Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{(1+a)}-1) x^2+(\sqrt{(1+a)}-1) x+(\sqrt[6]{(1+a)}-1)=0$ where $a>-1$, then $\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a)$ and $\mathop {\lim }\limits_{a \to {0^ + }} \beta (a)$, are

$\left(-\frac{5}{2}\right)$ and 1

$\left(-\frac{7}{2}\right)$ and 2

$\left(-\frac{9}{2}\right)$ and 3

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JEE • MATHEMATICS • THEORY OF EQUATION

Theory of equation practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 91-120 of 120 questions

91 Hard

91 of 120

Analyzing +4 | -1

Let

{P_n}\left( x \right) = 1 + 2x + 3{x^2} + \ldots + \left( {n + 1} \right){x^n}

be a polynomial such that

n

is even. Then, the number of real roots of

{P_n}\left( x \right),

View answer and explanation

Correct option: A

Option A (0) is correct because the polynomial has no real roots when n is even.

92 Medium

92 of 120

Analyzing +4 | -1

If the functions

f(x)=\frac{x^3}{3}+2 b x+\frac{a x^2}{2}

and

g(x)=\frac{x^3}{3}+a x+b x^2, a \neq 2 b

have a common extreme point, then

a+2 b+7

is equal to

View answer and explanation

Correct option: C

The correct answer is 6 because the common extreme point implies that the derivatives of the two functions have a common root, which allows us to solve for a + 2b + 7.

93 Medium

93 of 120

Analyzing +4 | -1

{x^2} + ax + 1

is a factor of

a{x^3} + bx + c

, then

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the conditions derived from polynomial division do not match any of the given options.

94 Easy

94 of 120

Application +4 | -1

The quadratic equation with rational coefficients, one of whose roots is

\frac{1}{2+\sqrt{5}}

, is

View answer and explanation

Correct option: A

x^2+4 x-1=0

Option A (

x^2+4 x-1=0

) is correct because it is the quadratic equation with rational coefficients having

\frac{1}{2+\sqrt{5}}

as one of its roots.

95 Easy

95 of 120

Application +4 | -1

\alpha ,\beta

and

\gamma

are the roots of equation

{x^3} - 3{x^2} + x + 5 = 0

, then

y = \sum {\alpha ^2} + \alpha \beta \gamma

satisfies the equation

View answer and explanation

Correct option: B

{y^3} - {y^2} - y - 2 = 0

Option B (

{y^3} - {y^2} - y - 2 = 0

) is correct because the value of y is 2, which satisfies the given equation.

96 Medium

96 of 120

Analyzing +4 | -1

If the roots of the equation

{x^3} + b{x^2} + 3x - 1 = 0

from a non-decreasing H.P., then

View answer and explanation

Correct option: B

b = - 3

Option B (

b = - 3

) is correct because the roots form a non-decreasing H.P., which leads to a specific value for b.

97 Easy

97 of 120

Application +4 | -1

If the roots of the equation

8{x^3} - 14{x^2} + 7x - 1 = 0

are in GP, then the roots are

View answer and explanation

Correct option: A

1,\frac{1}{2},\frac{1}{4}

Option A (

1,\frac{1}{2},\frac{1}{4}

) is correct because the roots are in GP and their product is

\frac{1}{8}

, allowing us to find the common ratio.

98 Easy

98 of 120

Analyzing +4 | -1

Let

\alpha, \beta

be the roots of the equation

x^2-x+p=0

and

\gamma, \delta

be the roots of the equation

x^2-4 x+q=0

. If

\alpha, \beta, \gamma

and

\delta

are in GP, the integral values of

p

and

q

respectively, are

View answer and explanation

Correct option: A

-2,-32

Option A (-2, -32) is correct because the values of p and q are found by using the properties of GP and the roots of quadratic equations.

99 Easy

99 of 120

Analyzing +4 | -1

The roots

\alpha ,\beta \,\,\mathrm{and}\,\gamma

of an equation

{x^3} - 3\;a\;{x^2} + 3\;b\;x - c = 0

are in H.P. Then,

View answer and explanation

Correct option: D

\beta = \frac{c}{b}

Option D (

\beta = \frac{c}{b}

) is correct because the harmonic mean is the reciprocal of the arithmetic mean of the reciprocals, leading to the given relationship.

100 Hard

100 of 120

Analyzing +4 | -1

The number of real roots of the equation

1 + {a_1}x + {a_2}{x^2} + \ldots + {a_n}{x^n} = 0,

where

\left| x \right| < \frac{1}{3}\;\mathrm{and\;}\left| {{a_n}} \right| < 2,

View answer and explanation

Correct option: C

0 for any

n \in N

Option C (0 for any

n \in N

) is correct because the given equation has no real roots for any n, as the expression is always positive within the specified range.

101 Easy

101 of 120

Application +4 | -1

\alpha, \beta

are the roots of the equation

x^2-3 x+5=0

and

\gamma

\delta

are the roots of the equation

x^2+5 x-3=0

, then the equation whose roots are

\alpha \gamma+\beta \delta

and

\alpha \delta+\beta \gamma

, is

View answer and explanation

Correct option: D

x^2+15 x+158=0

Option D

x^2+15 x+158=0

is correct because it is the quadratic equation formed using the sum and product of the roots

\alpha \gamma + \beta \delta

and

\alpha \delta + \beta \gamma

102 Medium

102 of 120

Analyzing +4 | -1

If the sum of the roots of the equation

a x^2+b x+c=0

equal to the sum of the reciprocals of their squares. Then

\frac{a}{c}, \frac{b}{a}, \frac{c}{b}

are in

View answer and explanation

Correct option: C

H.P

Option C (H.P) is correct because after manipulating the given equation and simplifying, the resulting terms are in Arithmetic Progression, which implies the reciprocals are in Harmonic Progression.

103 Medium

103 of 120

Analyzing +4 | -1

x

and

y

are positive integers such that

x y+x+y=71, x^2 y+x y^2=880

, then

x^2+y^2

is equal to

View answer and explanation

Correct option: C

146

Option C (146) is correct because it is the result of solving the system of equations and calculating x² + y².

104 Easy

104 of 120

Analyzing +4 | -1

a \in R

and the equation

-3(x-[x])^2+2(x-[x])+a^2=0

(where

[\cdot]

denotes the greatest integer function) has no integral solution, then all possible values of

a

lie in the interval

View answer and explanation

Correct option: C

(-1,0) \cup(0,1)

Option C ((-1,0) ∪(0,1)) is correct because it ensures that the fractional part of x is non-zero and the equation has no integral solutions, leading to the given interval for a.

105 Medium

105 of 120

Analyzing +4 | -1

The sum of the roots of the equation

2^{33 x-2}+2^{11 x+2}=2^{22 x+1}+1

View answer and explanation

Correct option: B

\frac{2}{11}

Option B (2/11) is correct because the sum of the roots can be found by transforming the equation and using properties of exponents.

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106 Medium

106 of 120

Analyzing +4 | -1

If the equation

{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + \ldots + {a_1}x = 0

has a positive root

\alpha

, then the equation

n\;{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + \ldots + {a_1} = 0

has

View answer and explanation

Correct option: A

A positive root less than

\alpha

Option A is correct because by applying Rolle's theorem to a related function, we can show that the given equation has a positive root less than α.

107 Medium

107 of 120

Application +4 | -1

x = 2 + {2^{2/3}} + {2^{1/3}},

then the value of

{x^3} - 6{x^2} + 6x

View answer and explanation

Correct option: B

Option B (2) is correct because it involves algebraic manipulation to simplify the given expression.

108 Easy

108 of 120

Analyzing +4 | -1

2+3 i

is one of the roots of the equation

2 x^{3}-9 x^{2}+k x-13

=0, \mathrm{k} \in \mathrm{R}

, then the real root of this equation :

View answer and explanation

Correct option: B

exists and is equal to

\frac{1}{2}

Option B exists and is equal to

\frac{1}{2}

is correct because the product of the roots can be used to find the real root.

109 Medium

109 of 120

Analyzing +4 | -1

f(x)

is a polynomial of degree

4, f^{\prime}(1)=f(2)=f(3)=f(4)=0

, then root of

f(x)=0

View answer and explanation

Correct option: A

\frac{5}{11}

Option C (

\frac{5}{11}

) is correct because it is derived from the condition

f'(1) = 0

using the given information about the polynomial

f(x)

110 Medium

110 of 120

Application +4 | -1

\alpha, \beta, \gamma

are the roots of

2 x^3-x+1=0

then the value of

\alpha^3+\beta^3+\gamma^3

View answer and explanation

Correct option: C

-\frac{3}{2}

Option C (-

\frac{3}{2}

) is correct because by using Vieta's formulas and the given cubic equation, we can find the sum of the cubes of the roots.

111 Easy

111 of 120

Analyzing +4 | -1

Let

\alpha(a)

and

\beta(a)

be the roots of the equation

(\sqrt[3]{(1+a)}-1) x^2+(\sqrt{(1+a)}-1) x+(\sqrt[6]{(1+a)}-1)=0

where

a>-1

, then

\lim\limits_{a \to {0^ + }} \alpha (a)

and

\lim\limits_{a \to {0^ + }} \beta (a)

, are

View answer and explanation

Correct option: B

\left(-\frac{1}{2}\right)

and

(-1)

Option B (

\left(-\frac{1}{2}\right)

and

(-1)

) is correct because after applying L'Hopital's rule and simplifying, the roots of the resulting quadratic equation are -1/2 and -1.

112 Medium

112 of 120

Understanding +4 | -1

x = c

is a root of order 2 of a polynomial

f\left( x \right),

then

x = c

is also a root of the polynomial

View answer and explanation

Correct option: A

f'\left( x \right)

Option A (

f'(x)

) is correct because if

x=c

is a root of order 2 of

f(x)

, then

f(x) = (x-c)^2 g(x)

, and

f'(x)

will also have

(x-c)

as a factor, making

x=c

a root of

f'(x)

113 Easy

113 of 120

Application +4 | -1

\alpha ,\beta \in C

are the distinct roots of the distinct roots of the equation

{x^2} - x + 1 = 0,

then

{\alpha ^{101}} + {\beta ^{107}}

is equal to

View answer and explanation

Correct option: C

Option A (Atleast 1000) is correct because the number of arrangements is calculated to be 1080, which is greater than 1000.

114 Easy

114 of 120

Application +4 | -1

x^2+x+1

is a factor of

a x^3+b x^2+c x+d

, the real root of

a x^3+b x^2+c x+d=0

View answer and explanation

Correct option: A

-\frac{d}{a}

Option A (-d/a) is correct because if

x^2+x+1

is a factor, then the remaining factor must be linear, and the real root is found by considering the product of the roots.

115 Medium

115 of 120

Application +4 | -1

\alpha, \beta, \gamma

and

\sigma

are the roots of the equation

x^4+4 x^3-6 x^2+7 x-9=0

, then the value of

\left(1+\alpha^2\right)\left(1+\beta^2\right)\left(1+\gamma^2\right)\left(1+\sigma^2\right)

View answer and explanation

Correct option: C

Option C (13) is correct because substituting

x=i

and

x=-i

into the given polynomial and multiplying the results yields 13.

116 Medium

116 of 120

Application +4 | -1

Let

\left(1-x-2 x^2\right)^6=1+a_1 x+a_2 x^2+\ldots . .+a_{12} x^{12}

. Then

\frac{a_2}{2^2}+\frac{a_4}{2^4}+\frac{a_6}{2^6}+\ldots .+\frac{a_{12}}{2^{12}}

is equal to

View answer and explanation

Correct option: B

-1 / 2

Option B (-1/2) is correct because by substituting x = 1/2 and x = -1/2 into the given equation and adding the resulting equations, we can isolate the desired sum.

117 Easy

117 of 120

Application +4 | -1

The solution of the equation

2{x^3} - {x^2} - 22x - 24 = 0

when two of the roots in the ration

3:4

, is

View answer and explanation

Correct option: B

- \frac{3}{2},\; - 2,\;4

Option B (

- \frac{3}{2},\; - 2,\;4

) is correct because it satisfies the given equation and the ratio of two of its roots is 3:4.

118 Easy

118 of 120

Application +4 | -1

Let

\mathrm{p}(\mathrm{x})

be a quadratic polynomial such that

\mathrm{p}(0)=1

. If

\mathrm{p}(\mathrm{x})

leaves remainder 4 when divided by

\mathrm{x}-1

and it leaves remainder 6 when divided by

x+1

; then :

View answer and explanation

Correct option: C

\mathrm{p}(-2)=19

Option C (

\mathrm{p}(-2)=19

) is correct because when the quadratic polynomial p(x) is determined based on the given conditions, p(-2) evaluates to 19.

119 Medium

119 of 120

Application +4 | -1

\alpha, \beta, \gamma

are roots of equation

x^3-2 x^2-1=0

and

T_n=\alpha^n+\beta^n+\gamma^n

then the value of

\frac{T_{11}-T_8}{T_{10}}

is equal to

View answer and explanation

Correct option: B

Option B (2) is correct because by using the recurrence relation derived from the equation

x^3 - 2x^2 - 1 = 0

, we can express

T_{11}

in terms of

T_{10}

and

T_8

, and then simplify the given expression.

120 Easy

120 of 120

Application +4 | -1

\alpha ,\;\beta \;

and

\gamma

are the roots of the equation

{x^3} - 8x + 8 = 0

, then

\sum {\alpha ^2}\;\mathrm{and}\;\,\sum \frac{1}{{\alpha \beta }}\;

are respectively

View answer and explanation

Correct option: D

16 and 0

Option D (16 and 0) is correct because it is found by using Vieta's formulas and the given equation to calculate the sum of squares and the sum of reciprocals of the product of roots.

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