Eamcet Mathematics -Theory of Equations Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
In a triangle PQR, \(\angle R=\frac { \pi }{ 2 } \).If \(tan\left( \frac { P }{ 2 } \right) \) and \(tan\left( \frac { Q }{ 2 } \right) \)are the roots of the equation \({ ax }^{ 2 }+bx+c=0(a\neq 0)\) then
- (a)
a+b=c
- (b)
b+c=a
- (c)
a+c=b
- (d)
b=c
\(\therefore \quad \quad P+Q+R=\pi \)
\(\Rightarrow \quad P+Q=\pi /2\)
\(\Rightarrow \left( \frac { P+Q }{ 2 } \right) =\pi /4\)
\(\therefore \quad tan\quad \left( \frac { P+Q }{ 2 } \right) =tan\left( \frac { \pi }{ 4 } \right) \)
\(\Rightarrow \quad \frac { tan\left( \frac { P }{ 2 } \right) +tan\left( \frac { Q }{ 2 } \right) }{ 1-tan\left( \frac { P }{ 2 } \right) tan\left( \frac { Q }{ 2 } \right) } =1\)
\(\Rightarrow \frac { -\frac { b }{ a } }{ 1-\frac { c }{ a } } =1\)
\(\Rightarrow \) -b=a-c
or a+b=c
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)]
\(\therefore \mu =16\)
Number of real roots of the equation \(\sqrt { x } +\sqrt { x-\sqrt { (1-x) } } =1\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
Given \(\sqrt { x } +\sqrt { x-\sqrt { (1-x) } } =1\) ...(i)
\(\Rightarrow \quad \quad \sqrt { x-\sqrt { (1-x) } } =1-\sqrt { x } \)
Squaring both sides
Then \(x-\sqrt { 1-x } =1+x-2\sqrt { x } \)
\(\Rightarrow -\sqrt { (1-x) } =1-2\sqrt { x } \)
Again, squaring both sides
\(\Rightarrow \quad \quad 1-x=1+4x-4\sqrt { x } \)
\(\Rightarrow \quad 4\sqrt { x } =5x\)
\(\Rightarrow \quad { 25 }x^{ 2 }-16x=0\)
\(\therefore \quad x=0,\frac { 16 }{ 25 } \)
\(x=\frac { 16 }{ 25 } \) (∵ x=0 does not satisfy Eq (i)
If x2-x-2 is a factor x4-λx2-μ ,then \(\sqrt { (\lambda ^{ 2 }-\mu ^{ 2 }) } \) equals
- (a)
1
- (b)
3
- (c)
5
- (d)
7
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x=2,-1\)
Then x=2,-1 are the roots of x4-λx2-μ=0
Then \((2)^{ 4 }-\lambda ({ 2) }^{ 2 }-\mu =0\)
\(\Rightarrow 4\lambda +\mu =16\)
and \((-1)^{ 4 }-\lambda ((-1)^{ 2 }-\mu =0\)
\(\Rightarrow \lambda +\mu =1\)
We get \(\lambda =5\quad and\quad \mu =-4\)
\(\therefore \quad \sqrt { (\lambda ^{ 2 }-\mu ^{ 2 }) } =\sqrt { 25-16 } =\sqrt { 9 } =3\)
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
If λ be an integer and α,β be the roots of x2-16x+λ=0 such that 1<∝<2 and 2<β3,then possible values of λ are
- (a)
{60,64,68}
- (b)
{61,62,63}
- (c)
{49,50,..62,63}
- (d)
{62,65,68,71,75}
\(\because \quad \quad \quad { x }^{ 2 }-16x+\lambda =0\)
\(\Rightarrow \quad x=\frac { 16\pm \sqrt { (256-4\lambda ) } }{ 8 } \)
\(=\frac { 8\pm \sqrt { (64-2\lambda ) } }{ 4 } \)
\(\alpha ,\beta =2\pm \frac { \sqrt { (64-2\lambda ) } }{ 4 } \)
Here \(64-\lambda >0\)
\(\therefore \quad \quad \lambda <64\)
Also \(1<\alpha <2\) and \(2<\beta <3\)
\(\Rightarrow \quad 1<2-\frac { \sqrt { 64-\lambda } }{ 4 } <2\quad and\quad 2<2+\frac { \sqrt { 64-\lambda } }{ 4 } <3\)
\(\Rightarrow \quad -1<-\frac { \sqrt { 64-\lambda } }{ 4 } <0\quad and\quad 0<\frac { \sqrt { 64-\lambda } }{ 4 } <1\)
\(\Rightarrow \quad 1>\frac { \sqrt { 64-\lambda } }{ 4 } >0\quad and\quad 0<\frac { \sqrt { 64-\lambda } }{ 4 } <1\)
i.e \(0<\frac { \sqrt { 64-\lambda } }{ 4 } <1\)
\(\Rightarrow \quad 0<\sqrt { (64-\lambda ) } <4\)
\(\Rightarrow \quad 0<(64-\lambda )<16\)
\(\Rightarrow \quad \quad \lambda >48\quad or\quad 48<\lambda <64\)
\(\therefore \quad \quad \lambda =\{ 49,50,51,52,....,63\} \)
If s={a ∈ N,1≤ a≤100}and [tan2 x]-tan x-a=0 has real roots, where [.] denotes the greatest integer function, then number of elements in set A equals
- (a)
2
- (b)
5
- (c)
6
- (d)
9
∵ [tan 2 x]=Integer, ∝ =Integer
∴ tan x is also an integer
Then tan 2 x-tan x-∝ =0
∴ ∝ =tan x(tan x-1)
=I(I-1_
=product of two consecutive integers
∴ a=2,6,12,20,30,42,56,72,90 Hence set S has 9 elements
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
∵ f(x) =x3+x2+100x+7 sin x
∴ 3x2+2x+100+7 cos x
=3x2+2x+93+7 (1+cos x)
= \(3\left\{ \left( x+\frac { 1 }{ 3 } \right) ^{ 2 } \right\} +\frac { 278 }{ 3 } +7(1+cos\quad x)>0\)
∴f (x) is increasing function
⇒ f(1)=a,f(2)=b and f(3)=c
ie, a<b<c
Now,given equatoion is \(\frac { 1 }{ y-a } +\frac { 2 }{ y-b } +\frac { 3 }{ y-c } =0\) or (y-b)(y-c)+2(y-c)(y-a)+3(y-a)(y-b)=0
Let g(y)=(y-b)(y-c)+2(y-c)(y-a)+3(y-a)(y-b)
∴ g(a)>0,g(b)<0 and g(c)>0
Hence one real root between a and b other between b and c
If the roots of the equation ax2+bx+c=0 are of the form ∝/(∝-1) and (∝+1)/∝ then the value of (a+b+c)2 is
- (a)
2b2-ac
- (b)
b2-2ac
- (c)
b2-4ac
- (d)
4b2-2 ac
By hypothesis \(\frac { \alpha }{ \alpha -1 } +\frac { \alpha +1 }{ \alpha } =-\frac { b }{ a } and\quad \frac { \alpha }{ \alpha -1 } .\frac { \alpha +1 }{ \alpha } =\frac { c }{ a } \)
\(\Rightarrow \quad \frac { 2\alpha ^{ 2 }-1 }{ \alpha ^{ 2 }-\alpha } =-\frac { b }{ a } \quad and\quad \frac { c+a }{ c-a } \)
\(\Rightarrow (c+a)^{ 2 }+2b(c+a)+{ b }^{ 2 }={ b }^{ 2 }-4ac\)
\(\Rightarrow \quad (a+b+c)^{ 2 }={ b }^{ 2 }-4ac\)
The real roots of the equation\(5^{ log\quad 5({ x }^{ 2 }-4x+5) }=x-1\) are
- (a)
1 and 2
- (b)
2 and 3
- (c)
3 and 4
- (d)
4 and 5
\(5^{ log\quad 5({ x }^{ 2 }-4x+5) }=x-1\)
\(\Rightarrow 5^{ log\quad 5\{ (x-2)^{ 2 }+1\} }=x-1\)
\(\Rightarrow (x-2)^{ 2 }-(x-2)=0\)
\(\Rightarrow (x-2)(x-3)=0\)
\(\therefore \quad x=2,3\)
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
a cos 2x +b sin2x+c=0
⇒ a(1-2 sin2x)+b sin2 x+c=0
⇒ (b-2a)sin2 x+(c+a)=0
It is an identity
∴ b-2a=0 and a+c=0
∴ \(\frac { a }{ 1 } =\frac { b }{ 2 } =\frac { c }{ -1 } =\lambda (\lambda \epsilon I)\)
∴Triplets (λ,2λ,-λ)λ∈ I
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
∵ \(\sqrt { (x+1) } -\sqrt { (x-1) } =\sqrt { (4x-1) } \)
∴ x+1≥ 0,x-1≥0,4x-1≥0
∴ x≥-1,x≥1,\(x\ge \frac { 1 }{ 4 } \)
Hence x≥1
and squaring both sides Eq(i) we get
\(x=\frac { 5 }{ 4 } \)
From Eq (i) and (ii) we get x=5/4 does satisfies Eq(i) hence no solution
If tan and tan are the roots of the equation ax2+bx+c=0,then the value of tan(∝+β) is
- (a)
b/(a-c)
- (b)
b/(c-a)
- (c)
a/(b-c)
- (d)
a/(c-a)
tan ∝ + tan β =-b/a,tan ∝,tan β=c/a
∴ tan (∝ + β) =c/a
\(\therefore \quad tan(\alpha +1)=\frac { tan\quad \alpha +tan\quad \beta }{ 1-tan\quad \alpha \quad tan\quad \beta } =\frac { -b/a }{ 1-c/a } =\frac { b }{ c-a } \)
If ∝,β are the roots of the equation 8x2-3x+27=0 then the value of \([(\alpha ^{ 2 }/\beta )^{ 1/3 }+(\beta ^{ 2 }/\alpha )^{ 1/3 }]\) is
- (a)
1/3
- (b)
1/4
- (c)
1/5
- (d)
1/6
\(\alpha +\beta =-3/8,\alpha \beta =27/8\)
\(\therefore \quad \quad \left( \frac { \alpha ^{ 2 } }{ \beta } \right) ^{ 1/3 }+\left( \frac { \beta ^{ 2 } }{ \alpha } \right) ^{ 1/3 }=\left( \frac { \alpha ^{ 3 } }{ \alpha \beta } \right) ^{ 1/3 }+\left( \frac { \beta ^{ 3 } }{ \alpha \beta } \right) ^{ 1/3 }\)
\(=\frac { \alpha +\beta }{ (\alpha \beta )^{ 1/3 } } \)
\(=\frac { 3/8 }{ (27/8)^{ 1/3 } } =\frac { 3/8 }{ 3/2 } =\frac { 1 }{ 4 } \)
The value of p for which both the roots of the equation 4x2-20px++(25p2+15p-66)=0 are less than 2 lies in
- (a)
(4/5,2)
- (b)
(2,∝)
- (c)
(-1,-4/5)
- (d)
(-∝,-1)
Discriminant ≥0
∴ p≤ 22/5
Roots less than 2
∴ f(2)>0
∴ p2-p-2>0
⇒ p>2 or p<-1
Combine both cases, we get p ∈ (-∝ ,-1) ∈ (.\(\frac { 22 }{ 5 } \)]
Let a,b,c ∈ R and a≠ 0 .If ∝ is a root of a2x2+bx+c=0,β is a root of a2x2+bx-c=0 and 0<∝<β then the equation,a2x2+2bx+2c=0 has a root Υ that always satisfies
- (a)
γ=∝
- (b)
γ=β
- (c)
γ=(∝+β)/2
- (d)
∝<γ<β
Let f (x)=a2x2+2bx+2c
From the question, a2∝2+b∝+c=0 and a2β2-bβ-c=0
Now f(∝)a2∝2+b∝+2c=b∝+c=-a2∝2
f(β)=a2β2+2bβ+2c=3(bβ+c)=3a2β2
but 0<∝<⋎<β are real
∴ f(∝)<0,f(β)>0
Hence ∝<⋎<β
The number of real solutions of the equation cos (ex) =2x+2-x is
- (a)
0
- (b)
1
- (c)
2
- (d)
infinitely many
cos (ex) =2x+2-x
∵ AM≥GM
\(\Rightarrow \frac { 2^{ x }+2^{ -x } }{ 2 } \ge \sqrt { { 2 }^{ x }.2^{ -x } } \)
\(\Rightarrow \quad { 2 }^{ x }+2^{ -x }\ge 2\)
\(\Rightarrow \quad cos(e^{ x })\ge 2\) [From Eq.(i)]
impossible ie,zero solutions
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
Let roots be ∝,β
∝+β=-2a and ∝β=b
By hypothesis, |∝-β| ≤ 2 m
⇒ |∝2 - β2|≤(2m)2
(∝+β)2 -4∝β ≤ 4m2 ⇒ 4a2-4b≤ 4m2
⇒ a2-m2 =b and discriminant >0
4a2-4b>0
b<a2
hence, b ∈ [a2-m2 ,a2)
∵ ax3+bx3+c=(x2+px+1)(ax+c) ...(i)
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
∵ ax3+bx2+c=(x2+px+1)(ax+c) ..(i)
Comparing coefficient of x2 and coeffieient of x in Eq (i),then
\(b=c+ap\Rightarrow p=\frac { b-c }{ a } \) ....(ii)
and \(0=pc+a\Rightarrow p=-\frac { a }{ c } \) ...(iii)
From Eqs(ii) and (iii)
\(\frac { b-c }{ a } =-\frac { a }{ c } \Rightarrow { a }^{ 2 }-c^{ 2 }=-bc\)
If ∝ and β are the roots of ax2 + bx + c = 0, then the equation ax2 - bx (x - 1) + e (x - 1)2 = 0 has roots
- (a)
\(\frac { \alpha }{ 1-\alpha } ,\frac { \beta }{ 1-\beta } \)
- (b)
\(\frac { 1-\alpha }{ \alpha } ,\frac { 1-\beta }{ \beta } \)
- (c)
\(\frac { \alpha }{ \alpha +1 } ,\frac { \beta }{ \beta +1 } \)
- (d)
\(\frac { \alpha +1 }{ \alpha } ,\frac { \beta +1 }{ \beta } \)
\(\because ax^{ 2 }-bx(x-1)+c(x-1)^{ 2 }=0\)
\(\Rightarrow a\left( \frac { x }{ 1-x } \right) ^{ 2 }+b\left( \frac { x }{ 1-x } \right) +c=0\)
Now, ∵ ax2+bx+c=0
Then, \(\alpha =\frac { x }{ 1-x } \) and \(\beta =\frac { x }{ 1-x } \)
\(\Rightarrow \quad x=\frac { \alpha }{ \alpha +1 } \) and \(\Rightarrow \quad x=\frac { \beta }{ \beta +1 } \)
Hence, \(\frac { \alpha }{ \alpha +1 } ,\frac { \beta }{ \beta +1 } \) are the roots of Eq (i)
If x2 + x + 1 is a factor of a x3 + bx2 + cx + d, then the real root of a x3 + bx2 + ex + d = 0 is
- (a)
-d/a
- (b)
d/a
- (c)
a/d
- (d)
none of these
x2 + x + 1=(x-ω)(x-ω2)
If real root is ∝ ,then
product ω.ω2.∝=-d/a
∴ ∝ =-d/a
The values of x which satisfy the equation \(\sqrt { (5x^{ 2 }-8x+3) } -\sqrt { (5x^{ 2 }-9x+4) } =\sqrt { (2x^{ 2 }-2x) } -\sqrt { (2x^{ 2 }-3x+1) } \) are
- (a)
3
- (b)
2
- (c)
1
- (d)
0
\(\sqrt { (5x^{ 2 }-8x+3) } -\sqrt { (5x^{ 2 }-9x+4) } \)
\(=\sqrt { (2x^{ 2 }-2x) } -\sqrt { (2x^{ 2 }-3x+1) } \)
\(\because \quad (5x^{ 2 }-8x+3)-(5x^{ 2 }-9x+4)\)
\(=(2x^{ 2 }-2x)-(2x^{ 2 }-3x+1)\)
\(\Rightarrow \quad \quad (x-1)=(x-1)\)
Hence, x=1 is the solution
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
x2-xy+y2-4x-4y+16=0
⇒ x2-x(y+4)+y2-4y+16=0
Discriminent ≥ 0
∴ (y-4)2 ≤ 0⇒(y-4)2=0
∴ y=4 then x=4
∴ pair(4,4)
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 \(\log _{ 2 }{ x } =\pm \sqrt { 2 } \)
\(x={ 2 }^{ \pm \sqrt { 2 } }\)
\(\therefore \quad \quad x={ 2 }^{ -\sqrt { 2 } }.{ 2 }^{ \sqrt { 2 } }\)
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
\(\left( \frac { 9 }{ 10 } \right) ^{ x }=-3+x-{ x }^{ 2 }\)
\(\Rightarrow \quad \quad \left( \frac { 9 }{ 10 } \right) ^{ x }=-\left\{ \left( 1-\frac { 1 }{ 2 } \right) ^{ 2 }+\frac { 11 }{ 4 } \right\} \)
∵ LHS is always positive and RHS is always negative for all x
Hence ,no solution
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]
The number of negative integral solutions of \({ x }^{ 2 }.{ 2 }^{ x+1 }+{ 2 }^{ |x-3|+2 }={ x }^{ 2 }.{ 2 }^{ |x-3|+4 }+2^{ x-1 }\) is
- (a)
none
- (b)
only one
- (c)
two
- (d)
four
We have
\({ x }^{ 2 }.2^{ x+1 }+2^{ |x-3|+2 }={ x }^{ 2 }.2^{ |x-3|+4 }+2^{ x-1 }\)
If \(x-3\ge 0\) i.e x≥3 (impossible since x is negative)
Now, x-3<0
⇒ x<3
Then \({ x }^{ 2 }.{ 2 }^{ x+1 }+2^{ -x+5 }={ x }^{ 2 }.2^{ -x+7 }+2^{ x-1 }\)
\({ x }^{ 2 }.{ 2 }^{ x+7 }(2^{ 2x-6 }-1)=2^{ x+5 }(2^{ 2x-6 }-1)\)
\(\Rightarrow (2^{ 2x-6 }-1)({ x }^{ 2 }.4-1)=0\)
\(\therefore \quad \quad \quad x=\pm \frac { 1 }{ 2 } \) and x=3
∵ x<3 and x is negative integer ,then
\(x\neq \pm \frac { 1 }{ 2 } \) and \(x\neq 3\)
Hence, number of negative integral solutions is zero
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
∵ Ix2 - 2x - 3| = b
Alternate (a) : For b<0
Ix2 - 2x - 3| <0
Impossible,there are no solutions
Alternate (b) : For b=0
x2 - 2x - 3=0
∴ x=3,-1 there are two solutions
Alternate (c) : For 0<b<1
0<Ix2 - 2x - 3|<1
⇒ 0<x2 - 2x - 3<1
and -1<x2 - 2x - 3<0
⇒ \(x\epsilon (1-\sqrt { 5 } ,-1)\cup (3,1+\sqrt { 5 } )\)
Alternate (d) : For b=1
Ix2 - 2x - 3| = b1
⇒ x2 - 2x - 3=±1
\(\therefore \quad x=1\pm \sqrt { 5 } \) and \(x=1\pm \sqrt { 3 } \)
There are four solutions
If a,b,c are positive rational number such that a>b>c and the quadratic equation
(a+b-2c)a2+(b+c-2a)x+(c+a-2b)=0
has a root in the interval (-1,0), then
- (a)
b+c>a
- (b)
c+a<2a
- (c)
both roots of the given equation are rational
- (d)
the equation ax2+2bx+c=0 has both nagative real roots
∵ a>b>c .........(i)
and given equation is
(a+b-2c)x2+(b+c-2a)x+(c+a-2b)=0 .......(ii)
∵ Eq (ii) has a root in the interval (-1,0)
∴ f(-1)f(0)<0
⇒ (2a-b-c)(c+a-2b)<0 ...(iii)
From Eq (i)
a>b ⇒ a-b>0
and a>c ⇒ a-c>0
2a-b-c>0 .......(iv)
and Discriminant of Eq (ii)
=(b+c-2a)2-4(a+b-2c)(c+a-2b)
={(b-a)+(c-a)}2-4{(a-c)+(b-c)}(c-b+a-b)}
={(c-a)-(a-b)}-a{(b-c)-(c-a)} x {(a-b)-(b-c)}
If b-c=X, c-a=Y and a-b=Z
then, discriminant=(Y-Z)2-4(X_Y)(Z-X)
=(2X-Y-Z)2 (∵ X+Y+Z=0)
=(0+3X)2=9X2=9(b-c)2
∴ Both roots of the given equation are rational
Alternate (c) is correct:
If roots of ax2+2bx+c=0 are ∝ and β
Then ∝+β=-2b/a<0
∴ ∝+β<0 and ∝β=c/a>0 (∵ a>b>c)
and discriminant of
ax2+2bx+c=0
is 4b2-4ac>0 (∵ b>c and b<a)
Hence, roots of ax2+2bx+c=0 are negative and real
Alternate (d) is correct.
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
∵ Roots are real
∴ B2-4AC>0
⇒ a4>4b2, (a) is correct
If f(x)=x2+a2x+b2
(∵ c lie outside the roots)
∴ f(c)>0, then c2+a2c+b2>0
(b) is correct
Also (x-coordinate of vertex)>c
⇒ -a2/2>c
∴ (c) is correct
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
y=2[x]+3=3[x-2]+5
⇒ y=2[x]+3=3[x]-6+5
⇒ 2[x]+3=3[x]-1
⇒ [x]=4 ∴ y=11
Then [x+y]=y+[x]=11+4=15
The roots of the equation (3-x)4+(2-x)4=(5-2x)4 are
- (a)
all real
- (b)
all imaginary
- (c)
two real and two imaginary
- (d)
none of the above
Given equation is (3-x)4+(2-x)4=(5-2x)4 ...(i)
(3-x)4 +(2-x)4 = (5-2x)4
Let a=3-x and b=2-x
Eq. (i) becomes
a4+b4 =(a+b)4
⇒ 2ab(2a2+3ab+ab2)=0
∴ a=0⇒ 3-x=0⇒x=3
b=0⇒ 2-x=0⇒x=2
and 2a2+3ab+2b2=0
\(2\left( \frac { a }{ b } \right) ^{ 2 }+2\left( \frac { a }{ b } \right) +2=0\)
\(\quad \frac { a }{ b } =\frac { -3\pm \sqrt { 9-16 } }{ 4 } \)
\(\frac { 3-x }{ 2-x } =\frac { -3\pm i\sqrt { 7 } }{ 4 } \)
\(\Rightarrow \quad \frac { 3-x }{ 2-x } -1=\frac { -7\pm i\sqrt { 7 } }{ 4 } \)
\(\Rightarrow \quad \frac { 1 }{ 2-x } =\frac { -7\pm i\sqrt { 7 } }{ 4 } \)
\(2-x=\frac { 4 }{ -7\pm i\sqrt { 7 } } \)
\(x=2-\frac { 4 }{ -7\pm i\sqrt { 7 } } \)
\(x=\frac { -18\pm 2i\sqrt { 7 } }{ -7\pm i\sqrt { 7 } } \)
Taking '+'sign
\(x=\frac { (-18+2i\sqrt { 7 } )(-7-i\sqrt { 7 } ) }{ (-7+i\sqrt { 7 } )(-7-i\sqrt { 7 } ) } \)
\(=\frac { 140+4i\sqrt { 7 } }{ 56 } \)
\(x=\frac { 35+i\sqrt { 7 } }{ 14 } \)
Taking '-' sign ,x= \(x=\frac { 35-i\sqrt { 7 } }{ 14 } \)
Hence,roots of Eq(i) are
2,3,\(\\ \\ \frac { 35\pm i\sqrt { 7 } }{ 14 } \)
i.e Two real and two imaginary
The number of ordered 4-tuple (x,y,z,w),(x,y,z,w ∈ [1,10]) which satisfies the inequality \({ 2 }^{ sin^{ 2 }x }3^{ sin^{ 2 }y }4^{ sin^{ 2 }z }5^{ cos^{ 2 }w }\ge 120\) is
- (a)
0
- (b)
144
- (c)
81
- (d)
infinite
(x,y,z,w),(x,y,z,w ∈ [1,10])
Given \({ 2 }^{ sin^{ 2 }x }3^{ sin^{ 2 }y }4^{ sin^{ 2 }z }5^{ cos^{ 2 }w }\ge2.3.4.5\)
\(\Rightarrow 2^{ 1-{ sin }^{ 2 }x }.3^{ 1-{ cos }^{ 2 }\quad y }.{ 4 }^{ 1-{ sin }^{ 2 }z }.5^{ 1-cos^{ 2 }w }\le 0\)
\(\Rightarrow \quad 2^{ cos^{ 2 }x }.3^{ sin^{ 2 }y }.4^{ cos^{ 2 }z }.5^{ sin^{ 2 }w }\le 0\)
Then cos2x log2+sin2y log3+cos2Z log4+sin2w log 5≤0
It is possible only when
cos2x=0⇒mπ+π/2.m∈1
sin2y=0⇒nπ+m∈I
.cos2z=0⇒ω=pπ,p∈ I
x,y,x,w ∈ [0,10]
x= π/2,3π/2;5π/2,y=0,π,2π,3π
z=π/2,3π/2,5π/2;w=0,π,2π,3π
Total number of ordered 4 tiples (x,y,z,w) is
3.4.3.4=144
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.
If coefficient of x and constant term changes to each other, then
- (a)
no. of positive integral solutions of a is 4 in P
- (b)
no. of p~sitive integral solution of a is 3 in Q
- (c)
greatest integral value of a is 6 in P
- (d)
least integral value of a is 4 in Q
(a-5)x2+(a-4)x-2a=0
⇒ x2+\(\left( \frac { a-4 }{ a-5 } \right) x-\frac { 2a }{ (a-5) } \)=0
\(\therefore \quad \alpha \beta =-\frac { 2a }{ a-5 } >0\quad \Rightarrow \quad \frac { 2a }{ a-5 } <0\)
⇒ a \(\in \) (0,5)
If a \(\in \) I, then a={1,2,3,4}
if \(\alpha \beta \)<0 ⇒ a\(\in \)(-∞,0)U(5,∞)
If a\(\in \)I
a={..., -4,-3,-2,-1,6,7,....}
Let consider the quadratic equation (1+m)x2-2(1+3m)x+(1+8m)=0, where m∈R~{-1}
The number of real values of m such that the given quadratic equation has roots are in the ratio 2 : 3 are
- (a)
0
- (b)
2
- (c)
4
- (d)
infinite many
Roots are 2∝ and 3∝
2∝+3∝=\(\frac { 2(1+3m) }{ (1+m) } \)
∝=\(\frac { 2(1+3m) }{ 5(1+m) } \) ...(i)
2∝.3∝=\(\frac { (1+8m) }{ (1+m) } \)
⇒ ∝2=\(\frac { (1+8m) }{ 6(1+m) } \) ...(ii)
From Eqs. (i) and (ii), we get
81±√6225/32
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
(x)2+(x+1)2 =25
⇒ (x)2+{1+(x)}2 =25
⇒ 2(x)2 +2(x)-24=0
⇒ (x)2 +(x)-12=0
⇒ (x)=-4,3
If (x)=-4 then -5 <x ≤ -4
and if (x)=3,then 2<x≤3
∴ x ∊ (-5,-4] ∈ (2,3]
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:
Let \({ \alpha }_{ 1 },{ \alpha }_{ 2 }\) be the roots of x2 - x + p = 0 and \({ \alpha }_{ 3 },{ \alpha }_{ 4 }\) be the roots of x2 - 4x + q = 0. If \({ \alpha }_{ 1 },{ \alpha }_{ 2 },{ \alpha }_{ 3 },{ \alpha }_{ 4 }\) are in GP, then the integral values, of p and q respectively are
- (a)
- 2, - 32
- (b)
- 2, 3
- (c)
- 6, 3
- (d)
- 6, - 32
\({ \alpha }_{ 1 }{ +\alpha }_{ 2 }=1,{ \alpha }_{ 1 }{ \alpha }_{ 2 }=p\) .... (i)
and \({ \alpha }_{ 3 }{ +\alpha }_{ 4 }=4,{ \alpha }_{ 3 }{ \alpha }_{ 4 }=q\) .... (ii)
\({ \alpha }_{ 1 },{ \alpha }_{ 2 },{ \alpha }_{ 3 },{ \alpha }_{ 4 }\) are in GP
\(\therefore \quad { \alpha }_{ 2 }={ { \alpha } }_{ 1 }R,{ { \alpha } }_{ 3 }={ { \alpha } }_{ 1 }{ R }^{ 2 },{ { \alpha } }_{ 4 }={ { \alpha } }_{ 1 }{ R }^{ 3 }\)
From Eq. (i), \({ { \alpha } }_{ 1 }+{ { \alpha } }_{ 1 }R=1\)
\(\Rightarrow \quad { { \alpha } }_{ 1 }\left( 1+R \right) =1\) ... (iii)
and from Eq. (ii), \({ { \alpha } }_{ 1 }{ R }^{ 2 }+{ { \alpha } }_{ 1 }{ R }^{ 3 }=4\)
\(\Rightarrow \quad { { \alpha } }_{ 1 }{ R }^{ 2 }\left( 1+R \right) =4\) ... (iv)
Dividing Eq. (iv) by (iii), then
R2 = 4
or \(R=\pm 2\)
If R = 2, then \({ { \alpha } }_{ 1 }=\frac { 1 }{ 3 } \) [from Eq. (iii)]
and if R = - 2, then \({ { \alpha } }_{ 1 }=-1\) [from Eq. (iii)]
p and q are integer.
\(\therefore \quad R=-2\quad and\quad { { \alpha } }_{ 1 }=-1\)
\(\Rightarrow \quad { \alpha }_{ 2 }=2,{ { \alpha } }_{ 3 }=-4,{ { \alpha } }_{ 4 }=8\)
\(p={ { \alpha } }_{ 1 }{ \alpha }_{ 2 }=-2\) and \(q={ { \alpha } }_{ 3 }{ \alpha }_{ 4 }=-32\) [from Eq. (i)]
p = - 2, q = - 32
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
\({ \beta }_{ 1 }+{ \beta }_{ 3 }=\frac { 4 }{ A } ,{ \beta }_{ 1 }{ \beta }_{ 3 }=\frac { 1 }{ A } \) ... (i)
and \({ \beta }_{ 2 }+{ \beta }_{ 4 }=\frac { 6 }{ B } ,{ \beta }_{ 2 }{ \beta }_{ 4 }=\frac { 1 }{ B } \) .... (ii)
From Eq. (i), \(\frac { { \beta }_{ 1 }{ \beta }_{ 3 } }{ \left( { \beta }_{ 1 }+{ \beta }_{ 3 } \right) } =\frac { 1 }{ 4 } \) ....(iii)
and from Eq. (ii), \(\frac { { \beta }_{ 2 }{ \beta }_{ 4 } }{ \left( { \beta }_{ 2 }+{ \beta }_{ 4 } \right) } =\frac { 1 }{ 6 } \) ...(iv)
Also, \({ \beta }_{ 1 },{ \beta }_{ 2 },{ { \beta }_{ 3 },\beta }_{ 4 }\) are in HP
Then \({ \beta }_{ 2 }=\frac { 2{ \beta }_{ 1 }{ \beta }_{ 3 } }{ { \beta }_{ 1 }+{ \beta }_{ 3 } } \quad and\quad { \beta }_{ 3 }=\frac { 2{ \beta }_{ 2 }{ \beta }_{ 4 } }{ { \beta }_{ 2 }+{ \beta }_{ 4 } } \)
\(\Rightarrow \quad { \beta }_{ 2 }=\frac { 1 }{ 2 } \quad and\quad { \beta }_{ 3 }=\frac { 1 }{ 3 } \) [from Eqs. (iii) and (iv)]
\({ \beta }_{ 3 }\) is a root of Ax2 - 4x + 1 = 0
Then \(A{ \left( \frac { 1 }{ 3 } \right) }^{ 2 }-4\left( \frac { 1 }{ 3 } \right) +1=0\)
\(\Rightarrow \quad \frac { A }{ 9 } -\frac { 1 }{ 3 } =0\)
A = 3
and \({ \beta }_{ 2 }\) is a root of Bx2 6x +1 = 0
then \(B{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }-6\left( \frac { 1 }{ 2 } \right) +1=0\)
B = 8
Hence, A = 3, B = 8
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
\(\because \quad \alpha +\beta =\frac { \left( 4+\sqrt { 5 } \right) }{ \left( 5+\sqrt { 2 } \right) } \quad and\quad \alpha \beta =\left( \frac { 8+2\sqrt { 5 } }{ 5+\sqrt { 2 } } \right) =2\left( \alpha +\beta \right) \)
\(\therefore \quad HM=\frac { 2\alpha \beta }{ \left( \alpha +\beta \right) } =\frac { 2\times 2\left( \alpha +\beta \right) }{ \left( \alpha +\beta \right) } =4\)
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 \(\left( 4+\sqrt { 15 } \right) ^{ [x] }+\left( 4-\sqrt { 15 } \right) ^{ [x] }=62\) , where [.] denotes the greatest integer function, then
- (a)
\(x\varepsilon [-3,-2)\cup [1,2)\)
- (b)
\(x\varepsilon [-3,2)\cup [-2,1)\)
- (c)
\(x\varepsilon [-2,-1)\cup [2,3)\)
- (d)
\(x\varepsilon [-2,3)\cup [-1,2)\)
\(\because \quad \left( 4+\sqrt { 15 } \right) ^{ 2 }+\left( 4-\sqrt { 15 } \right) ^{ 2 }=62\)
Then \(\frac { \left[ x \right] }{ 2 } =\pm 1\)
\(\therefore \quad \left[ x \right] =\pm 2\)
\(\Rightarrow \quad \left[ x \right] =-2\quad and\quad \left[ x \right] =2\)
\(\therefore \quad -2\le x<-1\quad and\quad 2\le x<3\)
\(\Rightarrow \quad x\epsilon [-2,-1)\cup [2,3)\)
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)
\(1\pm \sqrt { 2} ,2\)
\(\left( 2+\sqrt { 3 } \right) ^{ { x }^{ 2 }-2x+1 }+\left( 2-\sqrt { 3 } \right) ^{ { x }^{ 2 }-2x-1 }=\frac { 4 }{ 2-\sqrt { 3 } } \)
\(\left( 2+\sqrt { 3 } \right) ^{ { x }^{ 2 }-2x+1 }+\left( 2-\sqrt { 3 } \right) ^{ { x }^{ 2 }-2x-1 }=4\)
\(\because \quad \left( 2+\sqrt { 3 } \right) +\left( 2-\sqrt { 3 } \right) =4\)
\(\therefore \quad { x }^{ 2 }-2x=\pm 1\)
or \({ \left( x-1 \right) }^{ 2 }=1\pm 1\)
or \({ \left( x-1 \right) }^{ 2 }=2,0\)
\(\therefore \quad x-1=\pm \sqrt { 2 } ,0\)
\(x=1\pm \sqrt { 2 } ,1\)
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:
The number of real solutions of the equation \(\left( 15+4\sqrt { 14 } \right) ^{ t }+\left( 15-4\sqrt { 14 } \right) ^{ t }=30\) are where \(t={ x }^{ 2 }-2\left| x \right| \)
- (a)
0
- (b)
2
- (c)
4
- (d)
6
\(\because \quad \left( 15+4\sqrt { 14 } \right) +\left( 15-4\sqrt { 14 } \right) =30\)
\(\therefore \quad t=\pm 1\)
or \({ x }^{ 2 }-2\left| x \right| =\pm 1\)
or \(\left( \left| x \right| -1 \right) ^{ 2 }=1\pm 1\)
= 2, 0
\(\therefore \quad \left| x \right| -1=\pm \sqrt { 2 } ,0\)
or \(\left| x \right| =1\pm \sqrt { 2 } ,1\)
\(=1\pm \sqrt { 2 } ,1\) \(\left( \because \quad \left| x \right| \ge 0 \right) \)
\(\therefore \quad x=\pm \left( 1+\sqrt { 2 } \right) ,\pm 1\)
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]
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
For \(x\ge 4\) the last digit of the sum 1! + 2! + ... + x! is equal to 3 and therefore this sum cannot be equal to a square of a whole number k (because a square of a whole number cannot end with 3.)
For x > 4, the given equation has the only solutions
\(x=1,k=\pm 1\quad and\quad x=3,k=\pm 3\)
\(\therefore \quad \alpha =1,\beta =3\)
\(\therefore \quad 1+2\alpha +3{ \alpha }^{ 2 }+4{ \alpha }^{ 3 }+5{ \alpha }^{ 4 }=15\quad and\quad -5\beta =-15\)
Then \(\left( a+\sqrt { b } \right) ^{ { x }^{ 2 }-15 }+\left( a-\sqrt { b } \right) ^{ { x }^{ 2 }-15 }=2a\)
\(\because \quad \left( a+\sqrt { b } \right) +\left( a-\sqrt { b } \right) =2a\)
\(\therefore \quad { x }^{ 2 }-15=\pm 1\)
\(\Rightarrow \quad { x }^{ 2 }=16,4\quad \Rightarrow \quad x=\pm 4,\pm \sqrt { 14 } \)
\({ \alpha }_{ 1 }=-4,{ \alpha }_{ 2 }=4,{ \alpha }_{ 3 }=-\sqrt { 14 } ,{ \alpha }_{ 4 }=\sqrt { 14 } \)
\(\therefore \quad \left| { \alpha }_{ 1 }+{ \alpha }_{ 2 }+{ \alpha }_{ 3 }+{ \alpha }_{ 4 }-{ \alpha }_{ 1 }{ \alpha }_{ 2 }{ \alpha }_{ 3 }{ \alpha }_{ 4 } \right| =\left| 0-16\times 14 \right| =224\)
The number of solutions of the following inequality \({ 2 }^{ 1/{ sin }^{ 2 }x2 }.3^{ 1/sin^{ 2 } }{ x }_{ 3 }.{ 4 }^{ 1/{ sin }^{ 2 }\quad { x }_{ 4 } }...{ n }^{ 1/{ sin }^{ 2 }{ x }_{ n } }\le n!\) where
- (a)
1
- (b)
\({ 2 }^{ n-1 }\)
- (c)
\({ n }^{ n }\)
- (d)
infinite number of solutions
\(\because \quad \quad { 2 }^{ 1/{ sin }^{ 2 }{ x }_{ 2 } }.{ 2 }^{ 1/{ sin }^{ 2 }{ x }_{ 3 } }.4^{ 1/{ sin }^{ 2 }{ x }4 }....{ n }^{ 1/{ sin }^{ 2 }{ x }_{ 2 } }\quad \quad \le 1.2.3.4..n\)
\(\Rightarrow \quad { 2 }^{ cot^{ 2 }{ x }_{ 2 } }.{ 3 }^{ cot^{ 2 }{ x }_{ 3 } }.4^{ cot^{ 2 }{ x }_{ 4 } }...{ n }^{ cot^{ 2 }{ x }_{ n } }\le 1\)
\(\Rightarrow \quad { cot }^{ 2 }{ x }_{ 2 }.In2+{ cot }^{ 2 }{ x }_{ 3 }.In3+{ cot }^{ 2 }{ x }_{ 4 }.In4+....{ cot }^{ 2 }{ x }_{ n }In\quad n\le 0\)
which is possible only when
\({ cot }^{ 2 }{ x }_{ 2 }={ cot }^{ 2 }{ x }_{ 3 }={ cot }^{ 2 }{ x }_{ 4 }={ cot }^{ 2 }{ x }_{ n }=0\)
∴ Each x2,x3,x4 ......x4 take 2 values π/2,3π/2
∴ Number of solutions=2X2X2X.... ....X2=2n-1.
If b2≥4ac for the equation ax4+bx2+c=0, then all roots of the equation will be real, if
- (a)
b>0,a>0,c>0
- (b)
b<0,a>0,c>0
- (c)
b>0,a>0,c<0
- (d)
b>0,a<0,c<0
All roots of equation ax4+bx2+c=0 will be real if both roots of ay2+by+c=0 will be +ve (let x2=y)
ie, sum of roots=-b/a>0 and product of roots=c/a>0
Hence, a and b are of opposite sign, while a and c are of same sign.
Let f(x)=ax2+bx+c; a,b,c \(\in \) R and a≠0, Suppose f(x)>0 for all x\(\in \)R. Let g(x)=f(x)+f'(x)+f''(x) Then
- (a)
g(x)>0 ∀ x \(\in \) R
- (b)
g(x)<0 ∀ x \(\in \) R
- (c)
g(x)=0 has non real complex roots
- (d)
g(x)=0 has real roots
Since, f(x)>0, ∀ x\(\in \)R, a>0 and b2-4ac<0
we have, f'(x)=2ax+b and f''(x)=a
Thus, g(x)=ax2+bx+c+2ax+b+2a
=ax2+(2a+b)x+(2a+b+c)
we have a>0 and D=(2a+b)2-4a(2a+b+c)
=b2-4ac-4a2<0, since b2-4ac<0
Thus, g(x)>0, ∀ x\(\in \)R. Therefore g(x)=0 has non real complex roots
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,\(\gamma \) are the roots of the equation x3+3bx+c=0, then the value of \(\sum { (\alpha -\beta )^{ 2 }(\beta -\gamma ) } ^{ 2 }\) is equal to
- (a)
9b2
- (b)
27b2
- (c)
81b2
- (d)
243b2
Put y=(∝-β)2=(∝+β)2-4∝β
=(∝+β+\(\gamma \)--\(\gamma \))2-\(\frac { 4\alpha \beta \gamma }{ \gamma } \)
=(0-\(\gamma \))2-\(\frac { 4(-c) }{ \gamma } \) (∵ ∝+β+\(\gamma \)=0)
=\(=\frac { \gamma ^{ 3 }+4c }{ \gamma } \) (∝β\(\gamma \)=-c)
\(=\frac { ({ \gamma }^{ 3 }+c)+3c }{ \gamma } \) \((\because \quad \gamma ^{ 3 }+3b\gamma +c=0)\)
=\(\frac { -3b\gamma +3c }{ \gamma } \)
or \(\gamma =\frac { 3c }{ y+3b } \)
Thus, \(\left( \frac { 3c }{ y+3b } \right) ^{ 3 }+3b\left( \frac { 3c }{ y+3b } \right) \)+c=0
⇒ 27c3+9bc(y+3b)2+c(y+3b)3=0
or (y+3b)3+9b(y+3b)2+27c2=0
or y3+18by2+81b2y+(108b2+27c2)=0
∴ \(\sum { (\alpha -\beta )^{ 3 }(\beta -\gamma )^{ 2 } } \)
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,\(\gamma \) 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
Put y=(∝-β)(∝-\(\gamma \))
=∝2-(β+\(\gamma \))∝+β\(\gamma \)
=∝2-(-∝)∝+(-2/∝) (∝+β+\(\gamma \)=0)
\(=\frac { 2{ \alpha }^{ 3 }-2 }{ \alpha } \) (∝β\(\gamma \)=-2)
\(=\frac { 2(-\alpha -2)-2 }{ \alpha } \) (∵ ∝2+∝+2=0)
\(=-2-\frac { 6 }{ \alpha } \)
or \(\alpha =-\frac { 6 }{ (y+2) } \)
Thus, \(\left( \frac { 6 }{ (y+2) } \right) ^{ 3 }+\left( \frac { 6 }{ (y+2) } \right) \)+2=0
⇒ y3+3y2-112=0
or x3+3x2-112=0
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