maths qe teacher notes
TRANSCRIPT
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THEORY OF QUADRATIC EQUATIONS / INEQUATIONS /ALGEBRAIC EQUATIONS
Syllabus IIT JEE : Quadratic equations with real coefficients, relations between roots andcoefficients, formation of quadratic equations with given roots, symmetric functions of roots.
1ST LECTURE :QUADRATIC POLYNOMIAL :
A polynomial of degree two in one variable of the typey = ax2 + bx + c where a 0, a, b, c R is called a quadratic polynomial, wherea = leading coefficient of the trinomialc = absolute term of the trinomialIn case a = 0 , y = bx + c, is called a linear polynomial. (b 0 )If c = 0 then y = bx is called an odd linear polynomialThe standard appearance of a polynomial of degree n isy = f (x) = anxn + an – 1xn – 1 + an – 2 xn – 2 + ....... + a0 when an 0 and ai's = R(note that a polynomial of degree 3 is called a cubic and of degree 4 is called abiquadratic polynomial)Now for different values of a, b, c, if graph of y = ax2 + bx + c is plotted then thefollowing 6 different shapes are obtained. The graph is called a parabola.
EXPLANATION OF ABOVE GRAPHS :The first 3 figures are obtained when a > 0. Here the mouth of the parabola opensupwards. Shape resembles that of a cup, filling water or sign of union. Last 3 graphsare obtained when a < 0. Here the mouth of the parabola opens downwards. Shaperesembles that of a cup, spilling water or sign of intersection.Figure-1 is indicative that there are two values of x for which the value of y is zero.These values x = x1 or x = x2 are called the zeroes of the polynomial.Note that for x > x2 or x < x1, y is positive whereas for x1 < x < x2, y is negative. In thiscase y can take both positive and negative values.In figure-2 the curve touches the axis of x. Here both zeroes of the polynomial coincide.Note that in this case the value of y is always non negative for all x R.In figure-3 the curve completely lies above the x-axis. There is no real zero and thevalue of y is always greater than zero for all x R. This is an important case.Similar explanation can be given for figure-4, 5 and 6.
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Now a quadratic polynomial when equated to zero is called a quadratic equation i.e.ax2 + bx + c = 0, a 0, a, b, c R
Solving a quadratic equation would mean finding the value or values of x for whichax2 + bx + c vanishes and these values of x are also called the roots of the quadraticequation or zeroes of the corresponding quadratic polynomial. Two methods of solvinga quadratic equation and(1) graphical (not very useful) (2) algebraic
ALGEBRAIC METHOD :ax2 + bx + c = 0 a 0, a, b, c R
x2 + ab
x + ac
= 0 (a 0)
2
a2bx
= 2
2
a4b
– ac
= 2
2
a4ac4b
x + a2b
= ± a2
ac4b2
x = a2
ac4bb 2 (Vietta's theorem)
Hence = a2
Db and =
a2Db
where D = b2 – 4ac
The quantity D = b2 – 4ac is called the discriminant of the quadratic equation and playsa very vital role in deciding the nature of roots of the equation without actuallydetermining them. NowIf D > 0 then roots are real. (This corresponds to the figure-1 or figure-4 depending onthe sign of 'a')If D = 0 roots are coincident. (This corresponds to the figure-2 or figure-5)If D 0 roots are realIf D < 0 no real roots. Infact roots are complex conjugate. This corresponds tofigure-3 or figure-6.We therefore see that the condition for y = ax2 + bx + c to be + ve x R becomesa > 0 and D < 0. Similarly ax2 + bx + c < 0 x R if a < 0 and D < 0.
Note :(1) In case the coefficient of quadratic equation are rational then the roots are rational if
D > 0 and is a perfect square.(2) Irrational roots occur in pair of conjugate surd i.e. if one root is 2 + 3 the other is 2– 3 .(3) If coefficient of quadratic equation are real and one root is + i then other root is – i.
RELATION BETWEEN ROOTS AND COEFFICIENT OF QUADRATICEQUATION :
ax2 + bx + c = 0, a 0, a, b, c R
If , are the roots then + = – ab
; = ac
Hence we can form the quadratic equation if the sum and product of its roots are known
i.e. x2 + ab
x + ac
= 0
x2 – (– ab
)x + ac
= 0
x2 – (sum of roots)x + product of roots = 0
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GLS-IITC-M-QE-3Ex.1 Form a quadratic equation whose one root is
(a) cos36° (b) tan12
(c) tan 8
(d) cos28
Note :
(a) If exactly one root of quadratic equation is 0, then P = 0 ac
= 0 c = 0 and the
quadratic becomes ax2 + bx = 0.(b) If both roots of the quadratic equation are zero then S = 0 and P = 0 b = c = 0 and
the quadratic equation becomes x2 = 0.
(c) If one root is , put x = y1
in ax2 + bx + c = 0, we get
cy2 + by + a = 0 must have one root zero P = 0 i.e. ca
= 0
Hence , a = 0 and – cb
0 b 0.
original quadratic equation becomes bx + c = 0(d) When both roots of the quadratic equation are infinity then. The quadratic equation
cy2 + by + a = 0 must have both roots zero.
i.e. – cb
= 0 and ca
= 0 b = 0 ; a = 0 and c 0.
In this case the equation becomes y = c. e.g. if (2p – q)x2 + (p – 1)x + 5 = 0 has both roots infinite. Find p and q.
[Ans. p = 1 ; q = 2](e) If f (, ) = f (, ) then f (, ) denotes symmetric functions of roots.
e.g. f (, ) = 2 + 2 ; f (, ) = cos ( – )It is to be noted that every symmetric function in , can be expressed in terms oftwo symmetric functions + and .
ILLUSTRATION FOR THE FIRST LECTURE :1. Graph of y = ax2 + bx + c is as shown in the figure then
(i) a < 0 (ii) D > 0 (iii) S > 0 (iv) P < 0
(v) – ab
> 0 (b > 0) (vi) ac
< 0 (c > 0)
(vii) b and c have the same sign and different than a.
2. The quadratic equation ax2 + bx + c = 0 has no real root, then prove thatc (a + b + c) > 0
[Hint: both f (0) and f (1) have the same sign f (0) · f (1) > 0 result ]
3. Find the set of values of 'a' for which the quadratic polynomial(i) (a + 4)x2 – 2ax + 2a – 6 < 0 x R [Ans. (– , – 6)](ii) (a – 1)x2 – (a + 1)x + (a + 1) > 0 x R [Ans. (5/3, ) ]
4. If , are the roots of the quadratic equation x2 – 2x + 5 = 0 then form a quadraticequation whose roots are 3 + 2– + 22 and 3 + 42– 7 + 35.
[Hint : 2 – 2 + 5 = 0 3 + 2 – + 22 = 7 & 2 – 2 + 5 = 0 3 + 42 – 7 + 35=5]
5. If f (x) = ax2+bx+c > 0 xR then prove that g (x) =f (x)+ f '(x)+ f "(x)>0 ( x R)
6. If ax2 + bx + c = 0 then find the value of (ax1 + b)–3 + (ax2 + b)–3.
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7. If p (q – r)x2 + q(r – p)x + r (p – q) = 0 has equal root prove that r1
p1
q2
(cyclic
order) i.e. p, q, r are in H.P.
8. If a quadratic equation (in x or y) is formed from y2 = 4ax and y = mx + c and hasequal roots then prove that c = a/m.
9. If x = 53 find the value of x4 – 12x3 + 44x2 – 48 x + 17. [Ans. 1]
10. If the roots of the quadratic equation (x – a) (x – b) – k = 0are c and d then prove that a and b are the roots of thequadratic equation (x – c) (x – d) + K = 0.
[Sol. x2 – (a + b)x + ab – K = 0 c + d = a + b
cd = ab – k must be true
TPT (x – c)(x – d) + K = 0 x2 – (a + d)x + cd + K = 0
a + b = c + d ....(1)ab = cd + K ....(2)
Hence Proved ]
11. For what value of p the vertex of the parabola if x2 + 2px + 13 lies at a distance of5 unit from the origin.
12. Find the least value of the function f (x) = 2bx2 – x4 – 3b2 in [–2, 1] depending on theparameter b. [Ans. for b (– , 2] least value is f (4) = 8b – 3b2 – 16 ;
for b [2, ) least value is f (0) = – 3b2 ]
13. Find all numbers p for each of which the least value of the quadratic trinomial4x2 – 4px + p2 – 2p + 2 on the interval 0 x 2 is equal to 3.
[Ans. p = 1 – 2 or 5 + 10 ]
14. If the roots of ax2 + 2bx + c = 0 be possible and different, then the roots of(a + c) (ax2 + 2bx + c) = 2(ac – b2)(x2 + 1)
will be impossible, and vice versa.[Hint: 4(ac – b2) [(a– c)2 + 4b2] ]
Home work after 1st lecture :Q. No. 26 to 71 Page 34 Prilepko
2ND LECTUREIf a quadratic equation has more than 2 roots then it becomes an identity.
Proof : Let ax2 + bx + c = 0
Hence a2 + b + c = 0 ....(1) a2 + b + c = 0 ....(2) a2 + b + c = 0 ....(3)
From equation (1) and (2)a(2 – 2) + b( – ) = 0
,
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GLS-IITC-M-QE-5hence a ( + ) + b = 0 ....(4)similarly a ( + ) + b = 0 ....(5)by subtraction, (4) and (5)
a( – ) = 0 , hence a = 0if a = 0 b = 0 c = 0Hence QE becomes 0x2 + 0x + 0 = 0 which is an identity.
EXAMPLES :1. For what values of p, the equation
(p + 2)(p – 1)x2 + (p – 1) (2p + 1) x + p2 – 1 = 0 has more than two roots.
2. )bc)(ac()bx)(ax(
+ )ca)(ba()cx)(bx(
+ )ab)(cb()ax)(cx(
= 1
3. )ca)(ba()cx)(bx(a2
+ )ab)(cb()ax)(cx(b2
+ )bc)(ac()bx)(ax(c2
= x2
4. )ca)(ba()xa( 2
+ )ab)(cb()xb( 2
+ )bc)(ac()xc( 2
= 1
[Ans. x = – a ; – b ; – c are the solutions ]
Solving quadratic and rational inequalities. (Method of intervals)Type-1 : Quadratic inequality involving non-repeated linear factors.
(1) 3x2 – 7x + 6 < 0 [Ans. x ]
(2) (x2 – x – 6)(x2 + 6x) 0 (Asking)
(3) Solve f ' (x) g ' (x) where f (x) = 5 – 3x + 2x25
– 3x3
, g (x) = 3x – 7. [Ans. [2, 3] ]
Type-2 : Quadratic inequality involving Repeated linear factos.
(1) (x + 1)(x – 3)(x – 2)2 0. [Ans. (– , –1) (3, )]
(2) x(x + 6)(x + 2)2(x – 3) > 0 [Ans. (–6, 0)(x +1)3 – {2} ]
(3) (x – 1)2(x + 1)3(x – 4) < 0 [Ans. (–1, 4) – {1} ]
(4) Number of positive integral solution of 43
623
)8x3()3x()4x()3x2(x
0
(A) only one (B) 2 (C*) 3 (D) 4[Hint: x {1, 2, 4} ]
Type-3 : Quadratic / algebraic inequality of the type of )x(g)x(f
. (Rational inequality)
(1) 7x33x2
[(–, 23
) ( 37
, + )] (2) 5x4x
12x5x2
2
> 3 [ (21
, 3)]
(3) 1xx6x5x
2
2
< 0 [(2, 3)] (4) )2x(x)1x()1x(
4
32
< 0 [(–1, 2) – {0, 1}]
(5)1x5x
1x1x
[(– , – 1) (1, 3)] (6) 2x1
)7x)(1x()4x(2
[(1, 2) (7, + ) ]
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(7) |4x|7x6x2
< 0 [(–7, –4) (– 4, 1)]
(8) 1xx24x4x
2
2
> 0 [(– , – 2) (– 2, – 21
) (–1, ) ]
Type-4 : Double inequality and biquadratic inequality.
(1) 1 < 1x
8x7x32
2
2
(2) For what value of 'x' sin–1
1x8x7x3
2
2
is meaningful.
(3) (x2 + 3x + 1) (x2 + 3x – 3) 0 [Ans. (–, – 4] [–2, –1] [1, ) ]
(4) (x2 + 3x)(2x + 3) – 16 )x3x()3x2(
2
0.
CONDITION OF COMMON ROOTS :Let a1x2 + b1x + c1= 0 and a2x2 + b2x + c2 = 0 have a common root .Hence a1
2 + b1 + c1 = 0 a2
2 + b2 + c2 = 0by cross multiplication
1221
2
cbcb
= 2112 caca
=
1221 baba1
= 2112
1221cacacbcb
= 1221
2112babacaca
Which is the required condition.This is also the condition that the two quadratic functions a1x2 + b1x y + c1y2 anda2x2 + b2x y + c2y2 may have a common factor.
Note: If both roots of the given equations are common then 2
1
2
1
1
2cc
bb
aa
EXAMPLES :1. Find the value of k for which the equations 3x2 + 4kx + 2 = 0 and 2x2 + 3x – 2 = 0
have a common root. [Ans. k = 47
or – 811
]
[Hint: From the 2nd equation x =21
or x = – 2. Either x =21
or x =– 2 may be the common
root]
2. If the quadratic equation x2 + bx + c = 0 and x2 + cx + b = 0 (b c) have a commonroot then prove that their uncommon roots are the roots of the equation x2 + x+bc=0
[Hint: 2 + b + c = 02 + c + b = 0
22
2
cb
= bc
=
bc1
; Hence = 1 or = – (b + c)
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GLS-IITC-M-QE-7if = 1 then 1 = c 1 = cand 2 = b 2 = b
where 1 and 2 are the common root required equation x2 – (b + c)x + bc = 0but – (b + c) = 1 x2 + x + bc = 0 ]
Home Work after Q.No. 2.If the equations x2 + abx + c = 0 and x2 + acx + b = 0 have a common root then showthat the quadratic equation containing their other common roots is
a (b + c)x2 + (b + c)x – abc = 0
3.(a) Find the value of p and q if the equation px2 + 5x + 2 = 0 and 3x2 + 10x + q = 0 have
both roots is common. [Hint: 3p
= 105
= q2
] [Ans. p = 23
; q = 4]
(b) If the equation x2 – 4x + 5 = 0 and x2 + ax + b = 0 have a common root find a and b.[Hint:since roots of 1st are imaginary hence both roots must be common]
4. If the equation 4x2 sin2 – (4sin)x + 1 = 0 anda2(b2 – c2)x2 + b2(c2 – a2)x + c2(a2–b2) = 0 have a common root and the 2nd equationhas equal root find the possible values of in (0, ).
[Hint: x = 1 is the common ( = 6
or 6
5 ) put in equation to get sin=
21
]
5. If one root of the quadratic equation x2 – x + 3a = 0 is double the root of the equationx2 – x + a = 0 then find the value of 'a' (a 0)
[Hint: x2 – x + 3a = 0 ; x2 – x + a = 0 ,
2 – + a = 042 – 2+ 3a = 0
a2a3
2
= a3a4
=42
1
a
2
= a
= 21
= – 1 or = 2a
a = – 2 ]
6. If each pair of the equations
0qxpx0qxpx
0qxpx
332
222
112
has exactly one root in common then show that
(p1 + p2 + p3)2 = 4 (p1p2 + p2p3 + p3p1 – q1 – q2 – q3)
[Sol. Let x2 + p1x + q1 = 0
x2 + p2x + q2 = 0
x2 + p3x + q3 = 0
Now | ( + ) – ( + ) |2 = | – |2
2122
21 pp2pp = 3
23 q4p
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or 23
22
21 ppp = 2p1p2 – 4q3
|||ly 21
23
22 ppp = 2p3p4 – 4q1
and 22
21
23 ppp = 2p3p1 – 4q2
adding 23
22
21 ppp = 2 21 pp – 1q4
(p1 + p2 + p3)2 = 4 ][ 121 qpp ]
7. Find the value of '' for which the system of inequalityx2 + 2x + 0 and x2 – 4x – 6 0 has a unique solution.
[Ans. = 0, 1]
8. For what value of 'a' do the curves y = 1 + 3
2
ax
and y = x4 possesses only one point
in common. [Ans. a = 1/3 or a < 0]Home Work : General inequaliteis from Prilepko.
3RD LECTUREMAXIMUM AND MINMUM VALUES OF QUADRATIC ANDRATIONAL FUNCTIONS :
(1) y = ax2 + bx + c attains its maximum or minima at the point with abscissa x = – a2
b
according as a < 0 or a > 0.
(2) Maximum or minimum value can also be obtained by making a perfect square and thentaking an interpretation.
Ex. y = 2x2 – 3x + 1 find the minimum value.y = 7 + 5x – 2x2 find the maximum value.
(3) For computing the maximum or minimum values of rational function consider thefollowing examples :
EXAMPLES :
1. If x is real then 4x3x4x3x
2
2
lies from 71
and 7.
[Sol. dxdy
= 22
2
)4x3x()4x(6
which vanishes where x = 2 or – 2 ;
f (2) = 71
& f (–2) = 7
Note that y is always > 0 as both Nr & Dr > 0, x RThe graph is as follows
Note :
7,71
is also the range of the given function. ]
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2. If x is real, prove that the expression y = )3x(211x2x2
can have all numerical values except which lie between 2and 6.
[Sol. dxdy
= 21
2
2
)3x(5x6x
= 0 x = 1 & x = 5
y |x = 1 = 2 ; y |x = 5 = 6 ]
3. Prove that y = )3x(x)2x)(1x(
can have any
value in (– , ) for x R.
[Sol. dxdy
= 22
2
)x3x()3xx2(2
x R
y in (– , – 3) (– 3, 0) (0, ) ]
4. Find the maximum and minimum value of y = 3x2x9x14x
2
2
x R.
[Ans. Maximum = 4 ; minimum = – 5]
5. Find all possible values of 'a' for which the expression ax7x55x7ax
2
2
may be capable
of all values, x being any real quantity. [Ans. a (– 12, 2)][Hint: For common roots between Nr and Dr
ax2 – 7x + 5 = 05x2 – 7x + a = 0
35a7
2
= 2a25
= )5a(7
1
a 5
7
2 = 5a
= 71
5a7
= 7
5a a = 2 or a = – 12
Hence for a = 2 or a = – 12 we have a common factor in Nr & Dr.for a = 2
y = 2x7x55x7x2
2
2
= )2x5)(1x()1x)(5x2(
....(1)
for a = – 12
y = 12x7x5
5x7x122
2
= )12x5)(1x()x125)(1x(
....(2)
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for a = 2in (1) when x 1 y = – 1
y can take the value – 1 for any x.
in (2) when x –1 y = 1; x ± , y 52
y can not take the value 1 for any x.
The graph of y = 1x1x·
2x55x2
= 2x55x2
when x 1
dxdy
= 2)2x5(5)5x2(2)2x5(
= 2)2x5(21 > 0 y
Note that in this case y can take all values except – 1and 2/5. Similar would be thesituation when a = – 12. Hence the values of a = 2 and – 12 are to be excluded. ]
6. Show that the expression )dcx)(abx()cdx)(bax(
will be capable of all values when x is real,
if a2 – b2 and c2 – d2 have the same sign.[Sol. TPT (a2 – b2) (c2 – d2) > 0 (note that ad bc)
y = adx)acbd(xbcbcx)bdac(xad
2
2
ybcx2 – (bd + ac)y x + y ad = adx2 – (ac+bd)x + bc
(ybc – ad)x2 – [ y(bd + ac) – (ac + bd)] x + (y ad – bc) = 0 as x R[ y(bd + ac) – (ac + bd)]2 4(y bc – ad) (y ad – bc)
(bd + ac)2·y2 + (ac + bd)2 – 2(bd + ac)(ac + bd) y 4[(abcd)y2 – (b2c2 + a2d2)y + abcd](ac – bd)2·y2 –2[(ac + bd)2 – 2(b2c2 + a2d2)] y + (ac – bd)2 0(ac – bd)2·y2 –2[(ac – bd)2 – 2(bc – ad)2] y + (ac – bd)2 04[(ac – bd)2 – 2(bc – ad)2]2 – 4[(ac – bd)2]2 0[(ac – bd)2 – 2(bc – ad)2 + (ac – bd)2] [(ac – bd)2 – 2(bc – ad)2 – (ac – bd)2] 0
Home Work : Exercise 9 (b) of H & K.
4TH LECTUREA. To find the condition that a quadratic function of x, y of the type
f (x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c= 0 may be resolved into two linear factors.The required condition is
abc + 2fgh – af2 – bg2 – ch2 = 0.
EXAMPLES :1. If the expression 3x2 + 2pxy + 2y2 +2ax – 4y + 1 can be resolved into linear factors
then prove that p must be one of the roots of the equation t2 + 4at + 2a2 + 6 = 0.[Sol. a = 3, h = p, b = 2, g = a, f = – 2, c = 1
condition for this to be resolved into 2 linear factorsabc + 2fgh – af2 – bg2 – ch2 = 06 + (– 4ap) – 12 – 2a2 – p2 = 0– p2 – 4ap – 2a2 – 6 = 0p2 + 4ap + 2a2 + 6 = 0
p is one of the roots of t2 + 4at + 2a2 + 6 = 0]
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GLS-IITC-M-QE-112.(a) If the equation x2 + 16y2 – 3x + 2 = 0 is satisfied by real values of x and y then
prove that 1 x 2 and – 81
y 81
.
[Sol. x2 – 3x + 16y2 + 2 = 0 16y2 + x2 – 3x + 2 = 0x R D 0 y R D 09 – 4(16y2 + 2) 0 – 64(x2 – 3x + 2) > 09 – 64y2 – 8 0 x2 – 3x + 2 01 – 64y2 0 (x – 2)(x – 1) 0 64y2 – 1 0(8y + 1)(8y – 1) 0
– 81 y < 8
1 x [1, 2]
1 x 2 ] (b) Show that in the equation x2 – 3xy + 2y2 – 2x – 3y – 35 = 0, for every real value of
x there is a real value of y, and for every value of y there is a real value of x.
3. Prove that the expression 2x2 + 3xy + y2 + 2y + 3x + 1 can be factorised into twolinear factors. Find them.
[Sol. a = 2, b = 1, h = 3/2, g = 3/2, f = 1, c = 1abc + 2fgh – af2 – bg2 – ch2 = 0
2 + 29 – 2 – 4
9 – 49
29 – 2
9 = 0 ]
4. If (ax2 + bx + c)y + a' x2 + b'x + c' = 0 find the condition that x may be a rationalfunction of y. [Ans. (ac' – a'c)2 = (ab' – a'b) (bc' – b'c)]
[Hint: Solve for x and then D must be a perfect square. ][Sol. (ay + a')x2 + (by + b') x + cy + c' = 0
2(ay + a')x = – (by + b') ± sumperfect
2 )'ccy)('aay(4)'bby(
(b2y2 + b' 2 + 2bb' y) – 4[acy2 + (ac' + a'c)y + a'c' ](b2 – 4ac)y2 + 2[bb' – 2(ac' + a'c)] y + (b' 2 – 4a'c') = 04 [bb' – 2(ac' + a'c)]2 = 4(b2 – 4ac)(b' 2 – 4a'c')on simplifying, we get(ac' – a'c)2 = (ab' – a' b) (bc' – b'c) Ans. ]
B. THEORY OF EQUATIONS :ax3 + bx2 + cx + d a(x – x1) (x – x2)(x – x3)
= a [x3 – (x1) x2 + ( x1x2)x2 – x1x2x3]
x1 + x2 + x3 = – ab
x1x2 + x2x3 + x3x4 = ac
and x1 x2 x3 = – ad
Note: A polynomial equations of degree odd with real coefficient must have at least one realroot as imaginary roots always occur in pair of conjugates.
GLS-IITC-M-QE-12
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EXAMPLES :1. Solve the cubic 24x3
– 14x2 – 63x + 45 = 0, one root being double the other.
[Sol. 24x3 – 14x2 – 63x + 45 = 0
3 + = 127
....(1)
22 + 2 + = – 821
....(2) or 22 + 3 = – 821
22 = – 815
....(3) must be – ve
put = 127
– 3 in equation (2)
22 + 3
3127
= – 821
22 + 47 – 92 = – 8
21
– 72 + 47 + 8
21 = 0 82 – 6 + 4 – 3
– 2 + 4
+ 83
= 0 2(4 – 3) + 4 – 3
82 – 2 – 3 = 0 = 43
or = – 21
if = 43
then = 127
– 49
= 12
277 = –
1220
= – 35
i.e. = 43
or = – 35
If = – 21
then = 127
+ 23
= 24
187 =
2425
i.e. = 43
& = 2425
(does not satisfy (3) )
Hence = 43
& = – 35
roots are 43
, 23
, – 35
]
Similar problem of Home Work : 24x3 + 46x2 + 9x – 9 = 02. Find the
(i) sum of the squares and
(ii) sum of the cubes of the roots of the cubic equation x3 – px2 + qx – r = 0
[Sol. (i) 2 + 2 + 2 = ( + + )2 – 2 (ii) 3 + 3 + 3 – 3 = ( + + ) (2 + 2 + 2 – – – )
3 = ( + + ) [2 – ] + 3
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GLS-IITC-M-QE-133. Solve the cubic 4x3 + 16x2 – 9x – 36 = 0, the sum of its two roots being equal to zero.
[Ans.
4,
23,
23
]
4. , , , are the roots of the equation tan
x4 = 3 tan3x no two of which have
equal tangents, find the value of tan + tan + tan + tan . [Ans. Zero]
5. Find the cubic each of whose roots is greater by unity than a root of the equationx3 – 5x2 + 6x – 3 = 0. [Ans. y3 – 8y2 + 19y – 1 = 0]
[Sol. If y is one root of the required eqaution then y = x + 1 x = y – 1.Now put x = y – 1 in the given equation.]
6. Form a cubic whose roots are the cubes of the roots of x3 + 3x2 + 2 = 0.[Ans. y3 + 33y2 + 12y + 8 = 0]
[Sol. + + = – 3 = 0 ; = – 2y3 – (3 + 3 + 3)y2 + (33 + 33 + 33)y + 333 = 0a3 + b3 + c3 = (a + b + c)[a2 + b2 + c2 – (ab + bc + ca)] + 3abca3 + b3 + c3 = (a + b + c)[(a + b + c)2 – 3(ab + bc + ca)] + 3abc
= (–3) [9] + 3 (–2) = – 33
Similarly intercept 33 + 33 + 33 ]7. Given the product p of sines of the angles of a triangle & product q of their cosines,
find the cubic equation, whose coefficients are functions of p & q & whose rootsare the tangents of the angles of the triangle. [REE’92, 6]
[Ans: qx3 px2 + (1 + q) x p = 0][Sol. Given sinA sinB sinC = p ; cosA cosB cosC = q
Hence tanA tanB tanC = tanA + tanB + tanC = p/qHence equation of cubic is
x pq
x A B x pq
3 2 0 tan tand i ...(i)
now tan tan sin sin cos sin sin cos sin sin coscos cos cos
A B A B C B C A C A BA B C
We know that A + B + C = cos(A+B+C) = –1cos(A+B) cosC – sin(A+B) sinC = –1( cosA cosB – sinA sin B) cosC – sinC (sinA cosB + cosA sinB) = –11+ cosA cosB cosC= sinA sinB cosC + sinB sinC cosA + sinC sinA cosB
dividing by cosA cosB cosC1
A Btan tan
Hence (i) becomes qx px q x p3 2 1 0 ( ) Ans.]Home Work : 9 (c) of H & K.
5TH LECTURELOCATION OF ROOTS:This article deals with an elegant approach of solving problems on quadratic equationswhen the roots are located / specified on the number line with variety of constraints :Consider f (x) = ax2 + bx + c with a > 0.
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Type–1 : Both roots of the quadratic equation are greater than a specified number say (d).The necessary and sufficient condition for this are :
(i) a > 0 ; (ii) D 0 ; (iii) f (d) > 0 ; (iv) – a2b
> d
Note : If a < 0 then intercept accordingly.
EXAMPLES ON (TYPE–1) :1. Find all the values of the parameter 'd' for which both roots of the equation
x2 – 6dx + (2 – 2d + 9d2) = 0 exceed the number 3. [Ans. d > 9
11]
[Sol. (i) D 09t2 – 2 + 2t – 9t2 0t – 1 0t 1
(ii) – a2b
> 3; 3t > 3; t > 1
(iii) f (3) > 09 – 18t + 2 – 2t + 9t2 > 09t2 – 20t + 11 > 09t2 – 9t – 11t + 11 > 0(t – 1)(9t – 11) > 0
t (– , 1) (11/9, ) Intersection of (i), (ii) and (iii) is t > 11/9 ]
2. Find all the values of 'a' for which both roots of the equation x2 + x + a = 0 exceed thequantity 'a'. [Ans. (– , – 2)]
[Sol. (i) D 01 – 4a 04a 1a 1/4
(ii) – a2b
> a; – 21
> a; a < – 21
(iii) f (a) > 0a2 + 2a > 0a(a + 2) > 0
a (– , –2) (0, ) a (– , – 2) Ans. ]
3. Determine the values of 'a' for which both roots of the quadratic equation(a2 + a – 2)x2 – (a + 5)x – 2 = 0 exceed the number minus one.
[Ans. (– , –2) (–1, – 1/2) (1, )]
4. Find the values of a > 0 for which both the roots of equation ax2 – (a + 1)x + a – 2 = 0are greater than 3.
Type–2 : Both roots lie on either of a fixed number say (d). Alternatively one root isgreater than 'd' and other less than 'd' or 'd' lies between the roots of the givenequation.
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GLS-IITC-M-QE-15Conditions for this
(i) a > 0or
(i) a < 0
and (ii) f (d) < 0 (ii) f (d) > 0
Note that no consideration for discriminant will be useful here.
EXAMPLES ON (TYPE–2) :1. Find the value of k for which one root of the equation of x2 – (k + 1)x + k2 + k–8=0
exceed 2 and other is smaller than 2. [Ans. k (–2, 3)][Sol. since a > 0 f (0) < 0 f (2) < 0
4 – 2(k + 1) + k2 + k – 8 < 0k2 – K + 6 < 0(k + 2)(k – 3) < 0
k (–2, 3) ]
2. Find the set of values of 'a' for which zeroes of the quadratic polynomial(a2 + a + 1) x2 + (a – 1)x + a2 are located on either side of 3. [Ans. ]
[Sol. Leading co-efficient is always + ve f (3) < 0 9a2 + 9a + 9 + 3a – 3 + a2 < 0
10a2 + 12a + 6 < 05a2 + 6a + 3 < 0This is always + ve as 5 > 0
D < 0 a Ans.]
Type–3 : Exactly one root lies in the interval (d, e) when d < e.Conditions for this are :(i) a 0 ; (ii) f (d) · f (e) < 0
EXAMPLES ON (TYPE–3) :1. Find the set of values of m for which exactly one root of the equation
x2 + mx + (m2 + 6m) = 0 lie in (–2, 0) [Ans. (– 6, – 2) (–2, 0)]
2. Find all possible values of 'a' for which exactly one root of the quadratic equationx2 – (a + 1)x + 2a = 0 lie in the interval (0, 3). [Ans. (– , 0] (6, )]Note : In this case also check for end points. If interval is closed say [d, e] thenf (d) = 0 or f (e) = 0 no other root should lie in (d, e)
Type–4 : When both roots are confined between the number d and e (d < e). Conditionsfor this are(i) a > 0 ; (ii) D 0 ; (iii) f (d) > 0 ; (iv) f (e) > 0
d < – a2b
< e
EXAMPLES ON (TYPE–4) :1. If , are the roots of the quadratic equation
x2 + 2(k – 3)x + 9 = 0 ( ). If , (–6, 1) then find the values of k.
[Ans.
427,6 ]
Type–5 : One root is greater than e and the other roots is less than d.Conditions are :(i) f (d) < 0 and f (e) < 0 if (a > 0)
EXAMPLES ON (TYPE–5) :1. Find all the values of k for which one root of the quadratic equation
(k – 5)x2 – 2kx + k – 4 = 0 is smaller than 1 and the other root exceed 2. [Ans. (5, 24)]
GENERAL AND MIXED PROBLEM :(1) For y = f (x) = ax2 + bx + c
if f (p) < 0 and f (q) > 0i.e. f (p) f (q) < 0 then the equation ax2 + bx + c = 0 has one root lying betweenp and q.
EXAMPLES :1. Let be a real root of the equation ax2 + bx + c and be a real root of the equation
– ax2 + bx + c = 0. Show that there exists a root of the equation cbxx2a 2 = 0
that lie between and . (, 0).[Sol. is a root of equation ax2 + bx + c = 0
a2 + b + c = 0 ....(1)similarly – 2 + b + c = 0 ....(2)
Let f (x) = 2x2a
+ bx + c ....(3)
Now f () = 2a2 + b + c =
2a2 – a2 {From (1)}
= – 2a2
and f () = 2a2 + b + c =
2a2 + a2 {From (2)}
= 23
a2
Now f () f () = – 43
a22 2 < 0 [ , 0]
f () and f () have opposite signs, therefore equation f (x) = 0 will have exactlyone root between and if < or one root between and if < .
2. If a < b < c < d, then show that the quadratic equation (x – a)(x – c) + (x – b)(x – d)=0has real roots for all real values of .
[Sol. Let f (x) = (x – a) (x – c) + (x – b)(x – d)Given, a < b < c < dNow, f (b) = (b – a) (b – c) < 0 [ b – a > 0 and b – c < 0]and f (d) = (d – a)(d – c) > 0Since f (b) and f (d) have opposite signs therefore, equation f (x) = 0 has one realroot between b and d.Since one root is real and a, b, c, d, are real therefore, other root will also be real.Hence equation f (x) = 0 has real roots for all real values of . ]
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GLS-IITC-M-QE-173.(a) Prove that for any real value of a the inequality, (a2 + 3)x2 + (a + 2)x – 5 < 0 is true
for at least one negative x.[Sol. f (x) = (a2 + 3)x2 + (a + 2)x – 5
Case-I: when f (0) < 0obviously there is atleast one negative x for which f (x) < 0 f (0) = – 5 which is always true for any a RCase-II: If f (0) > 0
and D > 0
and – a2b
< 0
f (0) > 0 is not possible. Hence f a R]
3(b) If f (x) = 4x2 + ax + (a – 3) is negative for atleast one x, find all possible values of a.[Sol. Case-I: if f (0) < 0
a – 3 < 0 a < 3a (– , + 3)
Case-II: if f (0) 0 and
D > 0 and – a2b
< 0
f (0) > 0 gives a – 3 0 a 3 ....(1)D > 0 gives a2 – 16(a – 3) > 0
a2 – 16a – 48 > 0(a – 12)(a – 4) > 0 a > 12 or a < 4 ....(2)
– a2b
< 0 gives – 8a
< 0 a > 0 ....(3)
from (1), (2) and (3) a [3, 4) (12, )
finally a (– , 4) (12, ) ]
4. Find the values of a for which the equation x2 + 2(a – 1)x + a + 5 = 0 has at least onepositive root. [Ans. a – 1]
5. Let a, b, c R a 0. If and be the roots of equation ax2 +bx + c = 0,
where < – n and > n, then show that 1 + ab
n1
anc
2 < 0, n N.
[Sol. ax2 +bx + c = 0
f (x) = x2 + ab
x + ac
= 0
f (x) < 0 and f (–x) < 0
x2 + ab
x + ac
< 0
and x2 – ab
x + ac
< 0
hence 1 + axb
+ 2axc
< 0 ....(1)
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1 – axb
+ 2axc
< 0
This two inequalities must simultaneously be truenote that the prodction of the roots is < – n2
ac
< – n2 or 2axc
+ 1 < D
consider E = ab
n1
axc1
ve
2
....(2)
if b/a > 0 then E = 1 + 2axc
+ n1
ab
< 0
if b/a < 0 then E = 1 + 2axc
– n1
ab
< 0 from (1)
hence (2) simultaneously satisfied both (1) (1) is equivalent to
1 + 2axc
+ ab
< 0 ]
6TH LECTUREMISCELLANEOUS EQUATIONS INEQUATIONS AND LOGARITHMICINEQUALITIES :
A. LINEAR EQUATION / INEQUATIONS INVOLVING MODULUS :1. | x – 3 | + 2 | x + 1 | = 4 Ans { – 1}
2. | x + 2 | – | x – 1 | < x – 23
3. Find the least +ve integer satisfying | x + 1 | + | x – 4 | > 7.
4. Greater integer satisfying 31x2
– 2
1x3 > 1
B. QUADRATIC EQUATION / INEQUATION INVOLVINGMODULUS &EXPONENTIAL :
1. | x2 + 4x + 2 | = 316x5
Ans : { – 2, 1}
2. ( | x – 1 | – 3) (| x + 2 | – 5) < 0 Ans. (–7, 2) (3, 4)3. | x – 5 | > | x2 – 5x + 9 | Ans. (1, 3)4. 2| x + 2 | – | 2x + 1 – 1 | = 2x + 1 + 1
5. 4x4x5x
2
2
1 [(0, 58
)(25
, + )]
6. 1xx1x3x
2
2
< 3 [(– , – 2) (– 1, + )]
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GLS-IITC-M-QE-19C. LOGARITHMIC EQUATION :
1. x6x4log
51
0 Ans.
23,2
2. (a) log2x+3(x2) < log2x + 3(2x + 3) Ans. 3,11,23
(b) logx+3(x2 – x) < 1 Ans. (–3, 2) (–1, 0) (1, 3)
3.
1x26x2log7 > 0 Ans.
21,
4. log3 | 3 – 4x | > 2 Ans.
,3
23,
5. log0.2 (x2 – x – 2 ) > log0.2 (– x2 + 2x + 3) Ans.
25,2
6. 2x6x3loglog 22
31
)3.0(
> 1 Ans. (power < 0)
7. log0.5
4xxxlog
2
6 < 0 Ans. (– 4, – 3) (8, )
8.|5x|x3|x4x|log 2
2
3
0 Ans.
2,
21
32,
9.
x343x4log2
2 > – 21
Ans.
34,
43 or all domain
10. 8xlog3xlog2 323 6xlog3xlog2 3
23 3
[Hint : Put 6xlog3xlog2 323 = t ]