IISER Mathematics - Theory of Equations
Exam Duration: 45 Mins Total Questions : 30
If one root of the equation \({ x }^{ 2 }-\lambda x+12=0\) is even prime while \({ x }^{ 2 }+\lambda x+\mu =0\) has equal roots, then μ is
- (a)
8
- (b)
16
- (c)
24
- (d)
32
We know that only even prime is 2
then \((2)^{ 2 }-\lambda (2)+12=0\)
\(\Rightarrow \quad \quad \lambda =8\)
and \({ x }^{ 2 }+\lambda x+\mu =0\) ....(i)
has equal roots
\({ \lambda }^{ 2 }-4\mu =0\)
or \((8)^{ 2 }-4\mu =0\) [From Eq.(i)]
\(\therefore \mu =16\)
The value of a for which the equation x3+ax+1=0 and x4+ax2+1=0 have a common root is
- (a)
-2
- (b)
-1
- (c)
1
- (d)
2
Multiply first equation by x
then x4+ax2+x=0 ......(i)
and given x4+ax2+1=0 ....(ii)
Subtract Eq.(ii) from (i) then x-1=0⇒x=1 is a common root
then 1+a+1=0
∴ a=-2
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 ( λ+1)x2+2=λx+3 has only one solution.If discriminant=0
\(\Rightarrow { \lambda }^{ 2 }-4(\lambda +1)(-1)=0\)
\(\Rightarrow { \lambda }^{ 2 }+4\lambda +4=0\)
\(\Rightarrow \quad (\lambda +2)^{ 2 }=0\)
\(\therefore \quad \lambda =-2\)
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
Let \((x-3)^{ 9 }+(x-3^{ 2 })^{ 9 }+(x-3^{ 3 })^{ 9 }+...(x-3^{ 9 })^{ 9 }=0\)
\(\therefore \quad f'(x)=9\{ (x-3)^{ 8 }+(x-3^{ 2 })^{ 8 }+(x-3^{ 3 })^{ 8 }+...+(x-3^{ 9 })^{ 8 }\} >0\)
So, f'(x)=0 has no real roots
So,f (x)=0 has one real and 8 imaginary roots
The number of real solutions of the equation 2 |x|2-5|x|+2=0 is
- (a)
0
- (b)
2
- (c)
4
- (d)
infinite
2|x|2 -5|x|+2=0
⇒(2|x|-1)(|x|-2)=0
∴ |x| =1/2,2
∴ x=±1/2,±2e
Let ∝,β be the roots of the equation (x-a)(x-b)=c,c≠0.Then the roots of the equation (x-∝)(x-β)+c=0 are
- (a)
a,c
- (b)
b,c
- (c)
a,b
- (d)
a+c,b+c
∵ (x-1)(x-b)
=(x-∝)(x-β)
∴ (x-∝)(x-β)+c
=(x-a)(x-b)
Then the roots of (x-∝)(x-β=c are
x=1,b
The number of real solutions of \(1+|{ e }^{ x }-1|={ e }^{ x }({ e }^{ x }-2)\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
4
\(1+1+|{ e }^{ x }-1|={ e }^{ x }-2{ e }^{ x }+1\)
\(\Rightarrow \quad 2+|{ e }^{ x }-1|=|{ e }^{ x }-1|^{ 2 }\)
\(\Rightarrow \quad |{ e }^{ x }-1|^{ 2 }-|{ e }^{ x }-1|-2=0\)
\(\therefore \quad \quad |{ e }^{ x }-1|=2,-1\quad \) \((\because \quad |{ e }^{ x }-1|\neq -1)\)
\(\Rightarrow \quad |{ e }^{ x }-1|=2={ e }^{ x }-1=\pm 2\)
\(\therefore \quad { e }^{ x }=1\pm 2=3,-1\) \((\because \quad { e }^{ x }\neq -1)\)
\(\Rightarrow \quad { e }^{ x }=3\Rightarrow x={ log }_{ e }3\)
∴ No.of solutions is one
The roots of the equation , \({ 2 }^{ x+2 },3^{ 3x/(x-1) }=9\) are given by
- (a)
1-log2 3,2
- (b)
log2 (2/3),1
- (c)
2,-2
- (d)
-2,1-(log 3)/(log 2)
\({ 2 }^{ x+2 },3^{ 3x/(x-1) }=9\)
Taking logarithm
Then \((x+2)log2+\frac { 3 }{ (x-1) } log3=2log3\)
\(\Rightarrow \quad ({ x }^{ 2 }+x-2)log\quad 2+3x\quad log3=2(x-1)log3\)
\(\Rightarrow { x }^{ 2 }\quad log2+(log2+log3)x-2log2+2log3=0\)
\(x=\frac { -(log2+log3)\pm \sqrt { \{ (log2+log3)^{ 2 }-4log2(-2\quad log\quad 2+2\quad log\quad 3)\} } }{ 2log2 } \)
\(x=\frac { -(log2+log3)\pm \sqrt { \{ (3log\quad 2)^{ 2 }-6\quad log\quad 2\quad log3+(log3)^{ 2 }\} } }{ 2log2 } \)
\(=\frac { -(log\quad 2+log3)\pm (3\quad log\quad 2-log3) }{ 2\quad log\quad 2 } \)
\(\therefore \quad \quad \quad x=-2,1-\frac { (log\quad 3) }{ (log\quad 2) } \)
The solution of the equation \({ 3 }^{ { log }_{ a }x }+{ 3 }^{ { log }_{ a }3 }=2\) is given by
- (a)
\({ 3 }^{ \log _{ 2 }{ a } }\)
- (b)
\({ 3 }^{ -\log _{ 2 }{ a } }\)
- (c)
\({ 2 }^{ \log _{ 3 }{ a } }\)
- (d)
\({ 2 }^{ -\log _{ 3 }{ a } }\)
\({ 3 }^{ \log _{ a }{ x } }+{ 3 }x^{ \log _{ a }{ 3 } }=2\)
\(\Rightarrow \quad \quad \quad { 3 }^{ \log _{ a }{ x } }+{ 3.3 }x^{ \log _{ a }{ x } }=2\)
\(\Rightarrow \quad \quad \quad 4.{ 3 }^{ \log _{ a }{ x } }=2\)
\(\Rightarrow \quad \quad \quad \quad { 3 }^{ \log _{ a }{ x } }=1/2\Rightarrow log_{ a }x=-log_{ 3 }2\)
\(\therefore \quad \quad x=a^{ -\log _{ 3 }{ 2 } }=a^{ \log _{ 3 }{ 2(1/2) } }\)
\(=(1/2)^{ \log _{ 3 }{ a } }=2^{ -\log _{ 3 }{ a } }\)
If ∝,β,⋎,δ are the roots of x4 + x2 + 1 = 0, then the equation whose roots are ∝2,β2,⋎2,δ2 is
- (a)
(x2-x+1)2=0
- (b)
(x2-x+1)2=0
- (c)
x4-x2+1=0
- (d)
x2-x+1=0
Since ∝,β,⋎,δ are the roots of
x4-x2+1=0 ..... (i)
Now ,let y=∝2
Since ∝ is root of Eq (i)
∴ ∝4+∝2+1=0
or y4+y+1=0 [from Eq (ii)]
Hence,the equation whose roots are ∝2,β2,⋎2,δ2 is
(x2+x+1)2=0
The solution set of the equation \(\log _{ x }{ 2log_{ 2x } } 2=log_{ 4x }2\) is
- (a)
\(\{ { 2 }^{ -\sqrt { 2 } },{ 2 }^{ \sqrt { 2 } }\} \)
- (b)
\(\left\{ \frac { 1 }{ 2 } ,2 \right\} \)
- (c)
\(\left\{ \frac { 1 }{ 4 } ,2^{ 2 } \right\} \)
- (d)
none of these
\(\log _{ x }{ 2log_{ 2x } } 2=log_{ 4x }2\)
Here x>0 and x≠ 1, \(\frac { 1 }{ 2 } ,\frac { 1 }{ 4 } \)
\(\Rightarrow \quad \frac { 1 }{ \log _{ 2 }{ x } } .\frac { 1 }{ \log _{ 2 }{ 2x } } =\frac { 1 }{ \log _{ 2 }{ 4x } } \)
\(\Rightarrow \quad \log _{ 2 }{ x.(1+\log _{ 2 }{ x } )=(2+\log _{ 2 }{ x) } } \)
Put \(\log _{ 2 }{ x } =t\)
\(\therefore \quad \quad { t }^{ 2 }=2\)
\(\therefore \quad \quad t=\pm \sqrt { 2 } \)
Then \(\log _{ 2 }{ x } =\pm \sqrt { 2 } \)
\(x={ 2 }^{ \pm \sqrt { 2 } }\)
\(\therefore \quad \quad x={ 2 }^{ -\sqrt { 2 } }.{ 2 }^{ \sqrt { 2 } }\)
If 2a+3b+6c=0 (a,b,c \(\in \)R), then the quadratic equation ax2+bx+c=0 has
- (a)
at least one root in [0,1]
- (b)
at least one root in (-1,1]
- (c)
at least one root in [0.2]
- (d)
none of the above
Let f(x)=\(\frac { ax^{ 3 } }{ 3 } +\frac { bx^{ 2 } }{ 2 } \)+cx+d
Note that f is continuous and differentiable on R also
f(1)=\(\frac { a }{ 3 } +\frac { b }{ 2 } \)+c+d=\(\frac { 2a+3b+6c }{ 6 } \)+d
∴ f(0)=f(1)=d
f'(λ)=aλ2+bλ+c=0
aλ2+bλ+c=0
By Rolle's theorem λ\(\in \)(0,1)
Thus, ax2+bx+c=0 has at least one root in [0,1] but [0.1]⊆[-1,1] and [0.1]⊆[0,2]
If A,G and If an. the Arithmetic mean, Geometric mean and Harmonic mean between two unequal positive integers. Then the equation Ax2-|G|x-H=0 has
- (a)
both roots are fractions
- (b)
at least one root which is negative fraction
- (c)
exactly one positive root
- (d)
at least one root which is an integer
Given equation is
Ax2-|G|x-H=0
Let ∝,β are the roots
then ∝+β=\(\frac { |G| }{ A } \) and ∝β=-H/A
Since A>|G|>H
or \(1>\frac { G }{ A } >\frac { G }{ A } \)
Hence, A is positive
∴ ∝+β and ∝β has positive and negative fraction respectively.
Also |G|2=AH
Discriminant of Eq (i)=(-|G|)2-4.A.(H)
=|G|2+4AH
=5|G|2>0 [from Eq (ii)]
Hence, roots of Eq (i) are real and distinct
∵ ∝+β>0 and ∝β<0
One root is positive and other is negative and at least one root is a fraction.
If a be a positive integer, the number of values of a satisfying \(\int _{ 0 }^{ \pi /2 }{ \left[ { a }^{ 2 }\left( \frac { cos\quad 3x }{ 4 } +\frac { 3 }{ 4 } cos\quad x \right) +a\quad sin\quad x-20\quad cos\quad x \right] } dx\le -\frac { { a }^{ 2 } }{ 3 } \) is
- (a)
only one
- (b)
two
- (c)
three
- (d)
four
\(\int _{ 0 }^{ \pi /2 }{ \left[ { a }^{ 2 }\left( \frac { cos\quad 3x }{ 4 } +\frac { 3 }{ 4 } cos\quad x \right) +a\quad sin\quad x-20\quad cos\quad x \right] } dx\le -\frac { { a }^{ 2 } }{ 3 } \)
\(\left[ { a }^{ 2 }\left( -\frac { 1 }{ 12 } +\frac { 3 }{ 4 } \right) -0-20+a \right] \le -\frac { { a }^{ 2 } }{ 3 } \)
\(\Rightarrow \frac { 2 }{ 3 } { a }^{ 2 }+a-20+\frac { { a }^{ 2 } }{ 3 } \le 0\)
\(\Rightarrow \quad { a }^{ 2 }+a-20\le 0\)
\(\Rightarrow \quad (a+5)(a-4)\le 0\)
\(\\ \therefore \quad \quad \quad \quad -5\le a\le 4\)
\(a=1,2,3,4\)
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 regarding sets P, Q and R
- (a)
P∩Q=ф
- (b)
R⊆P
- (c)
PUQ \(\in \)R' ~{4,5}
- (d)
All of the above
P⋂Q=ф
R⊆P
PUQ=a \(\in \){(-∾,4)U(5,∾)}U(4,5)
=R'~{4,5}
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 regarding sets P, Rand T
- (a)
P⊆R
- (b)
R⊆T
- (c)
T⊆P
- (d)
P∩R-T
T⊆P
Let F (x) be a' function defined by \(F(x)=x-[x],0\neq x\epsilon R\), where [x] is the greatest integer less than or equal to x. Then the number of solutions of F(x) + F(1/x) = 1 is/are
- (a)
0
- (b)
infinite
- (c)
1
- (d)
2
∵ F(x)=x-[x],0≠x∈ R
Given \(F(x)+F\left( \frac { 1 }{ x } \right) =1\)
\(\therefore \quad \quad \quad \Rightarrow \quad x=|x|+\frac { 1 }{ x } -\left[ \frac { 1 }{ x } \right] =1\)
\(\Rightarrow \quad \quad \quad \left( x+\frac { 1 }{ x } \right) -\left( \left[ x \right] +\left[ \frac { 1 }{ x } \right] \right) =1\)
∵ RHS is an integer, hence LHS is also an integer
Let \(\left[ x \right] +\left[ \frac { 1 }{ x } \right] +1=A\) (Integer)
Then Eq (i) becomes
\(x+\frac { 1 }{ x } =A\)
\(\Rightarrow \quad \quad \quad { x }^{ 2 }-Ax+1=0\)
\(\therefore \quad \quad \quad x=\frac { A\pm \sqrt { ({ A }^{ 2 }-4) } }{ 2 } \)
For real x A2-4≥ 0
∴ A≥2 and A≤ -2
∵ A=2 and A=-2 does not satisfy Eq (i)
∴ A>2 Eq (i) has infinite many solutions
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
5 {x}=x+[x] and [x]-{x}= \(\frac { 1 }{ 2 } \)
Since x=[x]+{x}
After solving [x]=1 and {x}=1/2
∴ x=1+1/2=3/2
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
Put \(y=\frac { 2 }{ 2\alpha -3 } \Rightarrow \frac { 1 }{ 2 } \left( 3+\frac { 1 }{ y } \right) \)
\(\therefore \quad 2\left\{ \frac { 1 }{ 2 } \left( 3+\frac { 1 }{ y } \right) \right\} ^{ 2 }+4\left\{ \frac { 1 }{ 2 } \left( 3+\frac { 1 }{ y } \right) \right\} -5=0\)
⇒ 11y2+10y+1=0
or 11x2+10x+1=0
Let consider the quadratic equation (1+m)x2-2(1+3m)x+(1+8m)=0, where m∈R~{-1}
The set of values of m such that the given quadratic equation has both roots are positive is
- (a)
m∊R
- (b)
m∈(-1,3)
- (c)
m∈[3,∞)
- (d)
(-∞,-1)U[3,∞)
∝+β>0 and ∝β>0
⇒ (∝+β>0)∩(∝β>0)∩(D≥0)
⇒\(\left( \frac { 2(1+3m) }{ (1+m) } >0 \right) \cap \left( \frac { 1+8m }{ 1+m } >0 \right) \cap \left\{ 4m(m-3)\ge 0 \right\} \)
m∈ \(\left\{ \left( -\infty ,-1 \right) \cup \left( -\frac { 1 }{ 3 } ,\infty \right) \right\} \cap \left\{ \left( -\infty ,-1 \right) \cup \left( -\frac { 1 }{ 8 } ,\infty \right) \right\} \cap \{ m\in (-\infty ,0]\cup [3,\infty )\} \)
m∈(-∞,-1)U[3,∞)
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
∵ LHS is an integer
∴ RHS is must be an integer for which x is multiple of 30
x=30,60,90,120,...990
⇒ Number of possible values of x is 33
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:
If a, b, c, d and x are distinct real numbers such that (a2 + b2 + c2 )x2 - 2(ab + bc +cd)x + (b2 + c2 + d2) \(\le\) 0, then a, b,c, d
- (a)
are in AP
- (b)
are in GP
- (c)
are in HP
- (d)
satisfy ab = cd
(ax - b)2 + (bx - c)2 + (cx - d)2 \(\le\) 0
which is possible, (ax - b)2 + (bx - c)2 + (cx - d)2 = 0 or ax - b = 0, bx - c = 0 and cx - d = 0
\(\frac { b }{ a } =\frac { c }{ b } =\frac { d }{ c } =x\)
Hence, a,b,c,d are in GP
If \({ sin }^{ x }\theta +cos^{ x }\theta \ge \pi /2\), then
- (a)
x∈ [2,∞)
- (b)
x∈ (-∞ ,2 ]
- (c)
x∈ [-1,1]
- (d)
x ∈ [-2,2]
∵ \({ sin }^{ 2 }\theta +cos^{ 2 }\theta =1\)
and \({ sin }^{ x }\theta +cos^{ x }\theta >1\) if x<2
⇒ x≤ 2
∴ x ∈ (-∞,2]
If a,b,c \(\in \)R and a+b+c=0, then the quadratic equation 3ax2+2bx+c=0 has
- (a)
at least one root in [0,1]
- (b)
at least one root in [-1'1]
- (c)
at least one root in [0,2]
- (d)
none of the above
Let f(x)=ax3+bx2+cx+d
f is continuous and derivable on R. Also, f(0)=d and f(1)=a+b+c+d=0+d=d. By the Rolle's theorem, there exists at least one ∝\(\in \)(0,1) such that
f'(∝)=0 ⇒ 3a∝2+2b∝+c=0
Thus, 3ax2+2bx+c=0 hsa at least one root in [0,1]
Also, [0,1]⊆[-1,1] and [0,1]⊆[0,2]
If the roots of the equation x3+bx2+cx-1=0 form an increasing GP, then
- (a)
b+c=0
- (b)
b\(\in \)(-∞,-3)
- (c)
one of the roots is 1
- (d)
one root is smaller than one and one root is more than 1
Let the roots of the given equation be ∝/r,∝,∝r, where ∝>0 and r>1. Then
\(\frac { \alpha }{ r } +\alpha +\alpha r\)=-b ....(1)
\(\frac { \alpha }{ r } .\alpha +\alpha +\alpha r+\frac { \alpha }{ r } .\alpha r\)=c .....(ii)
and \(\left( \frac { \alpha }{ r } \right) (\alpha )(\alpha r)\)=1 .........(iii)
From Eq (iii), we get ∝3=1 or ∝=1
From Eq (i), we get 1/r+1+r=-b .....(iv)
\(\Rightarrow \quad \left( \frac { 1 }{ \sqrt { r } } -\sqrt { r } \right) ^{ 2 }\)+3=-b
⇒ -b-3>0 or b<-3
⇒ b\(\in \)(-∞,-3)
Also from Eq (ii) 1/r+r+1=c
From Eqs (iv) and (v) -b=c or b+c=0
As r>1, \(\frac { \alpha }{ r } =\frac { 1 }{ r } \)<1 and ∝r=r>1
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
\(\frac { p }{ 2x } =\frac { (a+b)x+c(b-a) }{ { x }^{ 2 }-{ c }^{ 2 } } \)
or p(x2-c2)=2(a+b)x2-2c(a-b)x
or (2a+2b-p)x2-2c(a-b)x+pc2=0
Now, c2(a-b)2-pc2(2a+2b-p)=0 (∵ equal roots)
⇒ (a-b)2-2p(a+b)+p2=0 (∵ c2≠0)
⇒ [p-(a+b)]2=(a+b)2-(a-b)2
⇒ p=a+b±2\(\sqrt { ab } =(\sqrt { a } \pm \sqrt { b } )^{ 2 }\)
Let consider the quadratic equation (1+m)x2-2(1+3m)x+(1+8m)=0, where m\(\in \)R〜({-1}
The number of integral values of m such that given quadratic equation has imaginary roots are
- (a)
0
- (b)
1
- (c)
2
- (d)
3
D<0
⇒ 4m(m-3)<0
⇒ 0<m<3
∴ m=1,2
Let consider the quadratic equation (1+m)x2-2(1+3m)x+(1+8m)=0, where m\(\in \)R〜({-1}
Tho set of values of m such that the given quadratic equation has at least one root is negative is
- (a)
m\(\in \)(-∞,-1)
- (b)
\(m\in \left( -\frac { 1 }{ 8 } ,\infty \right) \)
- (c)
\(m\in \left( -1,-\frac { 1 }{ 8 } \right) \)
- (d)
m\(\in \)R
At least one root is negative ie, one root is negative or both roots are negative, then {∝β<0)U(∝+β<0)}⋂(D≥0)
⇒ \(\left\{ \left( \frac { (1+8m) }{ (1+m) } <0 \right) \cup \left( \frac { 2(1+3m) }{ (1+m) } <0 \right) \right\} \cap m\in (-\infty ,0]\cup [3,\infty )\)
⇒ \(\left\{ m\in \left( -1,-\frac { 1 }{ 8 } \right) \cup m\in \left( -1,-\frac { 1 }{ 3 } \right) \right\} \cap m\in (-\infty ,0]\cup [3,\infty )\)
⇒ \(\left\{ m\in \left( -1,-\frac { 1 }{ 8 } \right) \right\} \cap m\in (-\infty ,0]\cup [3,\infty )\)
ie, \(m\in \left( -1,-\frac { 1 }{ 8 } \right) \)
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
Put y=\({ \alpha }^{ 2 }+\frac { r }{ p } \Rightarrow \alpha =\sqrt { \left( y-\frac { r }{ p } \right) } \)
\(\therefore \quad p\left( y-\frac { r }{ p } \right) -q\sqrt { \left( y-\frac { r }{ p } \right) } +r\)=0
⇒ py=\(\Rightarrow q\sqrt { \left( y-\frac { r }{ p } \right) } \)
⇒ p2y2=q2y-q2r/p
or p3y2-pq2y+q2r=0
or p3x2-pq2x+q2r=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
Put y=\(\beta \gamma +\frac { 1 }{ \alpha } =\frac { \alpha \beta \gamma +1 }{ \alpha } =\frac { r+1 }{ \alpha } \)
or \(\alpha =\left( \frac { r+1 }{ y } \right) \)
Thus, \(\left( \frac { r+1 }{ y } \right) ^{ 3 }+q\left( \frac { r+1 }{ y } \right) \)-r=0
⇒ ry3-q(r+1)y2-(r+1)3=0
or rx3-q(r+1)x2-(r+1)3=0