Mathematics - Theory of Equations
Exam Duration: 45 Mins Total Questions : 30
Number of real roots of the equation \(\sqrt { x } +\sqrt { x-\sqrt { (1-x) } } =1\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
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 positive integral solution of x4-y4=3789108
- (a)
0
- (b)
1
- (c)
2
- (d)
4
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}
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
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 values of for which is an identity in x is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
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 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)
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
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
The value of \(\sqrt { 7+\sqrt { 7- } \sqrt { 7+\sqrt { 7-....\infty } } } \) is
- (a)
5
- (b)
4
- (c)
3
- (d)
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 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
If the equation ax2-2bx+c=0 has real roots which are reciprocal of each other, then
- (a)
a=c
- (b)
b≤a
- (c)
|b|≥|a|
- (d)
|b|≥|c|
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
The largest interval in which x12- x9 + X4 - X + 1 > 0 is
- (a)
[0,∞)
- (b)
(-∞,0]
- (c)
(-∞,∞)
- (d)
none of these
The system of equation "x-1|+3y=4,x-|y-1|=2 has
- (a)
no solution
- (b)
a unique solution
- (c)
two solutions
- (d)
more than two solutions
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 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
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
The number of solutions of |[x] - 2x|= 4, where [x] denotes the greatest integer ≤ x, is
- (a)
infinite
- (b)
4
- (c)
3
- (d)
2
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
5x+\(\left( 2\sqrt { 3 } \right) ^{ 2x }\)-169≤0 is true in the interval
- (a)
(-∞,2)
- (b)
(0,2)
- (c)
(2,∞)
- (d)
(0,4)
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
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
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-x-1=0, then the value of \(\prod { \left( \frac { 1+\alpha }{ 1-\alpha } \right) } \) is equal to
- (a)
-7
- (b)
-5
- (c)
-3
- (d)
-1
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