 3x2 + ax + 1 = 0 and 2x2 + bx + 1 = 0 have a common root then the value of 5ab – 2a2 – 3b2, where a, b R, is equal to
(A) Zero (B) 1
(C) –1 (D) none of these
7. (B)
3x2 + ax +1 = 0, 2x2 + bx + 1 = 0 have a common root.
Subtracting these equations, we get
x2 + (a – b)x = 0
x = 0, x = (b – a)
Clearly the common root is (b – a)
3(b – a)2 + a(b – a) + 1 = 0
3b2 + 3a2 – 6ab + ab – a2 + 1 = 0
5ab – 2a2 – 3b2 = 1
8. If x2 + 5 = 2x – 4cos(a + bx), where a, b (0, 5), is satisfied for atleast one real x, then the maximum value of a + b is equal to
(A) 3 (B) 2
(C) (D) none of these
8. (A)
x2 + 5 = 2x – 4 cos(a + bx)
x2 – 2x + 1 + 4 = –4 cos(a + bx)
(x – 1)2 + 4(1 + cos(a + bx)) = 0
x = 1 and 1 + cos(a + bx) = 0
cos(a + b) = –1
a + b = , 3 ……
9. The set of values of a for which 1 lies between the roots of equation x2 – ax – a + 3 = 0 is
(A) (–, –6) (B) (–, –6]
(C) (–, –6) (2, ) (D) (2, )
9. (D) Let f(x) = – ax – (a – 3) = 0
Let and be the roots of equation
f(x) = 0
Since 1 lies between and
f(1) < 0 1 – a – a + 3 < 0 a > 2
10. The quadratic equation (2x –a) (2x –c) + (x –2b) (x –2d) = 0, (where 0 < 4a < 4b < c < 4d) has
(A) a root between b and d for all
(B) a root between b and d for all –ve
(C) a root between b and d for all +ve
(D) none of these
10. Let f(x) (2x – a) (2x – c) + (x – 2b)(x – 2d)
Clearly f(b) . f(d) < 0, .
Hence a root of given equation will lie between b and d.
11. If , are the roots of the equation ax2 + bx + c = 0, then the quadratic equation whose roots are and is
(A) ax2 – bx(1 – x) + c(1 – x)2 = 0 (B) ax2 – bx (x – 1) + c (x – 1)2 = 0
(C) ax2 + bx(1 – x) + c(1 – x)2 = 0 (D) ax2 + bx (x + 1) + c (1 + x)2 = 0
11. Since roots of ax2 bx c 0 are and . Hence roots of cx2 bx a 0
will be and . Now if we replace x x – 1, then roots of c(x – 1)2 b(x – 1) a 0
will be 1 and 1 . Now arg an replace x , we will get
c(1 – x)2 b(1 – x) ax2 0 whose roots are and .
12. If a, b, c R and x2 (a b)x c 0 has no real roots then
(A)c (a b c ) > 0 (B) c c (a b c ) > 0
(C)c c (a b – c ) > 0 (D) c (a b – c ) > 0
12. Since the roots of ax2 bx c 0 are non real. Thus
f(x) ax2 bx c will have same sign for every value of x.
f(0) c,
f(1) a b c,
f(–1) a – b c
f(2) 4a – 2b c
c. (a b c) > 0, c(a – b c) > 0,
c (4a – 2b c) > 0
13. If a + b + c > and equation ax2 + 2bx – 5c = 0 has non–real complex roots, then
(A) a > 0, c > 0 (B) a > 0, c < 0
(C) a < 0, c < 0 (D) a < 0, c > 0
13. (B)
Given, 4a + 4b – 5c > 0
Let f(x) = ax2 + 2bx – 5c, then
f(2) = 4a + 4b – 5c > 0
Since equation f(x) = 0 has imaginary roots, therefore f(x) will have same sign as that of a for all x R. Since f(2) > 0
14. If a, b, c R, a 0 and (b – 1)2 < 4ac, then the number of real roots of the system of equation (in three unknowns x1, x2, x3)
a + bx1 + c = x2, a + bx2 + c = x3, a + bx3 + c = x1 is
(A) 0 (B) 1
(C) 2 (D) 3
14. (A)
Let f(x) = ax2 + (b – 1)x + c
Given system of equations is equivalent to
f(x1) + f(x2) + f(x3) = 0
af(x1) + af(x2) + af(x3) = 0 (not possible)
as (b – 1)2 – 4ac < 0
af(x1), af(x2), af(x3) > 0
Hence given system of equations has no real root.
15. For the equation 3x2 + px + 3 = 0, p > 0 , if one of the roots is the square of the other, then p is equal to
(A) 1/3 (B) 1
(C) 3 (D) 2/3
15. (C)
+ 2 = – … (1)
3 = 1 = 1, , 2
From (1), p = –3 ( + 2) = –6, 3, 3
16. If the roots of equation ax2 + bx + 10 = 0 are not real and distinct, where a, b R and m and n are values of a and b respectively for which 5a + b is minimum, then the family of lines (4x + 2y + 3) + n (x – y – 1) = 0 are concurrent at
(A) (1, –1) (B) (–1/6, –7/6)
(C) (1, 1) (D) none of these
16. (B)
Let f(x) = ax2 + bx + 10
Since equation f(x) = 0 has no real and distinct roots
f(x) will have same sign for all real x
But f(0) = 10 > 0
f(x) 0, x R
f(5) 0 5(5a + b) + 10 0
5a + b – 2
Minimum value of 5a + b = –2
According to question
5m + n = – 2 n = – 5m – 2
Given family of lines is
m (4x + 2y +3) – (5m + 2)(x – y – 1) = 0
or 2(x – y – 1) + m (–x + 7y + 8) = 0
Clearly this family of lines pass through the fixed point
17. The values of a and b so that x4 + 12x3 + 46x2 + ax + b is square of quadratic expression are respectively
(A) 60, 25 (B) 45, 25
(C) 60, 30 (D) 25, 45
17. Given x4 12x3 46x2 ax b (x2 cx d)2
can paring coefficient of x both the sides, we get 12 2c, 46 2d c2, a 2cd
and b d2
c 6 and d 5
a 60 and b 25
18. If the equation x2 + 5bx + 8c = 0, does not have two distinct real roots, then minimum value of 5b + 8c is
(A) 1 (B) 2
(C) –2 (D) –1
18. Since the equation x2 5bx 8c 0 does not have two distinct real roots and coefficient of x2 is positive, hence x2 5bx 8c 0
5b 8c –1. Hence the minimum value is –1.
19. The number of real solutions of the equation –x2 + x – 1 = sin4 x is
(A) 1 (B) 2
(C) 0 (D) 4
19. Given equation is
LHS < 0 while RHS > 0 x
Hence the given equation has no solution.
20. The number of real roots of equation x8 – x5 + x2 – x + 2 = 0 is
(A) 2 (B) 4
(C) 6 (D) 0
20. (D) Given equation is x8 – x5 + x2 – x + 2 = 0 . Clearly given equation will have no negative root. Now given equation can be written as x5(x3 – 1) x(x – 1) 2 0.
Clearly no value of x will satisfy the given equation
21. has
(A) exactly one feal solution (B) two real solution
(C) 3 real solution (D) No solution
21 Since coeff. of x2 = – 1 < 0 and D = 1 – 4 = – 3 < 0
The –x2 + x – 1 < 0 But sin4 v x R.
L.H.S. is negative, while R.H.S. is positive.
Hence there is no real solution (C)
22. The value of k for which the equation x2 2(k – 1)x k 5 0 possess at least one positive root as
(A) [4, ) (B) (– , –1) [4, )
(C) [–1, 4] (D) ( – , –1)
22. (D) Let f(x) = x8 – x5 + x2 – x + 1
For x < 0, f(x) > 0
For 0 < x < 1, f(x) = x8 + (x2 – x5) + (1 – x) > 0
For x > 1, f(x) = (x8 – x5) + (x2 – x) + 1 > 0
Hence f(x) = 0 has no real root.
23. Let f(x) = (1 + b2)x2 + 2bx + 1 and let m(b) be the minimum value of f(x). As b varies, the range of m(b) is
(A) [0, 1] (B)
(C) (D) (0, 1]
23. 1 + 2log2 x log2x (log2 x 1) 2 log22 x (log2 x)3 1
t (t2 3t 3) 0
log2 x 0
x 1 rejected.
24: Given lx2 – mx + 5 = 0 does not have two negative real roots, the minimum value of 5l + m is
(A) 5 (B) –5
(C) 1 (D) –1
24. Case – I
D 0 S > 0 P > 0
(x – h)(k 1) 0 – 2 (k – 1) > 0 x 5 > 0
k (–5, –1]
Case – II
(k – h)(k 1) > 0 – union k (–, –1)
25: If expression x2 – 4cx + b2 > 0 for all x Î R and a2 + c2 < ab, then range of the
function is
(A) (–¥, 0) (B) (0, ¥)
(C) (–¥, ¥) (D) none of these
25. (D) f(x) = (1 + ) + 2bx + 1 …(1)
f ‘(x) = 2 ( 1 + )x + 2b = 0 x = –
f ‘’(x) = 2 ( 1 + ) > 0
f (x) has min. value at x = –
Min. value of f(x) i.e.
m(b) = + 1
or m(b) = 1 –
Clearly, 0 < m (b) 1
[ 0 max. value of m (b) = 1]
26: Number of positive integers n for which n2 + 96 is a perfect square is
(A) 4 (B) 8
(C) 12 (D) infinite
26.
x2 – 5x 5 g
(x – 3a)(x – (a 3)) < 0
3a 1, a 1/3 , a 3 3, a 0
a {0, 1/3}
27. If the larger root of equation x2 + (2 – a2)x + (1 – a2) = 0 is less than both the roots of the equation x2 – (a2 + 4a + 1)x + a2 + 4a = 0, then the range of a, is
(A) (B)
(C) (D) none of these
27. The roots of first equation are –1 and a2 –1. Now the roots of second equation are 1, a2 4a.
According to given condition a2 – 1 < 1 and a2 – 1 < a2 4a
a and a >
a
28. Let a, b, c, , e R and satisfy the relations and then which is true
(A) (B)
(C) (D)
28. We have,
a(b c)2 a1bc e 0 … (1)
and a(b c1)2 a1bc1 e 0 … (2)
(1) and (2) indicate that c and c1 are the roots of
a(b x)2 a1xb e 0
i.e., of ax2 bx(a1 2ba) ab2 e 0
c c1 and cc1
acc1 ab2 e
29. Let a, b, c be real numbers with a 0 and let , be the roots of the equation . Then the roots of in terms of , are given by
(A) (B)
(C) (D)
29. Given equation a3x2 abcx c c3 0
written as .
Clearly roots of this equation are ,
x
30. . Solve for x.
(A) (B)
(C) (D) none of these
30. Put x – 1 t2 in the given equation, we get
|t – 2| |t – 3| 1
t [2, 3]
x [5, 10]
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