polynomials
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polynomialsTRANSCRIPT
1. Remainder theorem:If a polynomial f(x) is divided by x-, the remainder obtained is f()
Factor theorem:A polynomial f(x) is divided by x-, if f()=0.
If a1 /a2 =a1 /a2= a1 /a2 , then each of these ratios is equal to
a) (ka1 +lb1 +mc1) / (ka2 +lb2 +mc2)
b) (ka1x +lb1x +mc1x) / (ka2x +lb2x+mc2x)1/x
c) (a1 b1/a2b2 )1/2 = (a1 b1 c1/a2b2 c2 )1/3
For example, a/b= 3/4, then
a/b= 3/4 = (a+3)/(b+4) = (a2 +9)1/2/(b2 +16)1/2
2. Condition for resolution into linear factors of a quadratic function:
The quadratic function ax2 +2fgh+by2 +2gx +2fy + c is resolvable into linear factors iff
abc + 2fgh - af2 -bg2 -ch2 =0 i.e.2. Q1. If P(x) = ax2 + bx + c , and Q(x) = -ax2 + bx + c, where ac not equal to zero.Show that the equation P(x).Q(x) =0 has at least two real roots. ( IIT 89)
Solution: Roots of equation P(x).Q(x) =0 will be the roots of equation P(x)=0 and Q(x)=0D1 be the discriminant of P(x)D2 be the discriminant of Q(x)
D1 +D2 = (b2 -4ac) + (b2 + 4ac) = 2b2 >=0
So at least one of D1 and D2 must be greater than or equal to zero.
So at least one of the equations P(x)=0, and Q(x)=0 has real roots.
Hence the equation P(x).Q(x)=0 has at least two real roots.
second method:
ac is not equal to zero.So ac>0, or ac0, D2 = b2 + 4ac >0If ac0
So at least one of D1 and D2 must be greater than or equal to zero.
So at least one of the equations P(x)=0, and Q(x)=0 has real roots.
Hence the equation P(x).Q(x)=0 has at least two real roots.
Q.2. Show that if p, q,r,s are real numbers and pr=2(q+s)then at least one of the equations x2 + px +q =0 and x2 + rx +s =0has real roots.
Solution:x2 + px +q =0 .........(1)
x2 + rx +s =0 ..........(2)
pr=2(q+s) ...........(3)
Let D1 and D2 be the discriminants of equations (1) and (2) respectively.
D1 +D2 = p2 -4q + r2 -4s
= p2 +r2 -4(q+s)
= p2 +r2 -2pr (using equation (3) )
= (p-r)2 >= 0
So at least one of D1 and D2 >=0
So at least one of the equations (1) and (2) have real roots.
3. Show that the equation esinx -e-sinx -4 =0 has no real solution. (IIT 1982)
Let y= esinx .Then the equation becomes y - 1/y -4 =0
or, y2 -4y -1 =0
Discriminant, D = 42 -4(1)(-1) = 20
Two roots are y = (4 + 20)/2 , (4 - 20)/2
or, y= (2+5) , (2-5)
or, esinx = (2+5) , (2-5)
Taking log on both sides, we get
sinx = log(2+5) , or sinx=log (2-5)
2-5 is negative. But log of negative number is not defined.
So only one equation is left.sinx= log(2+5) > logeor, sinx>1This is not possible.So the required equation has no real solution.3. 1. If and are the roots of the equation x2 +px +q =0 and x2n + pn xn + qn =0, where n is an even integer, prove that / , / are the roots of the equation xn +1 + (x+1)n =0.
Solution: and are the roots of the equation x2 +px +q =0So, + =-p.........(1) =q .................(2)Since and are the roots of the equation x2n + pn xn + qn =02n + pn n + qn =0 or,( n )2 + pn n + qn =0..........(3)
and,( n )2 + pn n + qn =0..............(4)From (3) and (4) we see that are the roots of y2 + pn y + qn =0So, n + n =-pn.........(5) nn =qn .................(6)
From (1), we have + =-por, ( + )n =(-p)n = pn (n is even)
or, ( + )n =-(-pn )= -(n + n ) (from 5)
or, n + n + ( + )n = 0 ........(7)
Dividing (7) by n , we get
(/ )n + 1 + ((/ ) +1 )n =0 .........(8)
Dividing (7) by n , we get
( / )n + 1 + (( /) +1 )n =0 .........(9)
From (8) and (9) we see that / and / are the roots of xn +1 + (x+1)n =0.
2. If sum of the roots of the equation ax2 + bx + c =0 is equal to the sum of the squares of their reciprocals, show that bc2 , ca2 , ab2 are in AP. ( IIT 76)
Solution:
Let and be the roots of the equation ax2 + bx + c =0 . Then, + = -b/a ...........(1)
= c/a .............(2)
Now, + = 1/ 2 + 1/2
or, + = (2 +2)/( )2
or, + =( ( +)2 - 2 ) / ( )2 ........(3)
Now put the values of (+ ) and ( ) from (1) and (2) in equation (3).
Thus, we get
-b/a = (b2 -2ac)/ c2
or, -bc2 = b2a -2ca2
or, bc2 + ab2 = 2ca2
Hence bc2 , ca2 , ab2 are in A.P.Proved.4. Q1. If one root of a quadratic equation ax2 + bx + c =0 is equal to nth power of the other, show that
(acn )1/n+1 + (an c)1/n+1 + b =0.
Solution:Let and n be the roots of the given equation.
Thus one root is the nth power of the other.
+ n = -b/a ............(1)
n = c/a
or, n+1 = c/a
or, = (c/a)1/n+1 .............(2)
From (1) we get,
a + a n +b =0
Putting the value of from (2) in the above equation, we get
a (c/a)1/n+1 + a (c/a)n/n+1 + b =0
or, (an+1 * c/a )1/n+1 + a(cn /an )1/n+1 + b =0
or, (anc)1/n+1 + (an+1 cn /an )1/n+1 + b =0
or, (anc)1/n+1 + (acn )1/n+1 + b =0
or, (acn )1/n+1 + (anc)1/n+1 + b =0
Proved.
Q 2. If r be the ratio of the roots of the equation ax2 + bx + c =0, show that
(r+1)2 /r = b2 /ac
Solution:
Let and be the roots of the equation ax2 + bx + c =0.
We know the value of the sum of the roots and the product of the roots.
+ = -b/a ........(1)
= c/a .........(2) (see proof)
Here we have another relation i.e the ratio of the roots.
/ = r .......(3)
LHS = (r+1)2 /r =(( /) +1)2/( /)
= ( +)2/( )
= (-b/a)2/(c/a) ........(using (1) and (2) )
= b2 /ac
= RHS
Proved.
Q1. If c,d are the roots of the equation(x- a)(x-b) -k =0, show thata, b are the roots of the equation (x-c)(x-d) +k =0. ( IIT 65)
Solution:(x- a)(x-b) -k =0
or, x2 - (a+b)x + ab -k =0
You know the relation for sum of roots and product of roots of a quadratic equation.If not, We saw that the roots of a quadratic equation ax + bx + c =0 is given by- x = (-b + D) /2a, and x = (-b - D) /2a Let these roots be denoted by and .Let = (-b + D) /2a, and = (-b - D) /2a + = (-b + D) /2a + (-b - D) /2a= (-b/2a ) + (D/2a) -(b/2a) - (D/2a)= -2b/2a= -b/a
1. * = (-b + D) /2a * (-b - D) /2a
= { (-b) - (D) } / (2a)
= (b -D)/4a
= ( b - b + 4ac ) /4a [ D= b - 4ac ]
= c/a
SOLVED PROBLEMS OF IIT JEE MATHS PAPERS USING THIS RELATION:
Question.1: If the sum of the roots of the equation ax2 + bx + c =0 is equal to the sum of the squares of their reciprocals, show that bc , ca, ab are in A.P. ( IIT-JEE 76)
SOLUTION: Let the roots of the equation be and . Then + = 1/ + 1/
or, + = (+) /
or, + = ( + ) - 2 /
Putting the values of + and we get
-b/a =( (b/a) - 2c/a) / (c/a)
or, -b/a = (b- 2ac)/ c
or, -bc = ab - 2ca
or bc + ab = 2ca
Hence bc , ca, ab are in A.P.Proved.c+d = (a+b) ...............(1)
cd = (ab-k) ...............(2)
Using (1) and (2) we get
a+ b = (c+d) ...........(3)
ab = (cd +k ) ............(4)
Now you to find the equation whose roots are a and b.You know the sum of roots and product of roots. So it can easily be found out.
The equation whose roots are a and b is given byax + bx + c= 0 -1
a{ x + (b/a) x + c/a } = 0
{ x + (b/a) x + c/a } = 0 , as a0
{ x - (-b/a) x + c/a } = 0
{ x - (+) x + (*)} = 0 , where & are the roots of the quadratic equation ax + bx + c= 0.
[ As proved in the previous page, + = -b/a, & * = c/a ]
Thus, the quadratic equation whose roots are & is given by -
x - (+) x + (*) = 0
or, x - (sum of roots) x + (product of roots) = 0
x2 - (a+b)x + ab =0 ( see proof )
or, x2 - (c+d)x + cd +k =0
or, (x-c)(x-d) + k= 0
Thus a, b are roots of the equation (x-c)(x-d) + k= 0
Q2. The coefficient of x in the equation x2 + px +q =0 was wrongly written as 17 in place of 13 and the roots thus found were -2 and -15. Find the roots of the correct equation. ( IIT 77)
Solution:
This question can be seen as
The roots of the equation x2 + 17x + q =0 are -2 and -15.
Find the roots of the equation x2 + 13x + q =0
We need to find the value of q.product of roots = q
or, (-2)(-15)= q
or, q= 30
Now it is a simple question where we have to find the roots of
x2 + 13x + 30 =0
or, x2 + (10+3) x + (10*3) =0
or, (x+10) (x+3) =0
or, x= -10, -3
So the roots of the correct equation are -10 and -3.Q.1. The real numbers x1, x2, x3 satisfying the equation x3 -x2 + bx + c =0 are in AP. Find the intervals in which b and c lie. ( IIT 96)
Solution:
x3 -x2 + bx + c =0 ...........(1)Let the roots of the equation be a-d , a , a+d
Sum of the roots =1 ( see proof)If 1, 2,3 ... n are the roots of the equation
f(x)= a0xn +a1xn-1 +a2xn-2 +...+an-1x + an =0, then
f(x)= a0 (x-1)(x-2)(x-3)... (x-n)
Equating both the RHS terms we get,
a0xn +a1xn-1 +a2xn-2 +...+an-1x + an = a0(x-1)(x-2)(x-3)... (x-n)
Comparing coefficients of xn-1 on both sides, we get
S1 = 1 + 2+3 +... + n = i = -a1/ a0
or, S1= - coeff. of xn-1/coeff. of xn
Comparing coefficients of xn-2 on both sides, we get
S2 = 1 2+ 13 +... = i j = (-1)2a2/ a0 i jor, S2= (-1)2 coeff. of xn-2/coeff. of xn
Comparing coefficients of xn-3 on both sides, we get
S3 = 1 23+ 234 +... = i j k = (-1)3a3/ a0 i j kor, S3= (-1)3 coeff. of xn-3/coeff. of xn
... ... ... ... ... ... ... ...
... ... ... ... ... ... ... ...
Sn=123... n =(-1)nan/ a0= (-1)n constant term/coeff. of xn
Here, Sk denotes the sum of the products of the roots taken k at a time.For example, S3 denotes the sum of the product of roots taken 3 at a time.
PARTICULAR CASES:
Quadratic Equation:If and are roots of the quadratic equation ax2 + bx + c=0, then
+ = -b/a
* = c/a
Cubic Equation:If , , are roots of a cubic equation ax3 + bx2 + cx + d=0, then
+ + = -b/a
+ + = c/a
= -d/a
Biquadratic equation :If , , , are roots of a cubic equation ax4+ bx3 + cx2 + dx +e=0, then
+ + + = -b/a
+ + + + + = c/a
+ + + = -d/a
= e/a
or, a-d + a + a+d =1
or, a = 1/3 ...........(2)
(a-d)a + a(a+d) + (a-d)(a+d) = b
or, 3a2 -d2 =b
or, 3(1/9) -d2 = b
or, 1/3 - d2 = b
or, d2 = 1/3 -b
Since d is real , 1/3 -b >= 0
or, b= -1/27
So, -1/270
Solution:
Let f(x) = ax2 + bx +c
Since the equation f(x)=0 has no real root, f(x) will have the same sign for all real values of x.
So, f(1) and f(0) will have same signf(0)= cf(1) = a+ b+ c
f(0).f(1) >0or, (a+b+c)c >0
Proved.Q.1. If a, b, c belong to R, a not equal to zero, and the quadratic equation ax2 + bx + c=0 has no real roots, then show that (a+b+c)c >0
Solution:
Let f(x) = ax2 + bx +c
Since the equation f(x)=0 has no real root, f(x) will have the same sign for all real values of x.
So, f(1) and f(0) will have same signf(0)= cf(1) = a+ b+ c
f(0).f(1) >0or, (a+b+c)c >0
Proved.Q.1. Solve for x:
( 5 +26)x2-3 + ( 5 -26)x2-3 =10
Solution:
( 5 +26)x2-3 * ( 5 -26)x2-3 = (25-24)x2-3 = 1
Let ( 5 +26)x2-3 = y
Then, ( 5 -26)x2-3 =1/y
Now the equation becomes y + 1/y =10
or, y2 -10y +1 =0
Using the quadratic formula , we get
y = ( 5 +26) , ( 5 -26)
The value of y which we took was ( 5 +26)x2-3
Now we get,
( 5 +26)x2-3 = ( 5 +26) .........(1)
( 5 +26)x2-3 = ( 5 -26) ............(2)
From (1) we get x2 -3 =1or, x2 = 4or, x =2, -2
From (2) we get,
x2 -3 = -1 ( 5 -26 = 1/( 5 +26) = ( 5 +26)-1 )
or, x2 =2 or, x= 2 , -2
Thus, x =2, -2, 2 , -2Q.1. Let f(x) be a quadratic expression which is positive for all real x.
If g(x) = f(x) + f '(x) + f ''(x) , then for real x, show that g(x) >0.
Solution:
Let f(x) = ax2 + bx + c
f '(x) = 2ax + b
f ''(x) = 2a
As f(x) is positive for all real values of x,a>0 , and b2 -4ac 0, and D < 0
The quantity (x + b/2a) in eq 1 is always greater than or equal to 0 since it is the square of a real number. As D 0. And as a>0, 4a will be greater than zero. Therefore-
sign of y=sign of a{ +ve + (+ve/+ve) }
or, sign ofy= sign of a( +ve)
or, sign ofy= sign of a
conclusion: y is always +ve.
Example: y = x + x + 1Here a=1 >0D= 1- 4*1*1 = -3