Quadratic Equation and Inequalities
304 Questions
Start JEE Mains Test
2022
Q101
JEE Advanced
Numerical
14 Mar 2026
The product of all positive real values of $x$ satisfying the equation
$ x^{\left(16\left(\log _{5} x\right)^{3}-68 \log _{5} x\right)}=5^{-16} $
is __________.
$ x^{\left(16\left(\log _{5} x\right)^{3}-68 \log _{5} x\right)}=5^{-16} $
is __________.
Correct Answer: 1
Explanation:
Taking log to the base 5 on both sides
$\left(16\left(\log _{5} x\right)^{3}-68\left(\log _{5} x\right)\right)\left(\log _{5} x\right)=-16$
Let $\left(\log _{5} x\right)=t$
$16 t^{4}-68 t^{2}+16=0$
$ \Rightarrow $ $4 t^{4}-16 t^{2}-t^{2}+4=0$
$ \Rightarrow $ $\left(4 t^{2}-1\right)\left(t^{2}-4\right)=0$
$ \Rightarrow $ $t=\pm \frac{1}{2}, \pm 2$
So $\log _{5} x=\pm \frac{1}{2}$ or $\pm 2$
$\Rightarrow x=5^{\frac{1}{2}}, 5^{\frac{-1}{2}}, 5^{2}, 5^{-2}$
$\left(16\left(\log _{5} x\right)^{3}-68\left(\log _{5} x\right)\right)\left(\log _{5} x\right)=-16$
Let $\left(\log _{5} x\right)=t$
$16 t^{4}-68 t^{2}+16=0$
$ \Rightarrow $ $4 t^{4}-16 t^{2}-t^{2}+4=0$
$ \Rightarrow $ $\left(4 t^{2}-1\right)\left(t^{2}-4\right)=0$
$ \Rightarrow $ $t=\pm \frac{1}{2}, \pm 2$
So $\log _{5} x=\pm \frac{1}{2}$ or $\pm 2$
$\Rightarrow x=5^{\frac{1}{2}}, 5^{\frac{-1}{2}}, 5^{2}, 5^{-2}$
2021
Q102
JEE Mains
MCQ
14 Mar 2026
The numbers of pairs (a, b) of real numbers, such that whenever $\alpha$ is a root of the equation x2 + ax + b = 0, $\alpha$2 $-$ 2 is also a root of this equation, is :
A.
6
B.
2
C.
4
D.
8
2021
Q103
JEE Mains
MCQ
14 Mar 2026
The sum of the roots of the equation
$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$, is :
$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$, is :
A.
log2 14
B.
log2 11
C.
log2 12
D.
log2 13
2021
Q104
JEE Mains
MCQ
14 Mar 2026
cosec18$^\circ$ is a root of the equation :
A.
x2 + 2x $-$ 4 = 0
B.
4x2 + 2x $-$ 1 = 0
C.
x2 $-$ 2x + 4 = 0
D.
x2 $-$ 2x $-$ 4 = 0
2021
Q105
JEE Mains
MCQ
14 Mar 2026
The set of all values of K > $-$1, for which the equation ${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$ has real roots, is :
A.
$\left( {1,{5 \over 2}} \right]$
B.
[2, 3)
C.
$\left[ { - {1 \over 2},1} \right)$
D.
$\left( {{1 \over 2},{3 \over 2}} \right] - \{ 1\} $
2021
Q106
JEE Mains
MCQ
14 Mar 2026
Let $\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $ and $\beta = \mathop {\min }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $. If $8{x^2} + bx + c = 0$ is a quadratic equation whose roots are $\alpha$1/5 and $\beta$1/5, then the value of c $-$ b is equal to :
A.
42
B.
47
C.
43
D.
50
2021
Q107
JEE Mains
MCQ
14 Mar 2026
Let $\alpha$, $\beta$ be two roots of the
equation x2 + (20)1/4x + (5)1/2 = 0. Then $\alpha$8 + $\beta$8 is equal to
equation x2 + (20)1/4x + (5)1/2 = 0. Then $\alpha$8 + $\beta$8 is equal to
A.
10
B.
100
C.
50
D.
160
2021
Q108
JEE Mains
MCQ
14 Mar 2026
If [x] be the greatest integer less than or equal to x,
then $\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]} $ is equal to :
then $\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]} $ is equal to :
A.
0
B.
4
C.
$-$2
D.
2
2021
Q109
JEE Mains
MCQ
14 Mar 2026
The number of real solutions of the equation, x2 $-$ |x| $-$ 12 = 0 is :
A.
2
B.
3
C.
1
D.
4
2021
Q110
JEE Mains
MCQ
14 Mar 2026
The number of real roots of the equation ${e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0$ is :
A.
2
B.
4
C.
6
D.
1
2021
Q111
JEE Mains
MCQ
14 Mar 2026
Let [x] denote the greatest integer less than or equal to x. Then, the values of x$\in$R satisfying the equation ${[{e^x}]^2} + [{e^x} + 1] - 3 = 0$ lie in the interval :
A.
$\left[ {0,{1 \over e}} \right)$
B.
[loge2, loge3)
C.
[1, e)
D.
[0, loge2)
2021
Q112
JEE Mains
MCQ
14 Mar 2026
If $\alpha$ and $\beta$ are the distinct roots of the equation ${x^2} + {(3)^{1/4}}x + {3^{1/2}} = 0$, then the value of ${\alpha ^{96}}({\alpha ^{12}} - 1) + {\beta ^{96}}({\beta ^{12}} - 1)$ is equal to :
A.
56 $\times$ 325
B.
56 $\times$ 324
C.
52 $\times$ 324
D.
28 $\times$ 325
2021
Q113
JEE Mains
MCQ
14 Mar 2026
The value of $3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}$ is equal to
A.
1.5 + $\sqrt 3 $
B.
2 + $\sqrt 3 $
C.
3 + 2$\sqrt 3 $
D.
4 + $\sqrt 3 $
2021
Q114
JEE Mains
MCQ
14 Mar 2026
The value of $4 + {1 \over {5 + {1 \over {4 + {1 \over {5 + {1 \over {4 + ......\infty }}}}}}}}$ is :
A.
2 + ${2 \over 5}\sqrt {30} $
B.
2 + ${4 \over {\sqrt 5 }}\sqrt {30} $
C.
5 + ${2 \over 5}\sqrt {30} $
D.
4 + ${4 \over {\sqrt 5 }}\sqrt {30} $
2021
Q115
JEE Mains
MCQ
14 Mar 2026
Let $\alpha$ and $\beta$ be the roots of x2 $-$ 6x $-$ 2 = 0. If an = $\alpha$n $-$ $\beta$n for n $ \ge $ 1, then the value of ${{{a_{10}} - 2{a_8}} \over {3{a_9}}}$ is :
A.
3
B.
2
C.
4
D.
1
2021
Q116
JEE Mains
MCQ
14 Mar 2026
The integer 'k', for which the inequality x2 $-$ 2(3k $-$ 1)x + 8k2 $-$ 7 > 0 is valid for every x in R, is :
A.
4
B.
2
C.
3
D.
0
2021
Q117
JEE Mains
MCQ
14 Mar 2026
Let p and q be two positive numbers such that p + q = 2 and p4+q4 = 272. Then p and q are
roots of the equation :
A.
x2 – 2x + 8 = 0
B.
x2 - 2x + 136=0
C.
x2 – 2x + 16 = 0
D.
x2 – 2x + 2 = 0
2021
Q118
JEE Mains
Numerical
14 Mar 2026
Let f(x) be a polynomial of degree 3 such that
$f(k) = - {2 \over k}$ for k = 2, 3, 4, 5. Then the value of 52 $-$ 10f(10) is equal to :
$f(k) = - {2 \over k}$ for k = 2, 3, 4, 5. Then the value of 52 $-$ 10f(10) is equal to :
Correct Answer: 26
Explanation:
$k\,f(k) + 2 = \lambda (x - 2)(x - 3)(x - 4)(x - 5)$ .... (1)
put x = 0
we get $\lambda = {1 \over {60}}$
Now, put $\lambda$ in equation (1)
$ \Rightarrow kf(k) + 2 = {1 \over {60}}(x - 2)(x - 3)(x - 4)(x - 5)$
Put x = 10
$ \Rightarrow 10f(10) + 2 = {1 \over {60}}(8)(7)(6)(5)$
$ \Rightarrow 52 - 10f(10) = 52 - 26 = 26$
put x = 0
we get $\lambda = {1 \over {60}}$
Now, put $\lambda$ in equation (1)
$ \Rightarrow kf(k) + 2 = {1 \over {60}}(x - 2)(x - 3)(x - 4)(x - 5)$
Put x = 10
$ \Rightarrow 10f(10) + 2 = {1 \over {60}}(8)(7)(6)(5)$
$ \Rightarrow 52 - 10f(10) = 52 - 26 = 26$
2021
Q119
JEE Mains
Numerical
14 Mar 2026
Let $\lambda$ $\ne$ 0 be in R. If $\alpha$ and $\beta$ are the roots of the equation x2 $-$ x + 2$\lambda$ = 0, and $\alpha$ and $\gamma$ are the roots of equation 3x2 $-$ 10x + 27$\lambda$ = 0, then ${{\beta \gamma } \over \lambda }$ is equal to ____________.
Correct Answer: 18
Explanation:
3$\alpha$2 $-$ 10$\alpha$ + 27$\lambda$ = 0 ..... (1)
$\alpha$2 $-$ $\alpha$ + 2$\lambda$ = 0 ...... (2)
(1) $-$ 3(2) gives
$-$7$\alpha$ + 21$\lambda$ = 0 $\Rightarrow$ $\alpha$ = 3$\lambda$
Put $\alpha$ = 3$\lambda$ in equation (1) we get
9$\lambda$2 $-$ 3$\lambda$ + 2$\lambda$ $-$ 0
9$\lambda$2 = $\lambda$ $\Rightarrow$ $\lambda$ = ${1 \over 9}$ as $\lambda$ $\ne$ 0
Now, $\alpha$ = 3$\lambda$ $\Rightarrow$ $\lambda$ = ${1 \over 3}$
$\alpha$ + $\beta$ = 1 $\Rightarrow$ $\beta$ = 2/3
$\alpha$ + $\gamma$ = ${10 \over 3}$ $\Rightarrow$ $\gamma$ = 3
${{\beta \gamma } \over \lambda } = {{{2 \over 3} \times 3} \over {{1 \over 9}}} = 18$
$\alpha$2 $-$ $\alpha$ + 2$\lambda$ = 0 ...... (2)
(1) $-$ 3(2) gives
$-$7$\alpha$ + 21$\lambda$ = 0 $\Rightarrow$ $\alpha$ = 3$\lambda$
Put $\alpha$ = 3$\lambda$ in equation (1) we get
9$\lambda$2 $-$ 3$\lambda$ + 2$\lambda$ $-$ 0
9$\lambda$2 = $\lambda$ $\Rightarrow$ $\lambda$ = ${1 \over 9}$ as $\lambda$ $\ne$ 0
Now, $\alpha$ = 3$\lambda$ $\Rightarrow$ $\lambda$ = ${1 \over 3}$
$\alpha$ + $\beta$ = 1 $\Rightarrow$ $\beta$ = 2/3
$\alpha$ + $\gamma$ = ${10 \over 3}$ $\Rightarrow$ $\gamma$ = 3
${{\beta \gamma } \over \lambda } = {{{2 \over 3} \times 3} \over {{1 \over 9}}} = 18$
2021
Q120
JEE Mains
Numerical
14 Mar 2026
The sum of all integral values of k (k $\ne$ 0) for which the equation ${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$ in x has no real roots, is ____________.
Correct Answer: 66
Explanation:
${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$
$x \in R - \{ 1,2\} $
$ \Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)$
$ \Rightarrow k(x - 3) = 2({x^2} - 3x + 2)$
for x $\ne$ 3, $k = 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right)$
$x - 3 + {2 \over {x - 3}} \ge 2\sqrt 2 ,\forall x > 3$
& $x - 3 + {2 \over {x - 3}} \le - 2\sqrt 2 ,\forall x < - 3$
$ \Rightarrow 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right) \in \left( { - \infty ,6 - 4\sqrt 2 } \right] \cup \left[ {6 + 4\sqrt 2 ,\infty } \right)$
for no real roots
$k \in (6 - 4\sqrt 2 ,6 + 4\sqrt 2 ) - \{ 0\} $
Integral k$\in${1, 2 ..... 11}
Sum of k = 66
$x \in R - \{ 1,2\} $
$ \Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)$
$ \Rightarrow k(x - 3) = 2({x^2} - 3x + 2)$
for x $\ne$ 3, $k = 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right)$
$x - 3 + {2 \over {x - 3}} \ge 2\sqrt 2 ,\forall x > 3$
& $x - 3 + {2 \over {x - 3}} \le - 2\sqrt 2 ,\forall x < - 3$
$ \Rightarrow 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right) \in \left( { - \infty ,6 - 4\sqrt 2 } \right] \cup \left[ {6 + 4\sqrt 2 ,\infty } \right)$
for no real roots
$k \in (6 - 4\sqrt 2 ,6 + 4\sqrt 2 ) - \{ 0\} $
Integral k$\in${1, 2 ..... 11}
Sum of k = 66
2021
Q121
JEE Mains
Numerical
14 Mar 2026
The number of real roots of the equation e4x $-$ e3x $-$ 4e2x $-$ ex + 1 = 0 is equal to ______________.
Correct Answer: 2
Explanation:
t4 $-$ t3 $-$ 4t2 $-$ t + 1 = 0, ex = t > 0
$ \Rightarrow {t^2} - t - 4 - {1 \over t} + {1 \over {{t^2}}} = 0$
$ \Rightarrow {\alpha ^2} - \alpha - 6 = 0,\alpha = t + {1 \over t} \ge 2$
$ \Rightarrow \alpha = 3, - 2$ (reject)
$ \Rightarrow t + {1 \over t} = 3$
$\Rightarrow$ The number of real roots = 2
$ \Rightarrow {t^2} - t - 4 - {1 \over t} + {1 \over {{t^2}}} = 0$
$ \Rightarrow {\alpha ^2} - \alpha - 6 = 0,\alpha = t + {1 \over t} \ge 2$
$ \Rightarrow \alpha = 3, - 2$ (reject)
$ \Rightarrow t + {1 \over t} = 3$
$\Rightarrow$ The number of real roots = 2
2021
Q122
JEE Mains
Numerical
14 Mar 2026
If a + b + c = 1, ab + bc + ca = 2 and abc = 3, then the value of a4 + b4 + c4 is equal to ______________.
Correct Answer: 13
Explanation:
(a + b + c)2 = 1
$ \Rightarrow $ a2 + b2 + c2 + 2(ab + bc + ca) = 1
$ \Rightarrow $ a2 + b2 + c2 = – 3 ….(i)
$ \Rightarrow $ ab + bc + ca = 2 ….(ii)
Squaring of equation (ii),
$ \Rightarrow $ a2b2 + b2c2 + c2a2 + 2(ab2c + bc2a + ca2b) = 4
$ \Rightarrow $ a2b2 + b2c2 + c2a2 + 2abc(a + b + c) = 4
$ \Rightarrow $ a2b2 + b2c2 + c2a2 + 6 = 4
$ \Rightarrow $ a2b2 + b2c2 + c2a2 = – 2 ….(iii)
Squaring of equation (i),
$ \Rightarrow $ a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2) = 9
$ \Rightarrow $ a4 + b4 + c4 – 4 = 9
$ \Rightarrow $ a4 + b4 + c4 = 13
$ \Rightarrow $ a2 + b2 + c2 + 2(ab + bc + ca) = 1
$ \Rightarrow $ a2 + b2 + c2 = – 3 ….(i)
$ \Rightarrow $ ab + bc + ca = 2 ….(ii)
Squaring of equation (ii),
$ \Rightarrow $ a2b2 + b2c2 + c2a2 + 2(ab2c + bc2a + ca2b) = 4
$ \Rightarrow $ a2b2 + b2c2 + c2a2 + 2abc(a + b + c) = 4
$ \Rightarrow $ a2b2 + b2c2 + c2a2 + 6 = 4
$ \Rightarrow $ a2b2 + b2c2 + c2a2 = – 2 ….(iii)
Squaring of equation (i),
$ \Rightarrow $ a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2) = 9
$ \Rightarrow $ a4 + b4 + c4 – 4 = 9
$ \Rightarrow $ a4 + b4 + c4 = 13
2021
Q123
JEE Mains
Numerical
14 Mar 2026
If $\alpha$, $\beta$ are roots of the equation ${x^2} + 5(\sqrt 2 )x + 10 = 0$, $\alpha$ > $\beta$ and ${P_n} = {\alpha ^n} - {\beta ^n}$ for each positive integer n, then the value of $\left( {{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}} \right)$ is equal to _________.
Correct Answer: 1
Explanation:
${x^2} + 5\sqrt 2 x + 10 = 0$
& ${P_n} = {\alpha ^n} - {\beta ^n}$ (Given)
Now, ${{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}}$ = ${{{{P_{17}}({P_{20}} + 5\sqrt 2 {P_{19}})} \over {{P_{18}}({P_{19}} + 5\sqrt 2 P_{18}^{})}}}$
${{{P_{17}}({\alpha ^{20}} - {\beta ^{20}} + 5\sqrt 2 ({\alpha ^{19}} - {\beta ^{19}}))} \over {{P_{18}}({\alpha ^{19}} - {\beta ^{19}} + 5\sqrt 2 ({\alpha ^{18}} - {\beta ^{18}}))}}$
${{{P_{17}}({\alpha ^{19}}(\alpha + 5\sqrt 2 ) - {\beta ^{19}}(\beta + 5\sqrt 2 ))} \over {{P_{18}}({\alpha ^{18}}(\alpha + 5\sqrt 2 ) - {\beta ^{18}}(\beta + 5\sqrt 2 ))}}$
Since, $\alpha + 5\sqrt 2 = - 10/\alpha $ ..... (1)
& $\beta + 5\sqrt 2 = - 10/\beta $ ....... (2)
Now, put there values in above expression $ = - {{10{P_{17}}{P_{18}}} \over { - 10{P_{18}}{P_{17}}}} = 1$
& ${P_n} = {\alpha ^n} - {\beta ^n}$ (Given)
Now, ${{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}}$ = ${{{{P_{17}}({P_{20}} + 5\sqrt 2 {P_{19}})} \over {{P_{18}}({P_{19}} + 5\sqrt 2 P_{18}^{})}}}$
${{{P_{17}}({\alpha ^{20}} - {\beta ^{20}} + 5\sqrt 2 ({\alpha ^{19}} - {\beta ^{19}}))} \over {{P_{18}}({\alpha ^{19}} - {\beta ^{19}} + 5\sqrt 2 ({\alpha ^{18}} - {\beta ^{18}}))}}$
${{{P_{17}}({\alpha ^{19}}(\alpha + 5\sqrt 2 ) - {\beta ^{19}}(\beta + 5\sqrt 2 ))} \over {{P_{18}}({\alpha ^{18}}(\alpha + 5\sqrt 2 ) - {\beta ^{18}}(\beta + 5\sqrt 2 ))}}$
Since, $\alpha + 5\sqrt 2 = - 10/\alpha $ ..... (1)
& $\beta + 5\sqrt 2 = - 10/\beta $ ....... (2)
Now, put there values in above expression $ = - {{10{P_{17}}{P_{18}}} \over { - 10{P_{18}}{P_{17}}}} = 1$
2021
Q124
JEE Mains
Numerical
14 Mar 2026
Let $\alpha$ and $\beta$ be two real numbers such that $\alpha$ + $\beta$ = 1 and $\alpha$$\beta$ = $-$1. Let pn = ($\alpha$)n + ($\beta$)n, pn$-$1 = 11 and pn+1 = 29 for some integer n $ \ge $ 1. Then, the value of p$_n^2$ is ___________.
Correct Answer: 324
Explanation:
Given, $\alpha$ + $\beta$ = 1, $\alpha$$\beta$ = $-$ 1
$ \therefore $ Quadratic equation with roots $\alpha$, $\beta$ is x2 $-$ x $-$ 1 = 0
$ \Rightarrow $ $\alpha$2 = $\alpha$ + 1
Multiplying both sides by $\alpha$n$-$1
$\alpha$n$+$1 = $\alpha$n + $\alpha$n$-$1 ......(1)
Similarly,
$\beta$n + 1 = $\beta$n + $\beta$n + 1 ..... (2)
Adding (1) & (2)
${\alpha ^{n + 1}} + {\beta ^{n + 1}} = ({\alpha ^n} + {\beta ^n}) + ({\alpha ^{n - 1}} + {\beta ^{n - 1}})$
$ \Rightarrow $ Pn+1 = Pn + Pn$-$1
$ \Rightarrow $ 29 = Pn + 11 (Given, Pn + 1 = 29, Pn $-$ 1 = 11)
$ \Rightarrow $ Pn = 18
$ \therefore $ $P_n^2$ = 182 = 324
$ \therefore $ Quadratic equation with roots $\alpha$, $\beta$ is x2 $-$ x $-$ 1 = 0
$ \Rightarrow $ $\alpha$2 = $\alpha$ + 1
Multiplying both sides by $\alpha$n$-$1
$\alpha$n$+$1 = $\alpha$n + $\alpha$n$-$1 ......(1)
Similarly,
$\beta$n + 1 = $\beta$n + $\beta$n + 1 ..... (2)
Adding (1) & (2)
${\alpha ^{n + 1}} + {\beta ^{n + 1}} = ({\alpha ^n} + {\beta ^n}) + ({\alpha ^{n - 1}} + {\beta ^{n - 1}})$
$ \Rightarrow $ Pn+1 = Pn + Pn$-$1
$ \Rightarrow $ 29 = Pn + 11 (Given, Pn + 1 = 29, Pn $-$ 1 = 11)
$ \Rightarrow $ Pn = 18
$ \therefore $ $P_n^2$ = 182 = 324
2021
Q125
JEE Mains
Numerical
14 Mar 2026
The sum of 162th power of the roots of the equation x3 $-$ 2x2 + 2x $-$ 1 = 0 is ________.
Correct Answer: 3
Explanation:
x3 $-$ 2x2 + 2x $-$ 1 = 0
x = 1 satisfying the equation
$ \therefore $ x $-$ 1 is factor of
x3 $-$ 2x2 + 2x $-$ 1
= (x $-$ 1) (x2 $-$ x + 1) = 0
x = 1, ${{1 + i\sqrt 3 } \over 2},{{1 - i\sqrt 3 } \over 2}$
x = 1, $-$ $\omega$2, $-$$\omega$
Sum of 162th power of roots
= (1)162 + ($-$$\omega$2)162 + ($-$$\omega$)162
= 1 + ($\omega$)324 + ($\omega$)162
= 1 + 1 + 1 = 3
x = 1 satisfying the equation
$ \therefore $ x $-$ 1 is factor of
x3 $-$ 2x2 + 2x $-$ 1
= (x $-$ 1) (x2 $-$ x + 1) = 0
x = 1, ${{1 + i\sqrt 3 } \over 2},{{1 - i\sqrt 3 } \over 2}$
x = 1, $-$ $\omega$2, $-$$\omega$
Sum of 162th power of roots
= (1)162 + ($-$$\omega$2)162 + ($-$$\omega$)162
= 1 + ($\omega$)324 + ($\omega$)162
= 1 + 1 + 1 = 3
2021
Q126
JEE Mains
Numerical
14 Mar 2026
The number of the real roots of the equation ${(x + 1)^2} + |x - 5| = {{27} \over 4}$ is ________.
Correct Answer: 2
Explanation:
When $x > 5$
${(x + 1)^2} + (x - 5) = {{27} \over 4}$
$ \Rightarrow {x^2} + 3x - 4 = {{27} \over 4}$
$ \Rightarrow {x^2} + 3x - {{43} \over 4} = 0$
$ \Rightarrow 4{x^2} + 12x - 43 = 0$
$x = {{ - 12 \pm \sqrt {144 + 688} } \over 8}$
$x = {{ - 12 \pm \sqrt {832} } \over 8} = {{ - 12 \pm 28.8} \over 8}$
$ = {{ - 3 + 7.2} \over 2}$
$ = {{ - 3 + 7.2} \over 2},{{ - 3 - 7.2} \over 2}$
= 2.1, -5.1 [ both are rejected as x should be > 5 ]
(Therefore no solution)
For $x \le 5$
${(x + 1)^2} - (x - 5) = {{27} \over 4}$
${x^2} + x + 6 - {{27} \over 4} = 0$
$4{x^2} + 4x - 3 = 0$
$x = {{ - 4 \pm \sqrt {16 + 48} } \over 8}$
$x = {{ - 4 \pm 8} \over 8} \Rightarrow x = - {{12} \over 8},{4 \over 8}$
$ \therefore $ So, the equation have two real roots.
${(x + 1)^2} + (x - 5) = {{27} \over 4}$
$ \Rightarrow {x^2} + 3x - 4 = {{27} \over 4}$
$ \Rightarrow {x^2} + 3x - {{43} \over 4} = 0$
$ \Rightarrow 4{x^2} + 12x - 43 = 0$
$x = {{ - 12 \pm \sqrt {144 + 688} } \over 8}$
$x = {{ - 12 \pm \sqrt {832} } \over 8} = {{ - 12 \pm 28.8} \over 8}$
$ = {{ - 3 + 7.2} \over 2}$
$ = {{ - 3 + 7.2} \over 2},{{ - 3 - 7.2} \over 2}$
= 2.1, -5.1 [ both are rejected as x should be > 5 ]
(Therefore no solution)
For $x \le 5$
${(x + 1)^2} - (x - 5) = {{27} \over 4}$
${x^2} + x + 6 - {{27} \over 4} = 0$
$4{x^2} + 4x - 3 = 0$
$x = {{ - 4 \pm \sqrt {16 + 48} } \over 8}$
$x = {{ - 4 \pm 8} \over 8} \Rightarrow x = - {{12} \over 8},{4 \over 8}$
$ \therefore $ So, the equation have two real roots.
2021
Q127
JEE Advanced
Numerical
14 Mar 2026
For x $\in$ R, the number of real roots of the equation $3{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 0$ is ________.
Correct Answer: 4
Explanation:
Given,
$3{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 0$ .... (i)
For $-$1 $\le$ x $\le$ 1 i.e., x$\in$[$-$1, 1]
From Eq. (i), we get
$3{x^2} - 4( - {x^2} + 1) + x - 1 = 0$
$ \Rightarrow 3{x^2} + 4{x^2} - 4 + x - 1 = 0$
$ \Rightarrow 7{x^2} + x - 5 = 0$
$ \Rightarrow x = {{ - 1 \pm \sqrt {1 + 140} } \over {(2 \times 7)}}$
Here, both values of x are acceptable.
For | x | > | i.e. x $\in$($-$ $\infty$, $-$1) $\cup$ (1, $\infty$)
From Eq. (i), we get
$3{x^2} - 4({x^2} - 1) + x - 1 = 0$
$ \Rightarrow {x^2} - x - 3 = 0$
$ \Rightarrow x = {{1 \pm \sqrt {1 + 12} } \over 2}$
Again here, both values of x are acceptable.
Hence, total number of solutions is 4.
$3{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 0$ .... (i)
For $-$1 $\le$ x $\le$ 1 i.e., x$\in$[$-$1, 1]
From Eq. (i), we get
$3{x^2} - 4( - {x^2} + 1) + x - 1 = 0$
$ \Rightarrow 3{x^2} + 4{x^2} - 4 + x - 1 = 0$
$ \Rightarrow 7{x^2} + x - 5 = 0$
$ \Rightarrow x = {{ - 1 \pm \sqrt {1 + 140} } \over {(2 \times 7)}}$
Here, both values of x are acceptable.
For | x | > | i.e. x $\in$($-$ $\infty$, $-$1) $\cup$ (1, $\infty$)
From Eq. (i), we get
$3{x^2} - 4({x^2} - 1) + x - 1 = 0$
$ \Rightarrow {x^2} - x - 3 = 0$
$ \Rightarrow x = {{1 \pm \sqrt {1 + 12} } \over 2}$
Again here, both values of x are acceptable.
Hence, total number of solutions is 4.
2020
Q128
JEE Mains
MCQ
14 Mar 2026
If $\alpha $ and $\beta $ are the roots of the equation
2x(2x + 1) = 1, then $\beta $ is equal to :
2x(2x + 1) = 1, then $\beta $ is equal to :
A.
$ - 2\alpha \left( {\alpha + 1} \right)$
B.
$ 2\alpha \left( {\alpha + 1} \right)$
C.
$2{\alpha ^2}$
D.
$ 2\alpha \left( {\alpha - 1} \right)$
2020
Q129
JEE Mains
MCQ
14 Mar 2026
If $\alpha $ and $\beta $ be two roots of the equation
x2 – 64x + 256 = 0. Then the value of
${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$ is :
x2 – 64x + 256 = 0. Then the value of
${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$ is :
A.
1
B.
3
C.
2
D.
4
2020
Q130
JEE Mains
MCQ
14 Mar 2026
If $\alpha $ and $\beta $ are the roots of the equation,
7x2 – 3x – 2 = 0, then the value of
${\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}$ is equal to :
7x2 – 3x – 2 = 0, then the value of
${\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}$ is equal to :
A.
${1 \over {24}}$
B.
${{27} \over {32}}$
C.
${{27} \over {16}}$
D.
${3 \over 8}$
2020
Q131
JEE Mains
MCQ
14 Mar 2026
The product of the roots of the
equation 9x2 - 18|x| + 5 = 0 is :
equation 9x2 - 18|x| + 5 = 0 is :
A.
${{5} \over {9}}$
B.
${{5} \over {27}}$
C.
${{25} \over {81}}$
D.
${{25} \over {9}}$
2020
Q132
JEE Mains
MCQ
14 Mar 2026
Let $\lambda \ne 0$ be in R. If $\alpha $ and $\beta $ are the roots of the
equation, x2 - x + 2$\lambda $ = 0 and $\alpha $ and $\gamma $ are the roots of
the equation, $3{x^2} - 10x + 27\lambda = 0$, then ${{\beta \gamma } \over \lambda }$ is equal to:
equation, x2 - x + 2$\lambda $ = 0 and $\alpha $ and $\gamma $ are the roots of
the equation, $3{x^2} - 10x + 27\lambda = 0$, then ${{\beta \gamma } \over \lambda }$ is equal to:
A.
36
B.
9
C.
27
D.
18
2020
Q133
JEE Mains
MCQ
14 Mar 2026
Let $\alpha $ and $\beta $ be the roots of x2 - 3x + p=0 and $\gamma $ and $\delta $ be the roots of x2 - 6x + q = 0. If $\alpha, \beta, \gamma, \delta $
form a geometric progression.Then ratio (2q + p) : (2q - p) is:
A.
9 : 7
B.
5 : 3
C.
3 : 1
D.
33 :31
2020
Q134
JEE Mains
MCQ
14 Mar 2026
Let [t] denote the greatest integer $ \le $ t. Then the equation in x,
[x]2 + 2[x+2] - 7 = 0 has :
[x]2 + 2[x+2] - 7 = 0 has :
A.
no integral solution.
B.
exactly two solutions.
C.
exactly four integral solutions.
D.
infinitely many solutions.
2020
Q135
JEE Mains
MCQ
14 Mar 2026
The set of all real values of $\lambda $ for which the
quadratic equations,
($\lambda $2 + 1)x2 – 4$\lambda $x + 2 = 0 always have exactly one root in the interval (0, 1) is :
($\lambda $2 + 1)x2 – 4$\lambda $x + 2 = 0 always have exactly one root in the interval (0, 1) is :
A.
(–3, –1)
B.
(2, 4]
C.
(0, 2)
D.
(1, 3]
2020
Q136
JEE Mains
MCQ
14 Mar 2026
If $\alpha $ and $\beta $ are the roots of the equation
x2 + px + 2 = 0 and ${1 \over \alpha }$ and ${1 \over \beta }$ are the
roots of the equation 2x2 + 2qx + 1 = 0, then
$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right)$ is equal to :
x2 + px + 2 = 0 and ${1 \over \alpha }$ and ${1 \over \beta }$ are the
roots of the equation 2x2 + 2qx + 1 = 0, then
$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right)$ is equal to :
A.
${9 \over 4}\left( {9 - {q^2}} \right)$
B.
${9 \over 4}\left( {9 + {q^2}} \right)$
C.
${9 \over 4}\left( {9 - {p^2}} \right)$
D.
${9 \over 4}\left( {9 + {p^2}} \right)$
2020
Q137
JEE Mains
MCQ
14 Mar 2026
Let f(x) be a quadratic polynomial such that
f(–1) + f(2) = 0. If one of the roots of f(x) = 0
is 3, then its other root lies in :
f(–1) + f(2) = 0. If one of the roots of f(x) = 0
is 3, then its other root lies in :
A.
(–3, –1)
B.
(1, 3)
C.
(–1, 0)
D.
(0, 1)
2020
Q138
JEE Mains
MCQ
14 Mar 2026
Let
$\alpha $ and
$\beta $ be the roots of the equation
5x2 + 6x – 2 = 0. If Sn = $\alpha $n + $\beta $n, n = 1, 2, 3...., then :
5x2 + 6x – 2 = 0. If Sn = $\alpha $n + $\beta $n, n = 1, 2, 3...., then :
A.
5S6
+ 6S5
= 2S4
B.
5S6
+ 6S5
+ 2S4 = 0
C.
6S6
+ 5S5
+ 2S4 = 0
D.
6S6
+ 5S5
= 2S4
2020
Q139
JEE Mains
MCQ
14 Mar 2026
Let a, b $ \in $ R, a $ \ne $ 0 be such that the equation,
ax2 – 2bx + 5 = 0 has a repeated root $\alpha $, which
is also a root of the equation, x2 – 2bx – 10 = 0.
If $\beta $ is the other root of this equation, then
$\alpha $2 + $\beta $2 is equal to :
A.
28
B.
24
C.
26
D.
25
2020
Q140
JEE Mains
MCQ
14 Mar 2026
The number of real roots of the equation,
e4x + e3x – 4e2x + ex + 1 = 0 is :
e4x + e3x – 4e2x + ex + 1 = 0 is :
A.
1
B.
2
C.
3
D.
4
2020
Q141
JEE Mains
MCQ
14 Mar 2026
Let $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$.
If $a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $ and
$b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}} $, then a and b are the roots of the quadratic equation :
If $a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $ and
$b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}} $, then a and b are the roots of the quadratic equation :
A.
x2 + 101x + 100 = 0
B.
x2 + 102x + 101 = 0
C.
x2 – 102x + 101 = 0
D.
x2 – 101x + 100 = 0
2020
Q142
JEE Mains
MCQ
14 Mar 2026
Let S be the set of all real roots of the equation,
3x(3x – 1) + 2 = |3x – 1| + |3x – 2|. Then S :
3x(3x – 1) + 2 = |3x – 1| + |3x – 2|. Then S :
A.
contains exactly two elements.
B.
is an empty set.
C.
is a singleton.
D.
contains at least four elements.
2020
Q143
JEE Mains
MCQ
14 Mar 2026
Let $\alpha $ and $\beta $ be the roots of the equation x2
- x - 1 = 0.
If pk = ${\left( \alpha \right)^k} + {\left( \beta \right)^k}$ , k $ \ge $ 1, then which one of the following statements is not true?
If pk = ${\left( \alpha \right)^k} + {\left( \beta \right)^k}$ , k $ \ge $ 1, then which one of the following statements is not true?
A.
(p1 + p2 + p3 + p4 + p5) = 26
B.
p5 = 11
C.
p3 = p5 – p4
D.
p5 = p2 · p3
2020
Q144
JEE Mains
MCQ
14 Mar 2026
Let $\alpha $ and $\beta $ be two real roots of the equation
(k + 1)tan2x - $\sqrt 2 $ . $\lambda $tanx = (1 - k), where k($ \ne $ - 1) and $\lambda $ are real numbers. if tan2 ($\alpha $ + $\beta $) = 50, then a value of $\lambda $ is:
(k + 1)tan2x - $\sqrt 2 $ . $\lambda $tanx = (1 - k), where k($ \ne $ - 1) and $\lambda $ are real numbers. if tan2 ($\alpha $ + $\beta $) = 50, then a value of $\lambda $ is:
A.
5$\sqrt 2 $
B.
10
C.
5
D.
10$\sqrt 2 $
2020
Q145
JEE Mains
Numerical
14 Mar 2026
The least positive value of 'a' for which the
equation
2x2 + (a – 10)x + ${{33} \over 2}$ = 2a has real roots is
2x2 + (a – 10)x + ${{33} \over 2}$ = 2a has real roots is
Correct Answer: 8
Explanation:
For real roots Discriminate $ \ge $ 0.
(a – 10)2 – 4$\left( {{{33} \over 2} - 2a} \right).2$ $ \ge $ 0
$ \Rightarrow $ a2 + 100 – 20a – 132 + 16a $ \ge $ 0
$ \Rightarrow $ a 2 – 4a – 32 $ \ge $ 0
$ \Rightarrow $ (a – 8) (a + 4) $ \ge $ 0
$ \Rightarrow $ a $ \le $ -4 $ \cup $ a $ \ge $ 8
$ \therefore $ least positive a = 8
(a – 10)2 – 4$\left( {{{33} \over 2} - 2a} \right).2$ $ \ge $ 0
$ \Rightarrow $ a2 + 100 – 20a – 132 + 16a $ \ge $ 0
$ \Rightarrow $ a 2 – 4a – 32 $ \ge $ 0
$ \Rightarrow $ (a – 8) (a + 4) $ \ge $ 0
$ \Rightarrow $ a $ \le $ -4 $ \cup $ a $ \ge $ 8
$ \therefore $ least positive a = 8
2020
Q146
JEE Advanced
MCQ
14 Mar 2026
Suppose a, b denote the distinct real roots of the quadratic polynomial x2 + 20x $-$ 2020 and suppose c, d denote the distinct complex roots of the quadratic polynomial x2 $-$ 20x + 2020. Then the value of
ac(a $-$ c) + ad(a $-$ d) + bc(b $-$ c) + bd(b $-$ d) is
ac(a $-$ c) + ad(a $-$ d) + bc(b $-$ c) + bd(b $-$ d) is
A.
0
B.
8000
C.
8080
D.
16000
2019
Q147
JEE Mains
MCQ
14 Mar 2026
If $\alpha $, $\beta $ and $\gamma $ are three consecutive terms of a non-constant G.P. such that the equations $\alpha $x
2
+ 2$\beta $x + $\gamma $ = 0 and
x2
+ x – 1 = 0 have a common root, then $\alpha $($\beta $ + $\gamma $) is equal to :
A.
$\alpha $$\gamma $
B.
0
C.
$\beta $$\gamma $
D.
$\alpha $$\beta $
2019
Q148
JEE Mains
MCQ
14 Mar 2026
The number of real roots of the equation
5 + |2x – 1| = 2x (2x – 2) is
5 + |2x – 1| = 2x (2x – 2) is
A.
2
B.
1
C.
3
D.
4
2019
Q149
JEE Mains
MCQ
14 Mar 2026
All the pairs (x, y) that satisfy the inequality
${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$
also satisfy the equation
${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$
also satisfy the equation
A.
sin x = |sin y|
B.
sin x = 2sin y
C.
2 sin x = sin y
D.
2 |sin x | = 3 sin y
2019
Q150
JEE Mains
MCQ
14 Mar 2026
If $\alpha $ and $\beta $ are the roots of the quadratic equation,
x2 + x sin $\theta $ - 2 sin $\theta $ = 0, $\theta \in \left( {0,{\pi \over 2}} \right)$, then
${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}$ is equal to :
x2 + x sin $\theta $ - 2 sin $\theta $ = 0, $\theta \in \left( {0,{\pi \over 2}} \right)$, then
${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}$ is equal to :
A.
${{{2^{12}}} \over {{{\left( {\sin \theta - 8} \right)}^6}}}$
B.
${{{2^6}} \over {{{\left( {\sin \theta + 4} \right)}^{12}}}}$
C.
${{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}$
D.
${{{2^{12}}} \over {{{\left( {\sin \theta - 4} \right)}^{12}}}}$