JEE Mathematics - Theory of Equations Sections Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
If tn denotes nth term of an AP and \({ t }_{ p }=\frac { 1 }{ q } \) and \({ t }_{ q }=\frac { 1 }{ q } \) then which of the following is necessarily a root of the equation \((p+2q-3r){ x }^{ 2 }+(q+2r-3p)x+(r+2p-3q)=0\) is
- (a)
tp
- (b)
tq
- (c)
tpq
- (d)
tp+q
The curve y=(λ+1)x2+2 intersects the curve y=λx+3 in exact one point if λ equals
- (a)
{-2}
- (b)
{2}
- (c)
{-2,2}
- (d)
{1}
If the equation ax2+bx+c=0 and x3+3x2+3x+2=0 have two common roots,then
- (a)
a=b≠c
- (b)
a≠b=c
- (c)
a=b=c
- (d)
a=-b=c
The equation \((x-3)^{ 9 }+(x-3^{ 2 })^{ 9 }+(x-3^{ 3 })^{ 9 }+...(x-3^{ 9 })^{ 9 }=0\) has
- (a)
all the roots are real
- (b)
1 real and 8 imaginary roots
- (c)
real roots namely x=3 32,....39
- (d)
five real and 4 imaginary roots
Number of ordered pair (x,y) satisfying x2+1=y and y2+1=x is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
the number of roots of the equation \({ 2 }^{ x }+{ 2 }^{ x-1 }+{ 2 }^{ x-2 }=7^{ x }+7^{ x-1 }+7^{ x-2 }\) is
- (a)
4
- (b)
2
- (c)
1
- (d)
0
The value of k for which the expression x2+2xy+ky2+2x+k=0 can be resolved into linear factors given by
- (a)
{0,2}
- (b)
{0}
- (c)
{-2,0}
- (d)
{2}
Let \(f(x)={ x }^{ 3 }+{ x }^{ 2 }+100x+7\quad sinx\) then equation \(\frac { 1 }{ y-f(1) } +\frac { 2 }{ y-f(2) } +\frac { 3 }{ y-f(3) } =0\) has
- (a)
no real root
- (b)
one real root
- (c)
two real roots
- (d)
more than two real roots
If ∝ and β are the roots of the equation x2-](x+1)-q=0, then the value of \(\frac { \alpha ^{ 2 }+2\alpha +1 }{ \alpha ^{ 2 }+2\alpha +q } +\frac { \beta ^{ 2 }+2\beta +1 }{ \beta ^{ 2 }+2\beta +q } \) is
- (a)
2
- (b)
1
- (c)
0
- (d)
none of these
Let f(x)=ax2+bx+c and f(-1)<1,f(1)>-1,f(3)<-4 and a≠0 then
- (a)
a>0
- (b)
a<0
- (c)
sign of a cannot be determined
- (d)
none of the above
The number of real solution of \(x-\frac { 1 }{ { x }^{ 2 }-4 } =2-\frac { 1 }{ { x }^{ 2 }-4 } \) is
- (a)
0
- (b)
1
- (c)
2
- (d)
infinite
The number of values of triplet (a,b,c) for which a cos 2x+b sin2 x+c=0 is satisfied by all real x is
- (a)
2
- (b)
4
- (c)
6
- (d)
infinite
The numbe rof real solution of the equation \({ 2 }^{ x/2 }+(\sqrt { 2 } +1)^{ x }=(5+2\sqrt { 2 } )^{ x/2 }\) is
- (a)
infinite
- (b)
six
- (c)
four
- (d)
one
The equation \(\sqrt { (x+1) } -\sqrt { (x-1) } =\sqrt { (4x-1) } \) has
- (a)
no solution
- (b)
one solution
- (c)
two solution
- (d)
more than two solution
If ∝,β are the roots of the equation x2+x+1=0 ,then the equation whose roots are ∝19 and β7 is
- (a)
x2+x+1=0
- (b)
x2-x-1=0
- (c)
x2+x+2=0
- (d)
x2+19x+7=0
The number of real solutions of the equation cos (ex) =2x+2-x is
- (a)
0
- (b)
1
- (c)
2
- (d)
infinitely many
If the roots of the equation, X2 + 2ax + b = 0, are real and distinct and they differ by at most 2m, then b lies in the interval
- (a)
(a2-m2,a2)
- (b)
[a2-m2,a2]
- (c)
(a2,a2,m2)
- (d)
none of these
If x2 + px + 1 is a factor of the expression ax3 + bx2 + c, then
- (a)
a2+c2=-ab
- (b)
a2-c2=-ab
- (c)
a2-c2=-bc
- (d)
none of these
Given that, for all x∈ R ,the expression \(\frac { { x }^{ 2 }-2x+4 }{ { x }^{ 2 }+2x+4 } \) lies between \(\frac { 1 }{ 3 } \) and 3, the value between which the expression\(\frac { 9.3^{ 2x }+6.3^{ x }+4 }{ 9.3^{ 2x }-6.3^{ x }+4 } \) lies are
- (a)
3-1 and 3
- (b)
-2 and 0
- (c)
-1 and 1
- (d)
0 and 2
If ∝ is one root of the equation 4x2+2x-1=0, then its other root is given by
- (a)
4∝3-3∝
- (b)
4∝3+3∝
- (c)
∝-(1/2)
- (d)
-∝-(1/2)
The number of number-pairs (x, y) which will satisfy the equation x2 - xy + y2 = 4 (x + y - 4) is
- (a)
1
- (b)
2
- (c)
4
- (d)
none of these
For any real x the expression 2(k-x) \([+\sqrt { { x }^{ 2 }+{ k }^{ 2 } } ]\\ \) can not exceed
- (a)
k2
- (b)
2k2
- (c)
3k2
- (d)
none of these
The number of real solution of the equation \(\left( \frac { 9 }{ 10 } \right) ^{ x }=-3+x-{ x }^{ 2 }\) is
- (a)
none
- (b)
one
- (c)
two
- (d)
more than two
The equation \(|x+1|^{ \log _{ (x+1) }{ (3+2x-{ x }^{ 2 }) } }=(x-3)|x|\) has
- (a)
unique solution
- (b)
two solutions
- (c)
no solution
- (d)
more than two solutions
For the equation Ix2 - 2x - 3| = b which statement or statements are true
- (a)
for b < 0 there are no solutions
- (b)
for b = 0 there are three solutions
- (c)
for 0 < b < 1 there are four solutions
- (d)
for b = 1 there are two solutions
The equation |x+1||x-1|=a2-2a-3 can have real solution for x, if a belongs x to
- (a)
(-∾,-1}U[3,∾)
- (b)
[1-\(\sqrt { 5 } \),1+\(\sqrt { 5 } \)]
- (c)
[1-\(\sqrt { 5 } \) ,-1]∪[3,1+\(\sqrt { 5 } \)]
- (d)
none of these
The equation x2+a2x+b2=0 has two roots each of which a number c, then
- (a)
a4>4b2
- (b)
c2+a2c+b2>0
- (c)
-a2/2>c
- (d)
none of these
If y = 2 [x] + 3 = 3 [x - 2] + 5, then [x + y] is ([x] denotes the integral part of x)
- (a)
10
- (b)
15
- (c)
12
- (d)
none of these
Let P, Q, R, Sand T are five sets about the quadratic equation (a - 5)x2-2ax+(a - 4) = 0, a≠-5 such that
P: All values of a for which the product of roots of given quadratic equation is positive.
Q: All values of a for which the product of roots of given quadratic equation is negative.
R: All values of a for which the product of real roots of given quadratic equation is positive.
S: All values of a for which the roots of given quadratic equation are real.
T: All values of a for which the given quadratic equation has complex roots.
Which statement is correct
- (a)
least positive integer for set R is 2
- (b)
least positive integer for set R is 3
- (c)
greatest positive integer for set T is 3
- (d)
none of the above
Let P, Q, R, Sand T are five sets about the quadratic equation (a - 5)x2-2ax+(a - 4) = 0, a≠-5 such that
P: All values of a for which the product of roots of given quadratic equation is positive.
Q: All values of a for which the product of roots of given quadratic equation is negative.
R: All values of a for which the product of real roots of given quadratic equation is positive.
S: All values of a for which the roots of given quadratic equation are real.
T: All values of a for which the given quadratic equation has complex roots.
Which statement is correct
- (a)
P U Q=S U T
- (b)
P U Q=S U T ~{4,5}
- (c)
P=T
- (d)
none of the above
The largest interval in which x12- x9 + X4 - X + 1 > 0 is
- (a)
[0,∞)
- (b)
(-∞,0]
- (c)
(-∞,∞)
- (d)
none of these
If 5 {x}=x+[x] and [x]-{x}= \(\frac { 1 }{ 2 } \) where {x} and [x] are fractional; and integral part of x then x is
- (a)
1/2
- (b)
3/2
- (c)
5/2
- (d)
7/2
If ∝,β,\(\gamma \) be the roots of the equation ax3+bx2+cx+d=0. To obtain the equation whose are f(∝),f(β),f(\(\gamma \)), where f is a function, we put y=f(∝) and obtain ∝=f-1(y)
Now, ∝ is a root of the equation ax3+bx2+cx+d=0, then we obtain the desired equation which is a {f-1(y)}3+b{f-1(y)}2+c{f-1(y)}+d=0
For example, if ∝,β,\(\gamma \) are the roots of ax3+bx2+cx+d=0. To find equation whose are ∝2,β2,\(\gamma \)2, we put y=∝2
⇒ ∝=\(\sqrt { y } \)
As ∝ is a root of ax3+bx2+cx+d=0
we get ay3/2+by+c\(\sqrt { y } \)+d=0
or \(\sqrt { y } \)(ay+c)=-(by+d)
On squaring both sides, then y(a2y2+2acy+c2)=b2y2+2bdy+d2 or a2y3+(2ac-b2)y2+(c2-2bd)y-d2=0 This is desired equation
If ∝,β are the roots of the equation 2x2+4x-5=0, the equation whose roots are the reciprocls of 2∝-3 and 2β-3 is
- (a)
x2+10x-11=0
- (b)
11x2+10x+1=0
- (c)
x2+10x+11=0
- (d)
11x2-10x+1=0
The values of a for which the equation 2 (log3x)2 -llog3 X I + a = a possess four real solutions
- (a)
-2<a<0
- (b)
\(0<a<\frac { 1 }{ 8 } \)
- (c)
0<a<5
- (d)
none of these
The solution set of (x)2 + (x + 1)2 = 25, where (x) is the nearest integer greater than or equal to x, is
- (a)
(2,4)
- (b)
[-5,-4] U [2,3)
- (c)
[-4,-3) U [3,4)
- (d)
none of these
If 0<x<1000 and \(\left[ \frac { x }{ 2 } \right] +\left[ \frac { x }{ 3 } \right] +\left[ \frac { x }{ 5 } \right] =\frac { 31 }{ 30 } x\) , where [x] is the greatest integer less than or equal to x, the number of possible values of x is
- (a)
34
- (b)
33
- (c)
32
- (d)
none of these
Let consider quadratic equation ax2 + bx + c = 0 .... (i)
where \(a,b,c\epsilon R\) and \(a\neq 0\). If Eq. (i) has roots, \(\alpha ,\beta \)
\(\therefore \quad \alpha +\beta =-\frac { b }{ a } ,\alpha \beta =\frac { c }{ a } \) and Eq. (i) can be written as ax2 + bx + c = a(x - \(\alpha \))(x - \(\beta \)).
Also, if a1 , a2 , a3, a4 , .... are in AP, then \({ a }_{ 2 }-{ a }_{ 1 }={ a }_{ 3 }-{ a }_{ 2 }={ a }_{ 4 }-{ a }_{ 3 }=....\neq 0\) and if b1 , b2 , b3 , b4 , ... are in GP, then \(\frac { { b }_{ 2 } }{ { b }_{ 1 } } =\frac { { b }_{ 3 } }{ { b }_{ 2 } } =\frac { { b }_{ 4 } }{ { b }_{ 3 } } =...\neq 1\) Now, if c1 , c2 , c3 , c4 ,.... are in HP, then \(\frac { 1 }{ { c }_{ 2 } } -\frac { 1 }{ { c }_{ 1 } } =\frac { 1 }{ { c }_{ 3 } } -\frac { 1 }{ { c }_{ 2 } } =\frac { 1 }{ { c }_{ 4 } } -\frac { 1 }{ { c }_{ 3 } } =...\neq 0\)
On the basis of above information, answer the following questions:
Let p and q be roots of the equation x2 - 2x + A = 0 and let r and s be the roots of the equation x2 - 18x + B = 0. If p < q < r < s are in arithmetic progression. Then the values of A and B respectively are
- (a)
-5, 67
- (b)
-3, 77
- (c)
67, -5
- (d)
77, -3
Let consider quadratic equation ax2 + bx + c = 0 .... (i)
where \(a,b,c\epsilon R\) and \(a\neq 0\). If Eq. (i) has roots, \(\alpha ,\beta \)
\(\therefore \quad \alpha +\beta =-\frac { b }{ a } ,\alpha \beta =\frac { c }{ a } \) and Eq. (i) can be written as ax2 + bx + c = a(x - \(\alpha \))(x - \(\beta \)).
Also, if a 1 , a 2 , a3, a 4 , .... are in AP, then \({ a }_{ 2 }-{ a }_{ 1 }={ a }_{ 3 }-{ a }_{ 2 }={ a }_{ 4 }-{ a }_{ 3 }=....\neq 0\) and if b 1 , b 2 , b 3 , b 4 , ... are in GP, then \(\frac { { b }_{ 2 } }{ { b }_{ 1 } } =\frac { { b }_{ 3 } }{ { b }_{ 2 } } =\frac { { b }_{ 4 } }{ { b }_{ 3 } } =...\neq 1\) Now, if c 1 , c 2 , c 3 , c 4 , .... are in HP, then \(\frac { 1 }{ { c }_{ 2 } } -\frac { 1 }{ { c }_{ 1 } } =\frac { 1 }{ { c }_{ 3 } } -\frac { 1 }{ { c }_{ 2 } } =\frac { 1 }{ { c }_{ 4 } } -\frac { 1 }{ { c }_{ 3 } } =...\neq 0\)
On the basis of above information, answer the following questions:
Given that \({ \beta }_{ 1 },{ \beta }_{ 3 }\) be roots of the equation Ax2 - 4x + 1 = 0 and \({ \beta }_{ 2 },{ \beta }_{ 4 }\) the roots of the equation Bx2 - 6x + 1 = 0. If \({ \beta }_{ 1 },{ \beta }_{ 2 },{ { \beta }_{ 3 },\beta }_{ 4 }\) are in HP; then the integral values of A and B respectively are
- (a)
-3, 8
- (b)
-3. 16
- (c)
3,8
- (d)
3,16
Let consider quadratic equation ax2 + bx + c = 0 .... (i)
where \(a,b,c\epsilon R\) and \(a\neq 0\). If Eq. (i) has roots, \(\alpha ,\beta \)
\(\therefore \quad \alpha +\beta =-\frac { b }{ a } ,\alpha \beta =\frac { c }{ a } \) and Eq. (i) can be written as ax2 + bx + c = a(x - \(\alpha \))(x - \(\beta \)).
Also, if a 1 , a 2 , a3, a 4 , .... are in AP, then \({ a }_{ 2 }-{ a }_{ 1 }={ a }_{ 3 }-{ a }_{ 2 }={ a }_{ 4 }-{ a }_{ 3 }=....\neq 0\) and if b 1 , b 2 , b 3 , b 4 , ... are in GP, then \(\frac { { b }_{ 2 } }{ { b }_{ 1 } } =\frac { { b }_{ 3 } }{ { b }_{ 2 } } =\frac { { b }_{ 4 } }{ { b }_{ 3 } } =...\neq 1\) Now, if c 1 , c 2 , c 3 , c 4 , .... are in HP, then \(\frac { 1 }{ { c }_{ 2 } } -\frac { 1 }{ { c }_{ 1 } } =\frac { 1 }{ { c }_{ 3 } } -\frac { 1 }{ { c }_{ 2 } } =\frac { 1 }{ { c }_{ 4 } } -\frac { 1 }{ { c }_{ 3 } } =...\neq 0\)
On the basis of above information, answer the following questions:
The harmonic mean of the roots of the equation \(\left( 5+\sqrt { 2 } \right) { x }^{ 2 }-\left( 4+\sqrt { 5 } \right) x+8+2\sqrt { 5 } =0\) is
- (a)
2
- (b)
4
- (c)
6
- (d)
8
Let \(\left( a+\sqrt { b } \right) ^{ Q(x) }+\left( a-\sqrt { b } \right) ^{ Q(x)-2\lambda }=A,\) where \(\lambda \epsilon N,A\varepsilon R\) and a2 - b = 1
\(\therefore \quad \left( a+\sqrt { b } \right) \left( a-\sqrt { b } \right) =1\quad \Rightarrow \quad \left( a+\sqrt { b } \right) =\left( a-\sqrt { b } \right) ^{ -1 }\quad and\quad \left( a-\sqrt { b } \right) =\left( a+\sqrt { b } \right) ^{ -1 }\)
ie, \(\left( a\pm \sqrt { b } \right) =\left( a+\sqrt { b } \right) ^{ \pm 1 }\quad or\quad \left( a-\sqrt { b } \right) ^{ \pm 1 }\)
On the basis of above information, answer the following questions:
Solution of \(\left( 2+\sqrt { 3 } \right) ^{ { x }^{ 2 }-2x+1 }+\left( 2-\sqrt { 3 } \right) ^{ { x }^{ 2 }-2x-1 }=\frac { 4 }{ 2-\sqrt { 3 } } \) are
- (a)
\(1\pm \sqrt { 3 } ,1\)
- (b)
\(1\pm \sqrt { 2 } ,1\)
- (c)
\(1\pm \sqrt { 3 } ,2\)
- (d)
\(1\pm \sqrt { 2} ,2\)
Let \(\left( a+\sqrt { b } \right) ^{ Q(x) }+\left( a-\sqrt { b } \right) ^{ Q(x)-2\lambda }=A,\) where \(\lambda \epsilon N,A\varepsilon R\) and a2 - b = 1
\(\therefore \quad \left( a+\sqrt { b } \right) \left( a-\sqrt { b } \right) =1\quad \Rightarrow \quad \left( a+\sqrt { b } \right) =\left( a-\sqrt { b } \right) ^{ -1 }\quad and\quad \left( a-\sqrt { b } \right) =\left( a+\sqrt { b } \right) ^{ -1 }\)
ie, \(\left( a\pm \sqrt { b } \right) =\left( a+\sqrt { b } \right) ^{ \pm 1 }\quad or\quad \left( a-\sqrt { b } \right) ^{ \pm 1 }\)
On the basis of above information, answer the following questions:
If \(\alpha ,\beta \) are the roots of the equation \(1!+2!+3!+...\left( x-1 \right) !+x!={ k }^{ 2 }\) and \(k\epsilon I\) , where \(\alpha <\beta \) and if \({ \alpha }_{ 1 },{ \alpha }_{ 2 },{ \alpha }_{ 3 },{ \alpha }_{ 4 }\) are the roots of the equation \(\left( a+\sqrt { b } \right) ^{ { x }^{ 2 }-\left[ 1+2\alpha +3{ \alpha }^{ 2 }+{ 4\alpha }^{ 3 }+{ 5\alpha }^{ 4 } \right] }+\left( a-\sqrt { b } \right) ^{ { x }^{ 2 }+\left[ -5\beta \right] }=2a\) where a2 - b = 1 and [.] denotes the greatest integer function, then the value of \(\left| { \alpha }_{ 1 }+{ \alpha }_{ 2 }+{ \alpha }_{ 3 }+{ \alpha }_{ 4 }-{ \alpha }_{ 1 }{ \alpha }_{ 2 }{ \alpha }_{ 3 }{ \alpha }_{ 4 } \right| \) is
- (a)
216
- (b)
221
- (c)
224
- (d)
209
If x2 +px+1 is a factor of 2 cos2 θ x3 + 2x + sinθ, then
- (a)
\(\theta =n\pi ,n\epsilon 1\)
- (b)
\(θ=nπ+\frac { \pi }{ 2 } ,n\epsilon 1\)
- (c)
\(\theta =2n\pi ,n\epsilon 1\)
- (d)
\(\theta =\frac { n\pi }{ 2 } ,n\epsilon 1\)
If roots of ax2+2bx+c=0 (a≠0) are non real complex and a+c<2b
- (a)
c>0
- (b)
c<0
- (c)
4a+c<4b
- (d)
4a+c>4b
If ax+by=1, cx2+dy2=1 have only one solution, then
- (a)
\(\frac { { a }^{ 2 } }{ c } +\frac { { b }^{ 2 } }{ d } \)=1
- (b)
x=a/c
- (c)
y=b/d
- (d)
none of these
If c≠0 and the equation \(\frac { p }{ 2x } =\frac { a }{ x+c } +\frac { b }{ x-c } \) has two equal roots, then p can be
- (a)
\((\sqrt { a } -\sqrt { b } )^{ 2 }\)
- (b)
\((\sqrt { a } +\sqrt { b } )^{ 2 }\)
- (c)
a+b
- (d)
a-b
Suppose a,b \(\in \) R and a≠0, b≠0. Let ∝,β be the roots of x2+ax+b=0 then
- (a)
1/∝,1/β are roots of bx2+ax+1=0
- (b)
-∝,-β are roots of x2-ax+b=0
- (c)
∝2,-β2 are roots of x2+(2b-a2)x+b2=0
- (d)
∝/β,β/∝ are roots of bx2+(2b-a2)x+b=0
If ∝ and β are the roots of the equation x2+pa+q=0 and ∝4,β4 are the roots of x2-rx+s=0, then the equation x2-4qx+2q2-r=0 has always
- (a)
two real roots
- (b)
two positive roots
- (c)
two negative roots
- (d)
one positive and one negative root
If ∝,β,\(\gamma \) be the roots of the equation ax3+bx2+cx+d=0. To obtain the equation whose are f(∝),f(β),f(\(\gamma \)), where f is a function, we put y=f(∝) and obtain ∝=f-1(y)
Now, ∝ is a root of the equation ax3+bx2+cx+d=0, then we obtain the desired equation which is a {f-1(y)}3+b{f-1(y)}2+c{f-1(y)}+d=0
For example, if ∝,β,\(\gamma \) are the roots of ax3+bx2+cx+d=0. To find equation whose are ∝2,β2,\(\gamma \)2, we put y=∝2
⇒ ∝=\(\sqrt { y } \)
As ∝ is a root of ax3+bx2+cx+d=0
we get ay3/2+by+c\(\sqrt { y } \)+d=0
or \(\sqrt { y } \)(ay+c)=-(by+d)
On squaring both sides, then y(a2y2+2acy+c2)=b2y2+2bdy+d2 or a2y3+(2ac-b2)y2+(c2-2bd)y-d2=0 This is desired equation
If ∝,β are the roots of the equation px2-qx+r=0, then the equation whose roots are ∝2+r/p and β2+r/p is
- (a)
p3x2+pq2x+r=0
- (b)
px2-qx+r=0
- (c)
p3x2-pq2x+q2r=0
- (d)
px2+qx-r=0
If ∝,β,\(\gamma \) be the roots of the equation ax3+bx2+cx+d=0. To obtain the equation whose are f(∝),f(β),f(\(\gamma \)), where f is a function, we put y=f(∝) and obtain ∝=f-1(y)
Now, ∝ is a root of the equation ax3+bx2+cx+d=0, then we obtain the desired equation which is a {f-1(y)}3+b{f-1(y)}2+c{f-1(y)}+d=0
For example, if ∝,β,\(\gamma \) are the roots of ax3+bx2+cx+d=0. To find equation whose are ∝2,β2,\(\gamma \)2, we put y=∝2
⇒ ∝=\(\sqrt { y } \)
As ∝ is a root of ax3+bx2+cx+d=0
we get ay3/2+by+c\(\sqrt { y } \)+d=0
or \(\sqrt { y } \)(ay+c)=-(by+d)
On squaring both sides, then y(a2y2+2acy+c2)=b2y2+2bdy+d2 or a2y3+(2ac-b2)y2+(c2-2bd)y-d2=0 This is desired equation
If ∝,β,\(\gamma \) are the roots of the equation x3+qx-r=0 then the equation whose roots are \(\beta \gamma +\frac { 1 }{ \alpha } ,\gamma \alpha +\frac { 1 }{ \beta } \quad \alpha \beta +\frac { 1 }{ \gamma } \) is
- (a)
rx3+q(r+1)x2-(r+1)3=0
- (b)
rx3-q(r+1)x2-(r+1)3=0
- (c)
rx3+q(r+1)x2+(r+1)3=0
- (d)
rx3+q(r+1)x2+(r+1)2=0
If ∝,β,\(\gamma \) be the roots of the equation ax3+bx2+cx+d=0. To obtain the equation whose are f(∝),f(β),f(\(\gamma \)), where f is a function, we put y=f(∝) and obtain ∝=f-1(y)
Now, ∝ is a root of the equation ax3+bx2+cx+d=0, then we obtain the desired equation which is a {f-1(y)}3+b{f-1(y)}2+c{f-1(y)}+d=0
For example, if ∝,β,\(\gamma \) are the roots of ax3+bx2+cx+d=0. To find equation whose are ∝2,β2,\(\gamma \)2, we put y=∝2
⇒ ∝=\(\sqrt { y } \)
As ∝ is a root of ax3+bx2+cx+d=0
we get ay3/2+by+c\(\sqrt { y } \)+d=0
or \(\sqrt { y } \)(ay+c)=-(by+d)
On squaring both sides, then y(a2y2+2acy+c2)=b2y2+2bdy+d2 or a2y3+(2ac-b2)y2+(c2-2bd)y-d2=0 This is desired equation
If ∝,β,\(\gamma \) are the roots of the equation x3+3bx+c=0, then the equation whose roots are (∝-β)(∝-\(\gamma \)),(β-\(\gamma \))(β-α).(\(\gamma \)-∝)(\(\gamma \)-β)
- (a)
x3-6x2+216=0
- (b)
x3-3x2+112=0
- (c)
x3+6x2-216=0
- (d)
x3+3x2-112=0