Quadratic Equation and Inequalities
299 Questions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Evening Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
The number of real solutions of the equation, x2 $-$ |x| $-$ 12 = 0 is :
A.
2
B.
3
C.
1
D.
4
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Morning Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
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
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
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
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
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
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
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
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
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
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
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
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
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
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
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
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
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
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
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
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
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
JEE Advanced
Numerical
JEE Advanced 2021 Paper 1 Online
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
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Morning Slot
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
JEE Advanced
MCQ
JEE Advanced 2020 Paper 1 Offline
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
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
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
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
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
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
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
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
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.
${{{2^{12}}} \over {{{\left( {\sin \theta - 4} \right)}^{12}}}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
If m is chosen in the quadratic equation
(m2 + 1) x2 – 3x + (m2 + 1)2 = 0
such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-
(m2 + 1) x2 – 3x + (m2 + 1)2 = 0
such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-
A.
$4\sqrt 3 $
B.
$8\sqrt 3 $
C.
$8\sqrt 5 $
D.
$10\sqrt 5 $
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
Let p, q $ \in $ R. If 2 - $\sqrt 3$ is a root of the quadratic
equation, x2 + px + q = 0, then :
A.
p2 – 4q – 12 = 0
B.
q2 – 4p – 16 = 0
C.
q2 + 4p + 14 = 0
D.
p2 – 4q + 12 = 0
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
The number of integral values of m for which the
equation
(1 + m2 )x2 – 2(1 + 3m)x + (1 + 8m) = 0 has no real root is :
(1 + m2 )x2 – 2(1 + 3m)x + (1 + 8m) = 0 has no real root is :
A.
2
B.
infinitely many
C.
1
D.
3
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
The sum of the solutions of the equation
$\left| {\sqrt x - 2} \right| + \sqrt x \left( {\sqrt x - 4} \right) + 2 = 0$
(x > 0) is equal to:
$\left| {\sqrt x - 2} \right| + \sqrt x \left( {\sqrt x - 4} \right) + 2 = 0$
(x > 0) is equal to:
A.
9
B.
12
C.
4
D.
10
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
The number of integral values of m for which the quadratic expression, (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), x $ \in $ R, is always positive, is :
A.
7
B.
8
C.
3
D.
6