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lv3ve4b4 | maths | quadratic-equation-and-inequalities | modulus-function | <p>The number of distinct real roots of the equation $$|x+1||x+3|-4|x+2|+5=0$$, is _______</p> | [] | null | 2 | <p>Let's analyze the equation $ |x+1||x+3|-4|x+2|+5=0 $ based on different intervals of $ x $.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4nlbkg/60bbafa8-f162-4063-8d33-6f852bb91330/ef943d00-10fb-11ef-aaa0-17ca36a32505/file-1lw4nlbkh.png?format=png" data-orsrc="https://app-content.cdn.ex... | integer | jee-main-2024-online-8th-april-evening-shift | 7,818 |
lv9s20ns | maths | quadratic-equation-and-inequalities | modulus-function | <p>The number of real solutions of the equation $$x|x+5|+2|x+7|-2=0$$ is __________.</p> | [] | null | 3 | <p>$$x|x+5|+2|x+7|-2=0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwexk63n/0315e6c4-33b6-4400-9e04-b71afe90ed53/c1552330-16a2-11ef-afc3-d53e649859cd/file-1lwexk63o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwexk63n/0315e6c4-33b6-4400-9e04-b71afe90ed53/c155233... | integer | jee-main-2024-online-5th-april-evening-shift | 7,820 |
FDe9CsgnfSoRtuy8 | maths | quadratic-equation-and-inequalities | nature-of-roots | If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals | [{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$3$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$1$$ "}] | ["D"] | null | <p>Let n and (n + 1) be the roots of x<sup>2</sup> $$-$$ bx + c = 0.</p>
<p>Then, n + (n + 1) = b and n(n + 1) = c</p>
<p>$$\therefore$$ b<sup>2</sup> $$-$$ 4c = (2n + 1)<sup>2</sup> $$-$$ 4n(n + 1)</p>
<p>= 4n<sup>2</sup> + 4n + 1 $$-$$ 4n<sup>2</sup> $$-$$ 4n = 1</p> | mcq | aieee-2005 | 7,821 |
qHu7OenJ4BgHkd9E | maths | quadratic-equation-and-inequalities | nature-of-roots | If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals | [{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$3$$ "}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$1$$"}] | ["D"] | null | Let $$\alpha ,\,\,\alpha + 1\,\,$$ be roots
<br><br>Then $$\alpha + \alpha + 1 = b = $$ sum of -
<br><br>roots $$\alpha \left( {\alpha + 1} \right) = c$$
<br><br>$$=$$ product of roots
<br><br>$$\therefore$$ $${b^2} - 4c $$
<br><br>$$ = {\left( {2\alpha + 1} \right)^2} - 4\alpha \left( {\alpha + 1} \right)$$
<b... | mcq | aieee-2005 | 7,822 |
3ow2JWohjhrBtHTY | maths | quadratic-equation-and-inequalities | nature-of-roots | The equation $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$ has: | [{"identifier": "A", "content": "infinite number of real roots "}, {"identifier": "B", "content": "no real roots "}, {"identifier": "C", "content": "exactly one real root "}, {"identifier": "D", "content": "exactly four real roots "}] | ["B"] | null | Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$
<br><br>Put $${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$$ in the given equation,
<br><br>we get $${t^2} - 4t - 1 = 0$$
<br><br>$$ \Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}$$
<br><br>$... | mcq | aieee-2012 | 7,824 |
Fl2e9gnl2jKIkobMZcMni | maths | quadratic-equation-and-inequalities | nature-of-roots | The sum of all the real values of x satisfying the equation
<br/>2<sup>(x$$-$$1)(x<sup>2</sup> + 5x $$-$$ 50)</sup> = 1 is : | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "$$-$$ 5"}] | ["C"] | null | We know, 2<sup>x</sup> = 1 only when x = 0.
<br><br>Similarly, 2<sup>(x$$-$$1)(x<sup>2</sup> + 5x $$-$$ 50)</sup> = 1 when
<br><br>(x$$-$$1)(x<sup>2</sup> + 5x $$-$$ 50) = 0
<br><br>$$ \Rightarrow $$ (x - 1)(x + 10)(x - 5) = 0
<br><br>$$ \therefore $$ x = 1, -10, 5
<br><br>Sum of real values of x = 1 + (-10) + 5 = -4 | mcq | jee-main-2017-online-9th-april-morning-slot | 7,825 |
wQG21GrIK7fXq8Sa2uU4f | maths | quadratic-equation-and-inequalities | nature-of-roots | The number of all possible positive integral values of $$\alpha $$ for which the roots of the quadratic equation, 6x<sup>2</sup> $$-$$ 11x + $$\alpha $$ = 0 are rational numbers is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "5"}] | ["A"] | null | For rational D must be perfect square
<br><br>D = 121 $$-$$ 24$$\alpha $$
<br><br>for 121 $$-$$ 24$$\alpha $$ to be perfect square a must be 3, 4, 5
<br><br>So, ans $$\alpha $$ = 3 | mcq | jee-main-2019-online-9th-january-evening-slot | 7,826 |
wXK8RulvIwkGLmPr1c6RX | maths | quadratic-equation-and-inequalities | nature-of-roots | The number of integral values of m for which the
equation
<br/><br/>(1 + m<sup>2</sup>
)x<sup>2</sup>
– 2(1 + 3m)x + (1 + 8m) = 0
has no real root is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "infinitely many"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["B"] | null | (1 + m<sup>2</sup>
)x<sup>2</sup>
– 2(1 + 3m)x + (1 + 8m) = 0
<br><br>Given equation has no real solution,
<br><br>$$ \therefore $$ Discriminant (D) < 0
<br><br>$$ \Rightarrow $$ 4(1 + 3m)<sup>2</sup> - 4(1 + m<sup>2</sup>)(1 + 8m) < 0
<br><br>$$ \Rightarrow $$ 4[9m<sup>2</sup> + 6m + 1 - 8m - 1 - 8m<sup>3</sup>... | mcq | jee-main-2019-online-8th-april-evening-slot | 7,827 |
r4cRgmuaQNaqnCyfRN18hoxe66ijvwvt235 | maths | quadratic-equation-and-inequalities | nature-of-roots | If m is chosen in the quadratic equation
<br/><br/>(m<sup>2</sup> + 1)
x<sup>2</sup> – 3x + (m<sup>2</sup> + 1)<sup>2</sup> = 0
<br/><br/>such that the sum of its
roots is greatest, then the absolute difference of
the cubes of its roots is :- | [{"identifier": "A", "content": "$$4\\sqrt 3 $$"}, {"identifier": "B", "content": "$$8\\sqrt 3 $$"}, {"identifier": "C", "content": "$$8\\sqrt 5 $$"}, {"identifier": "D", "content": "$$10\\sqrt 5 $$"}] | ["C"] | null | Given quadratic equation
<br><br>(m<sup>2</sup> + 1)
x<sup>2</sup> – 3x + (m<sup>2</sup> + 1)<sup>2</sup> = 0
<br><br>Let roots of the equation $$\alpha $$ and $$\beta $$.
<br><br>$$ \therefore $$ Sum of roots = $$\alpha $$ + $$\beta $$ = $${3 \over {{m^2} + 1}}$$
<br><br>Product of roots = $$\alpha $$$$\beta $$ = m<... | mcq | jee-main-2019-online-9th-april-evening-slot | 7,828 |
xt1r2MgSVHVAqiqUTS7k9k2k5h0294b | maths | quadratic-equation-and-inequalities | nature-of-roots | The least positive value of 'a' for which the
equation <br/><br/>2x<sup>2</sup> + (a – 10)x + $${{33} \over 2}$$
= 2a has real
roots is | [] | null | 8 | For real roots Discriminate $$ \ge $$ 0.
<br><br>(a – 10)<sup>2</sup>
– 4$$\left( {{{33} \over 2} - 2a} \right).2$$ $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ a<sup>2</sup>
+ 100 – 20a – 132 + 16a $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ a
<sup>2</sup>
– 4a – 32 $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ (a – 8) (a + 4) $$ \ge $$ ... | integer | jee-main-2020-online-8th-january-morning-slot | 7,829 |
V6ZrAEdn5TU3mCjJcDjgy2xukez5rybs | maths | quadratic-equation-and-inequalities | nature-of-roots | Let f(x) be a quadratic polynomial such that
<br/>f(–1) + f(2) = 0. If one of the roots of f(x) = 0
<br/>is 3, then its other root lies in : | [{"identifier": "A", "content": "(\u20133, \u20131)"}, {"identifier": "B", "content": "(1, 3)"}, {"identifier": "C", "content": "(\u20131, 0)"}, {"identifier": "D", "content": "(0, 1)"}] | ["C"] | null | Let the other root is $$\alpha $$.
<br><br>$$ \therefore $$ f(x)
=
a(x
–
3)
(x
–
$$\alpha $$)
<br><br>f(2) = a($$\alpha $$– 2)
<br><br>f(–1) = 4a(1 + $$\alpha $$)
<br><br>Given f(–1) + f(2) = 0
<br><br>$$ \Rightarrow $$a($$\alpha $$ – 2 + 4 + 4$$\alpha $$) = 0
<br><br>$$ \Rightarrow $$ 5$$\alpha $$ = -2 ... | mcq | jee-main-2020-online-2nd-september-evening-slot | 7,830 |
1ktbh3q6c | maths | quadratic-equation-and-inequalities | nature-of-roots | 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 ____________. | [] | null | 66 | $${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$$<br><br>$$x \in R - \{ 1,2\} $$<br><br>$$ \Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)$$<br><br>$$ \Rightarrow k(x - 3) = 2({x^2} - 3x + 2)$$<br><br>for x $$\ne$$ 3, $$k = 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right)$$<br><br>$$x - 3 + {2 \over {x - 3}} \ge... | integer | jee-main-2021-online-26th-august-morning-shift | 7,831 |
1ktg0otbe | maths | quadratic-equation-and-inequalities | nature-of-roots | The set of all values of K > $$-$$1, for which the equation $${(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$$ has real roots, is : | [{"identifier": "A", "content": "$$\\left( {1,{5 \\over 2}} \\right]$$"}, {"identifier": "B", "content": "[2, 3)"}, {"identifier": "C", "content": "$$\\left[ { - {1 \\over 2},1} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},{3 \\over 2}} \\right] - \\{ 1\\} $$"}] | ["A"] | null | $${(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$$<br><br>Let $$3{x^2} + 4x + 3 = a$$<br><br>and $$3{x^2} + 4x + 2 = b \Rightarrow b = a - 1$$<br><br>Given equation becomes<br><br>$$ \Rightarrow {a^2} - (k + 1)ab + k{b^2} = 0$$<br><br>$$ \Rightarrow a(a - kb) - b(a - kb) ... | mcq | jee-main-2021-online-27th-august-evening-shift | 7,832 |
1l567jtfa | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The number of real solutions of the equation $${e^{4x}} + 4{e^{3x}} - 58{e^{2x}} + 4{e^x} + 1 = 0$$ is ___________.</p> | [] | null | 2 | <p>Dividing by e<sup>2x</sup></p>
<p>$${e^{2x}} + 4{e^x} - 58 + 4{e^{ - x}} + {e^{ - 2x}} = 0$$</p>
<p>$$ \Rightarrow {({e^x} + {e^{ - x}})^2} + 4({e^x} + {e^{ - x}}) - 60 = 0$$</p>
<p>Let $${e^x} + {e^{ - x}} = t \in [2,\infty )$$</p>
<p>$$ \Rightarrow {t^2} + 4t - 60 = 0$$</p>
<p>$$ \Rightarrow t = 6$$ is only possib... | integer | jee-main-2022-online-28th-june-morning-shift | 7,834 |
1l57o4z2c | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The number of distinct real roots of x<sup>4</sup> $$-$$ 4x + 1 = 0 is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "0"}] | ["B"] | null | <p>$$f(x) = {x^4} - 4x + 1 = 0$$</p>
<p>$$f'(x) = 4{x^3} - 4$$</p>
<p>$$ = 4(x - 1)({x^2} + 1 + x)$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9sp3c/20ce78f3-fabd-4839-b609-84bc56bb070f/63f06880-0655-11ed-903e-c9687588b3f3/file-1l5q9sp3d.png?format=png" data-orsrc="https://app-content.... | mcq | jee-main-2022-online-27th-june-morning-shift | 7,835 |
1l58gxgm4 | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>Let p and q be two real numbers such that p + q = 3 and p<sup>4</sup> + q<sup>4</sup> = 369. Then $${\left( {{1 \over p} + {1 \over q}} \right)^{ - 2}}$$ is equal to _________.</p> | [] | null | 4 | <p>$$\because$$ $$p + q = 3$$ ...... (i)</p>
<p>and $${p^4} + {q^4} = 369$$ ...... (ii)</p>
<p>$${\{ {(p + q)^2} - 2pq\} ^2} - 2{p^2}{q^2} = 369$$</p>
<p>or $${(9 - 2pq)^2} - 2{(pq)^2} = 369$$</p>
<p>or $${(pq)^2} - 18pq - 144 = 0$$</p>
<p>$$\therefore$$ $$pq = - 6$$ or 24</p>
<p>But $$pq = 24$$ is not possible</p>
<p... | integer | jee-main-2022-online-26th-june-evening-shift | 7,836 |
1l5b7s8ay | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The sum of all the real roots of the equation <br/><br/>$$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$$ is</p> | [{"identifier": "A", "content": "$${\\log _e}3$$"}, {"identifier": "B", "content": "$$ - {\\log _e}3$$"}, {"identifier": "C", "content": "$${\\log _e}6$$"}, {"identifier": "D", "content": "$$ - {\\log _e}6$$"}] | ["B"] | null | <p>$$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$$</p>
<p>Let $${e^x} = t$$</p>
<p>$$\therefore$$ $$({t^2} - 4)(6{t^2} - 5t + 1) = 0$$</p>
<p>$$ \Rightarrow ({t^2} - 4)(2t - 1)(3t - 1) = 0$$</p>
<p>$$\therefore$$ t = 2, $$-$$2, $${1 \over 2}$$, $${1 \over 3}$$</p>
<p>$$\therefore$$ $${e^x} = 2 \Rightarrow x = \ln 2$$</p... | mcq | jee-main-2022-online-24th-june-evening-shift | 7,837 |
1l5vz59za | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>Let S be the set of all integral values of $$\alpha$$ for which the sum of squares of two real roots of the quadratic equation $$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$$ is minimum. Then S :</p> | [{"identifier": "A", "content": "is an empty set"}, {"identifier": "B", "content": "is a singleton"}, {"identifier": "C", "content": "contains exactly two elements"}, {"identifier": "D", "content": "contains more than two elements"}] | ["A"] | null | <p>Given quadratic equation,</p>
<p>$$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$$</p>
<p>Let, a and b are the roots of the equation,</p>
<p>$$\therefore$$ $$a + b = - {{\alpha - 6} \over 3}$$</p>
<p>and $$ab = {{\alpha + 3} \over 3}$$</p>
<p>For real roots,</p>
<p>$$D \ge 0$$</p>
<p>$$ \Rightarrow {(\alpha - 6)^2... | mcq | jee-main-2022-online-30th-june-morning-shift | 7,839 |
1l6m705hn | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The sum of all real values of $$x$$ for which $$\frac{3 x^{2}-9 x+17}{x^{2}+3 x+10}=\frac{5 x^{2}-7 x+19}{3 x^{2}+5 x+12}$$ is equal to __________.</p> | [] | null | 6 | <p>$${{3{x^2} - 9x + 17} \over {{x^2} + 3x + 10}} = {{5{x^2} - 7x + 19} \over {3{x^2} + 5x + 12}}$$</p>
<p>$$ \Rightarrow {{3{x^2} - 9x + 17} \over {5{x^2} - 7x + 19}} = {{{x^2} + 3x + 10} \over {3{x^2} + 5x + 12}}$$</p>
<p>$${{ - 2{x^2} - 2x - 2} \over {5{x^2} - 7x + 19}} = {{ - 2{x^2} - 2x - 2} \over {3{x^2} + 5x + 1... | integer | jee-main-2022-online-28th-july-morning-shift | 7,841 |
ldo7nrak | maths | quadratic-equation-and-inequalities | nature-of-roots | The equation $\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R}$ has : | [{"identifier": "A", "content": "two solutions and both are negative"}, {"identifier": "B", "content": "two solutions and only one of them is negative"}, {"identifier": "C", "content": "four solutions two of which are negative"}, {"identifier": "D", "content": "no solution"}] | ["A"] | null | $e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0$
<br/><br/>Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}$
<br/><br/>Now, $\mathrm{t}^{4}+8 \mathrm{t}^{3}+13 \mathrm{t}^{2}-8 \mathrm{t}+1=0$
<br/><br/>Dividing equation by $\mathrm{t}^{2}$
<br/><br/>$$
\begin{aligned}
& t^{2}+8 t+13-\frac{8}{t}+\frac{1}{t^{2}}=0 \\\\
& t^{2}+\fr... | mcq | jee-main-2023-online-31st-january-evening-shift | 7,842 |
1ldww04sg | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The number of real solutions of the equation $$3\left( {{x^2} + {1 \over {{x^2}}}} \right) - 2\left( {x + {1 \over x}} \right) + 5 = 0$$, is</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2"}] | ["C"] | null | $3\left(x^{2}+\frac{1}{x^{2}}\right)-2\left(x+\frac{1}{x}\right)+5=0$
<br/><br/>
$3\left[\left(x+\frac{1}{x}\right)^{2}-2\right]-2\left(x+\frac{1}{x}\right)+5=0$
<br/><br/>
Put $x+\frac{1}{x}=t \Rightarrow t \in(-\infty,-2] \cup[2, \infty)$
<br/><br/>
$$
\begin{aligned}
& 3 t^{2}-2 t-1=0 \\\\
& 3 t^{2}-3 t+t-1=0 \\\\
&... | mcq | jee-main-2023-online-24th-january-evening-shift | 7,844 |
1lgyonh22 | maths | quadratic-equation-and-inequalities | nature-of-roots | <p> Let m and $$\mathrm{n}$$ be the numbers of real roots of the quadratic equations $$x^{2}-12 x+[x]+31=0$$ and $$x^{2}-5|x+2|-4=0$$ respectively, where $$[x]$$ denotes the greatest integer $$\leq x$$. Then $$\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2}$$ is equal to __________.</p> | [] | null | 9 | The givne eqn is : $x^2-12 x+[x]+31=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow\{x\}-x=x^2-12 x+31 \\\\
& \Rightarrow\{x\}=x^2-11 x+31
\end{aligned}
$$
<br/><br/>So, $0 \leq x^2-11 x+31<1$
<br/><br/>$$
\begin{aligned}
& \Rightarrow x^2-11 x+30 \leq 0 \\\\
& \Rightarrow(x-5)(x-6)<0 \\\\
& \Rightarrow x \in(5,6) \\\\
&... | integer | jee-main-2023-online-8th-april-evening-shift | 7,845 |
p18BDSZMK3wfynvD | maths | quadratic-equation-and-inequalities | range-of-quadratic-expression | If $$x$$ is real, the maximum value of $${{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$ is | [{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$$41$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$${17 \\over 7}$$ "}] | ["B"] | null | $$y = {{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$
<br><br>$$3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0$$
<br><br>$$D \ge 0$$ as $$x$$ is real
<br><br>$$81{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \ge 0$$
<br><br>$$ \Rightarrow \left( {y - 1} \rig... | mcq | aieee-2006 | 7,846 |
cX7V3ZAuOS6Qua2u | maths | quadratic-equation-and-inequalities | range-of-quadratic-expression | The sum of all real values of $$x$$ satisfying the equation $${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}}\, = 1$$ is : | [{"identifier": "A", "content": "$$6$$"}, {"identifier": "B", "content": "$$5$$ "}, {"identifier": "C", "content": "$$3$$"}, {"identifier": "D", "content": "$$-4$$ "}] | ["C"] | null | Given equation,
$${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}} = 1$$
<br><br><b>Case 1 : </b>When x<sup>2</sup> - 5x + 5 = 1 and x<sup>2</sup> + 4x - 60 is any real no then this equation satisfy.
<br><b>Note :</b> When we put any real number as a power of 1 the value stays always 1 (1<sup> any real no</sup> = 1)... | mcq | jee-main-2016-offline | 7,847 |
jaoe38c1lsd52291 | maths | quadratic-equation-and-inequalities | range-of-quadratic-expression | <p>Let $$a, b, c$$ be the lengths of three sides of a triangle satistying the condition $$\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$$. If the set of all possible values of $$x$$ is the interval $$(\alpha, \beta)$$, then $$12\left(\alpha^2+\beta^2\right)$$ is equal to __________.</p> | [] | null | 36 | <p>$$\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow a^2 x^2-2 a b x+b^2+b^2 x^2-2 b c x+c^2=0 \\
& \Rightarrow(a x-b)^2+(b x-c)^2=0 \\
& \Rightarrow a x-b=0, \quad b x-c=0 \\
& \Rightarrow a+b>c \quad b+c>a \quad c+a>b
\end{aligned}$$</p>
<p>$$\begin{array}{l|l|l}
a+a x>b x & a x... | integer | jee-main-2024-online-31st-january-evening-shift | 7,849 |
3YsoBigo1uuZD4hJ | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Difference between the corresponding roots of $${x^2} + ax + b = 0$$ and $${x^2} + bx + a = 0$$ is same and $$a \ne b,$$ then | [{"identifier": "A", "content": "$$a + b + 4 = 0$$ "}, {"identifier": "B", "content": "$$a + b - 4 = 0$$ "}, {"identifier": "C", "content": "$$a - b - 4 = 0$$ "}, {"identifier": "D", "content": "$$a - b + 4 = 0$$ "}] | ["A"] | null | Let $$\alpha ,\beta $$ and $$\gamma ,\delta $$ be the roots of the equations $${x^2} + ax + b = 0$$
<br><br>and $${x^2} + bx + a = 0$$ respectively.
<br><br>$$\therefore$$ $$\alpha + \beta = - a,\alpha \beta = b$$
<br><br>and $$\gamma + \delta = - b,\gamma \delta = a.$$
<br><br>Given $$\left| {\alpha - \bet... | mcq | aieee-2002 | 7,850 |
WCMl8h2haO23S7FT | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$p$$ and $$q$$ are the roots of the equation $${x^2} + px + q = 0,$$ then | [{"identifier": "A", "content": "$$p = 1,\\,\\,q = - 2$$ "}, {"identifier": "B", "content": "$$p = 0,\\,\\,q = 1$$ "}, {"identifier": "C", "content": "$$p = - 2,\\,\\,q = 0$$ "}, {"identifier": "D", "content": "$$p = - 2,\\,\\,q = 1$$ "}] | ["A"] | null | $$p + q = - p$$ and $$pq = q \Rightarrow q\left( {p - 1} \right) = 0$$
<br><br>$$ \Rightarrow q = 0$$ or $$p=1.$$
<br><br>If $$q = 0,$$ then $$p=0.$$ i.e.$$p=q$$
<br><br>$$\therefore$$ $$p=1$$ and $$q=-2.$$ | mcq | aieee-2002 | 7,851 |
6tmOFTAoGjVrDmnk | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha \ne \beta $$ but $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3$$ then the equation having $$\alpha /\beta $$ and $$\beta /\alpha \,\,$$ as its roots is | [{"identifier": "A", "content": "$$3{x^2} - 19x + 3 = 0$$"}, {"identifier": "B", "content": "$$3{x^2} + 19x - 3 = 0$$ "}, {"identifier": "C", "content": "$$3{x^2} - 19x - 3 = 0$$ "}, {"identifier": "D", "content": "$${x^2} - 5x + 3 = 0$$ "}] | ["A"] | null | We have $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3;$$
<br><br>$$ \Rightarrow \alpha \,\,\& \,\,\beta $$ are roots of
<br><br>equation, $${x^2} = 5x - 3$$ or $${x^2} - 5x + 3 = 0$$
<br><br>$$\therefore$$ $$\alpha + \beta = 5$$ and $$\alpha \beta = 3$$
<br><br>Thus, the equation having $${\alp... | mcq | aieee-2002 | 7,852 |
QaXT2TPJqyZ9tKIE | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of '$$a$$' for which one root of the quadratic equation
$$$\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0$$$
<br/>is twice as large as the other is | [{"identifier": "A", "content": "$$ - {1 \\over 3}$$ "}, {"identifier": "B", "content": "$$ {2 \\over 3}$$"}, {"identifier": "C", "content": "$$ - {2 \\over 3}$$"}, {"identifier": "D", "content": "$$ {1 \\over 3}$$ "}] | ["B"] | null | Let the roots of given equation be $$\alpha $$ and $$2$$$$\alpha $$ then
<br><br>$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$
<br><br>and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$
<br><br>$$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$$
<br>... | mcq | aieee-2003 | 7,854 |
yQOuJuXRcD24ppfE | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation | [{"identifier": "A", "content": "$${x^2} - 18x - 16 = 0$$ "}, {"identifier": "B", "content": "$${x^2} - 18x + 16 = 0$$"}, {"identifier": "C", "content": "$${x^2} + 18x - 16 = 0$$"}, {"identifier": "D", "content": "$${x^2} + 18x + 16 = 0$$"}] | ["B"] | null | Let two numbers be a and b then $${{a + b} \over 2} = 9$$
<br><br>and $$\sqrt {ab} = 4$$
<br><br>$$\therefore$$ Equation with roots $$a$$ and $$b$$ is
<br><br>$${x^2} - \left( {a + b} \right)x + ab = 0$$
<br><br>$$ \Rightarrow {x^2} - 18x + 16 = 0$$ | mcq | aieee-2004 | 7,855 |
vg7MAc99xWYqkfya | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | In a triangle $$PQR,\;\;\angle R = {\pi \over 2}.\,\,If\,\,\tan \,\left( {{P \over 2}} \right)$$ and $$ \tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0,\,\,a \ne 0$$ then | [{"identifier": "A", "content": "$$a = b + c$$ "}, {"identifier": "B", "content": "$$c = a + b$$ "}, {"identifier": "C", "content": "$$b = c$$ "}, {"identifier": "D", "content": "$$b = a + c$$ "}] | ["B"] | null | $$\angle $$R = 90<sup>o</sup> $$ \therefore $$ $$\angle $$P + $$\angle $$Q = 90<sup>o</sup>
<br><br>$$ \Rightarrow $$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$<sup>o</sup>
<br><br>$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0$$
<br><br>$$ \there... | mcq | aieee-2005 | 7,857 |
j6vQurOyjyj7GX1T | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$a$$ for which the sum of the squares of the roots of the equation $${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$
assume the least value is : | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["A"] | null | $${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$
<br><br>$$ \Rightarrow \alpha + \beta = a - 2;\,\,\alpha \beta = - \left( {a + 1} \right)$$
<br><br>$${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$
<br><br>$$ = {a^2} - 2a + 6 = {\left( {a - 1} \right)^2} + 5$$
<br><br>For min. va... | mcq | aieee-2005 | 7,858 |
fSshcquXjrbG2JCO | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$a$$ for which the sum of the squares of the roots of the equation <br/>$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$ assume the least value is | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["A"] | null | Given quadratic equation,
<br><br>$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$
<br><br>Let $$\alpha $$ and $$\beta $$ are the roots of the equation.
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$a - 2$$
<br><br>and $$\alpha $$$$\beta $$ = $$ - a - 1$$
<br><br>Now $${\alpha ^2} + {\beta ^2} = {\left( {\alpha ... | mcq | aieee-2005 | 7,859 |
JPyMtvr5Uy9w4a1n | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If the roots of the quadratic equation $${x^2} + px + q = 0$$ are $$\tan {30^ \circ }$$ and $$\tan {15^ \circ }$$, respectively, then the value of $$2 + q - p$$ is | [{"identifier": "A", "content": "2 "}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "0 "}, {"identifier": "D", "content": "1 "}] | ["B"] | null | $${x^2} + px + q = 0$$
<br><br>Sum of roots $$ = \tan {30^ \circ } + \tan {15^ \circ } = - p$$
<br><br>Products of roots $$ = \tan {30^ \circ }.\tan {15^ \circ } = q$$
<br><br>$$\tan {45^ \circ } = {{\tan {{30}^ \circ } + \tan {{15}^ \circ }} \over {1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}}$$
<br><br>$$\,\,\,\,\,... | mcq | aieee-2006 | 7,860 |
TpLwVDBYA4fxcyBw | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If the difference between the roots of the equation $${x^2} + ax + 1 = 0$$ is less than $$\sqrt 5 ,$$ then the set of possible values of $$a$$ is | [{"identifier": "A", "content": "$$\\left( {3,\\infty } \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - \\infty , - 3} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { - 3,3} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( { - 3,\\infty } \\right)$$ "}] | ["C"] | null | Let $$\alpha $$ and $$\beta $$ are roots of the equation $${x^2} + ax + 1 = 0$$
<br><br>So, $$\alpha + \beta = - a$$ and $$\alpha \beta = 1$$
<br><br>given $$\left| {\alpha - \beta } \right| < \sqrt 5 $$
<br><br>$$ \Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5 $$
<br... | mcq | aieee-2007 | 7,861 |
RL48R3LIbXTYoq2f | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be the roots of equation $${x^2} - 6x - 2 = 0$$. If $${a_n} = {\alpha ^n} - {\beta ^n},$$ for $$n \ge 1,$$ then the value of $${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$ is equal to : | [{"identifier": "A", "content": "$$3$$"}, {"identifier": "B", "content": "$$ - 3$$ "}, {"identifier": "C", "content": "$$6$$ "}, {"identifier": "D", "content": "$$ - 6$$ "}] | ["A"] | null | Given equation, x<sup>2</sup> - 6x - 2 = 0
<br><br> Roots are $$\alpha $$ and $$\beta $$.
<br><br>So, $$\alpha + \beta = 6$$ and $$\alpha \beta = - 2$$
<br><br>In the question given, $${a_n} = {\alpha ^n} - {\beta ^n}$$
<br><br>$$\therefore$$ $${a_8} = {\alpha ^8} - {\beta ^8}$$
<br><br>and $${a_9} = {\alpha ^9} - ... | mcq | jee-main-2015-offline | 7,864 |
0unmlbvFhbWLTnKlsl7ae | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If x is a solution of the equation, $$\sqrt {2x + 1} $$ $$ - \sqrt {2x - 1} = 1,$$ $$\,\,\left( {x \ge {1 \over 2}} \right),$$ then $$\sqrt {4{x^2} - 1} $$ is equal to : | [{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$2\\sqrt 2 $$"}] | ["A"] | null | Given,
<br><br>$$\sqrt {2x + 1} - \sqrt {2x - 1} = 1$$
<br><br>$$ \Rightarrow $$ $$\sqrt {2x + 1} = 1 + \sqrt {2x - 1} $$
<br><br>Squaring both sides, we get
<br><br>2x + 1 $$=$$ 1 + 2x $$-$$ 1 + 2$$\sqrt {2x - 1} $$
<br><br>$$ \Rightarrow $$ 1 $$=$$ 2$$\sqrt {2x - 1} $$
<br><br>$$... | mcq | jee-main-2016-online-10th-april-morning-slot | 7,865 |
7IrvwEVArDgbLPCv | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If for a positive integer n, the quadratic equation
<br/><br/>$$x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right)$$$$ + .... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right)$$$$ = 10n$$
<br/><br/>has two consecutive integral solutions, then n is equal to : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "12"}] | ["C"] | null | $$\sum\limits_{r = 1}^n {\left( {x + r - 1} \right)\left( {x + r} \right)} = 10n$$
<br><br>$$ \Rightarrow $$ $$\sum\limits_{r = 1}^n {\left( {{x^2} + xr + \left( {r - 1} \right)x + {r^2} - r} \right)} = 10n$$
<br><br>$$ \Rightarrow $$ $$\sum\limits_{r = 1}^n {\left( {{x^2} + \left( {2r - 1} \right)x + r\left( {r - 1}... | mcq | jee-main-2017-offline | 7,866 |
0pxENP4lfSOmWKuYrR5Mm | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$$-$$ 1 and it leaves remainder 6 when divided by x + 1; then : | [{"identifier": "A", "content": "p(2) = 11"}, {"identifier": "B", "content": "p(2) = 19"}, {"identifier": "C", "content": "p($$-$$ 2) = 19"}, {"identifier": "D", "content": "p($$-$$ 2) = 11"}] | ["C"] | null | Let, P(x) = ax<sup>2</sup> + bx + c
<br><br>As, P(0) = 1,
<br><br>$$\therefore\,\,\,$$ a(0)<sup>2</sup> + b(0) + c = 1
<br><br>$$ \Rightarrow $$$$\,\,\,$$ c = 1
<br><br>$$\therefore\,\,\,$$ P(x) = ax<sup>2</sup> + bx + 1
<br><br>If P(x) is divided by x $$-$$ 1, remainder = 4
<br><br>$$ \Rightarrow $$$$\,\,... | mcq | jee-main-2017-online-8th-april-morning-slot | 7,867 |
XAqhFZbkmjsfKVvXYMvuA | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If tanA and tanB are the roots of the quadratic equation, 3x<sup>2</sup> $$-$$ 10x $$-$$ 25 = 0, then the value of 3 sin<sup>2</sup>(A + B) $$-$$ 10 sin(A + B).cos(A + B) $$-$$ 25 cos<sup>2</sup>(A + B) is : | [{"identifier": "A", "content": "$$-$$ 10"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "$$-$$ 25"}, {"identifier": "D", "content": "25"}] | ["C"] | null | As tan A and tan B are the roots of 3x<sup>2</sup> $$-$$ 10x $$-$$ 25 = 0,
<br><br>So, tan(A + B) = $${{\tan A + \tan B} \over {1 - \tan A\tan B}}$$
<br><br>= $${{{{10} \over 3}} \over {1 + {{25} \over 3}}}$$ = $${{10/3} \over {28/3}}$$ = $${5 \over {14}}$$
<br><br>Now, cos<sup>2</sup> (A + B) = $$-$$ 1 + 2 cos<sup>2... | mcq | jee-main-2018-online-15th-april-morning-slot | 7,868 |
DCEEzNPGePYIxhKKzY7yv | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If an angle A of a $$\Delta $$ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x<sup>2</sup> + 27x + 20 = 0 are : | [{"identifier": "A", "content": "secA, cotA"}, {"identifier": "B", "content": "sinA, secA"}, {"identifier": "C", "content": "secA, tanA"}, {"identifier": "D", "content": "tanA, cosA"}] | ["C"] | null | Here, 9x<sup>2</sup> + 27x + 20 = 0
<br><br>$$\therefore\,\,\,$$ x = $${{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x = $${{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x = $$-$$ $${4 \over 3}$$, $$-$$ $${5 \over 3}$$... | mcq | jee-main-2018-online-16th-april-morning-slot | 7,869 |
Vi5B6NQIKaeP2PUAHr7rQ | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let p, q and r be real numbers (p $$ \ne $$ q, r $$ \ne $$ 0), such that the roots of the equation $${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to : | [{"identifier": "A", "content": "$${{{p^2} + {q^2}} \\over 2}$$"}, {"identifier": "B", "content": "p<sup>2</sup> + q<sup>2</sup> "}, {"identifier": "C", "content": "2(p<sup>2</sup> + q<sup>2</sup>)"}, {"identifier": "D", "content": "p<sup>2</sup> + q<sup>2</sup> + r<sup>2</sup>"}] | ["B"] | null | Given,
<br><br>$${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{x + p + x + q} \over {\left( {x + p} \right)\left( {x + q} \right)}} = {1 \over r}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ (2x + p + q) r = x<sup>2</sup> + px + qx + pq
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x<sup... | mcq | jee-main-2018-online-16th-april-morning-slot | 7,870 |
PVuqvBx83lZ1nES9eOeNX | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If f(x) is a quadratic expression such that f (1) + f (2) = 0, and $$-$$ 1 is a root of f (x) = 0, then the other root of f(x) = 0 is : | [{"identifier": "A", "content": "$$-$$ $${5 \\over 8}$$"}, {"identifier": "B", "content": "$$-$$ $${8 \\over 5}$$"}, {"identifier": "C", "content": "$${5 \\over 8}$$"}, {"identifier": "D", "content": "$${8 \\over 5}$$"}] | ["D"] | null | Let $$\alpha $$ and $$\beta $$ = - 1 are the roots of the polynomial, then we get
<br><br>f(x) = x<sup>2</sup> + (1 - $$\alpha $$)x - $$\alpha $$
<br><br>$$ \therefore $$ f(1) = 2 - 2$$\alpha $$
<br><br>and f(2) = 6 - 3$$\alpha $$
<br><br>Also given,
<br><br> f (1) + f (2) = 0
<br><br>$$ \therefore $$ 2 - 2$$\alpha $$ ... | mcq | jee-main-2018-online-15th-april-evening-slot | 7,871 |
nmugPDXgqKsFPiI0tLcu4 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\lambda $$ $$ \in $$ <b>R</b> is such that the sum of the cubes of the roots of the equation,
<br/>x<sup>2</sup> + (2 $$-$$ $$\lambda $$) x + (10 $$-$$ $$\lambda $$) = 0 is minimum, then the magnitude of the difference of the roots of this equation is : | [{"identifier": "A", "content": "$$4\\sqrt 2 $$"}, {"identifier": "B", "content": "$$2\\sqrt 5 $$"}, {"identifier": "C", "content": "$$2\\sqrt 7 $$"}, {"identifier": "D", "content": "20"}] | ["B"] | null | Let $$\alpha $$, $$\beta $$ are the roots of the equation,
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 2 and $$\alpha $$$$\beta $$ = 10 $$-$$ $$\lambda $$
<br><br>$${\alpha ^3} + {\beta ^3}$$ = ($$\alpha $$ + $$\beta $$)<sup>3</sup> $$-$$ 3$$\alpha $$$$\beta $$ ($$\alpha $$ + $$\beta $$)
<br... | mcq | jee-main-2018-online-15th-april-morning-slot | 7,872 |
Z0kE4u5jpksipfnZxwdCg | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$\lambda $$ such that sum of the squares of the roots of the quadratic equation, x<sup>2</sup> + (3 – $$\lambda $$)x + 2 = $$\lambda $$ has the least value is - | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${{15} \\over 8}$$"}, {"identifier": "D", "content": "$${4 \\over 9}$$"}] | ["B"] | null | $$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 3
<br><br>$$\alpha $$$$\beta $$ = 2 $$-$$ $$\lambda $$
<br><br>$$\alpha $$<sup>2</sup> + $$\beta $$<sup>2</sup> = ($$\alpha $$ + $$\beta $$)<sup>2</sup> $$-$$ 2$$\alpha $$$$\beta $$ = ($$\lambda $$ $$-$$ 3)<sup>2</sup> $$-$$ 2$$\left( {2 - \lambda } \right)$$
<br><br>= $$\l... | mcq | jee-main-2019-online-10th-january-evening-slot | 7,873 |
cSCA6jlUnOtHa6FTlPLZK | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If one real root of the quadratic equation 81x<sup>2</sup> + kx + 256 = 0 is cube of the other root, then a value of k is | [{"identifier": "A", "content": "$$-$$ 81"}, {"identifier": "B", "content": "$$-$$ 300"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "144"}] | ["B"] | null | 81x<sup>2</sup> + kx + 256 = 0 ; x = $$\alpha $$, $$\alpha $$<sup>3</sup>
<br><br>$$ \Rightarrow $$ $$\alpha $$<sup>4</sup> = $${{256} \over {81}}$$ $$ \Rightarrow $$ $$\alpha $$ = $$ \pm $$ $${{4} \over {3}}$$
<br><br>Now $$-$$ $${k \over {81}}$$ = $$\alpha $$ + $$\alpha $$<sup>3<... | mcq | jee-main-2019-online-11th-january-morning-slot | 7,874 |
UrxNTlitqqL3D9jwUZty7 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be the roots of the quadratic equation x<sup>2</sup>
sin $$\theta $$ – x(sin $$\theta $$ cos $$\theta $$ + 1) + cos $$\theta $$ = 0 (0 < $$\theta $$ < 45<sup>o</sup>), and $$\alpha $$ < $$\beta $$. Then $$\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{{{\left( { - 1} \righ... | [{"identifier": "A", "content": "$${1 \\over {1 + \\cos \\theta }} + {1 \\over {1 - \\sin \\theta }}$$"}, {"identifier": "B", "content": "$${1 \\over {1 - \\cos \\theta }} + {1 \\over {1 + \\sin \\theta }}$$"}, {"identifier": "C", "content": "$${1 \\over {1 - \\cos \\theta }} - {1 \\over {1 + \\sin \\theta }}$$"}, {"id... | ["B"] | null | D = (1 + sin$$\theta $$ cos$$\theta $$)<sup>2</sup> $$-$$ 4sin$$\theta $$cos$$\theta $$ = (1 $$-$$ sin$$\theta $$ cos$$\theta $$)<sup>2</sup>
<br><br>$$ \Rightarrow $$ roots are $$\beta $$ = cosec$$\theta $$ and $$\alpha $$ = cos$$\theta $$
<br><br>$$\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{\left... | mcq | jee-main-2019-online-11th-january-evening-slot | 7,875 |
YIRrhh126EC0R8efYDcTv | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\lambda $$ be the ratio of the roots of the quadratic equation in x, 3m<sup>2</sup>x<sup>2</sup> + m(m – 4)x + 2 = 0, then the least value of m for which $$\lambda + {1 \over \lambda } = 1,$$ is | [{"identifier": "A", "content": "$$ - 2 + \\sqrt 2 $$"}, {"identifier": "B", "content": "4$$-$$3$$\\sqrt 2 $$"}, {"identifier": "C", "content": "2 $$-$$ $$\\sqrt 3 $$"}, {"identifier": "D", "content": "4 $$-$$ 2$$\\sqrt 3 $$"}] | ["B"] | null | 3m<sup>2</sup>x<sup>2</sup> + m(m $$-$$ 4) x + 2 = 0
<br><br>$$\lambda + {1 \over \lambda } = 1,{\alpha \over \beta } + {\beta \over \alpha } = 1,{\alpha ^2} + {\beta ^2} = \alpha \beta $$
<br><br>($$\alpha $$ + $$\beta $$)<sup>2</sup> = 3$$\alpha $$$$\beta $$
<br><br>$${\left( { - {{m\left( {m - 4} \right)} \over {... | mcq | jee-main-2019-online-12th-january-morning-slot | 7,876 |
ASH34YkiuFpNO2zeGRoYa | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let p, q $$ \in $$ R. If 2 - $$\sqrt 3$$ is a root of the quadratic
equation, x<sup>2</sup> + px + q = 0, then : | [{"identifier": "A", "content": "p<sup>2</sup> \u2013 4q \u2013 12 = 0"}, {"identifier": "B", "content": "q<sup>2</sup> \u2013 4p \u2013 16 = 0"}, {"identifier": "C", "content": "q<sup>2</sup> + 4p + 14 = 0"}, {"identifier": "D", "content": "p<sup>2</sup> \u2013 4q + 12 = 0"}] | ["A"] | null | If a quadratic equation with rational coefficient has one irrational root then other root will be the conjugate of the irrational root.
<br><br>Here x<sup>2</sup> + px + q = 0 has one root 2 - $$\sqrt 3$$.
<br><br>$$ \therefore $$ Other root will be 2 + $$\sqrt 3$$.
<br><br>Sum of the roots = -p = 4
<br><br>and product... | mcq | jee-main-2019-online-9th-april-morning-slot | 7,877 |
W9EBjBBLvv4KN4kY2B3rsa0w2w9jwxdqfnm | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ are the roots of the quadratic equation,
<br/>x<sup>2</sup> + x sin $$\theta $$ - 2 sin $$\theta $$ = 0, $$\theta \in \left( {0,{\pi \over 2}} \right)$$, then
<br/>$${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta ... | [{"identifier": "A", "content": "$${{{2^{12}}} \\over {{{\\left( {\\sin \\theta - 8} \\right)}^6}}}$$"}, {"identifier": "B", "content": "$${{{2^6}} \\over {{{\\left( {\\sin \\theta + 4} \\right)}^{12}}}}$$"}, {"identifier": "C", "content": "$${{{2^{12}}} \\over {{{\\left( {\\sin \\theta + 8} \\right)}^{12}}}}$$"}, {... | ["C"] | null | Given $$\alpha + \beta = - \sin \theta $$ and$$\alpha \beta = - 2\sin \theta $$<br><br>
$${{\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){\alpha ^{12}}{\beta ^{12}}} \over {\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {\alpha \beta } \right)}^{12}}} \ov... | mcq | jee-main-2019-online-10th-april-morning-slot | 7,878 |
fpfAhxHaFIjNWVXwdijgy2xukfw0t2zw | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ be two roots of the equation<br/> x<sup>2</sup> – 64x + 256 = 0. Then the value of
<br/>$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$ is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["C"] | null | x<sup>2</sup> – 64x + 256 = 0
<br><br>$$\alpha $$ + $$\beta $$ = 64, $$\alpha $$$$\beta $$ = 256
<br><br>$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$
<br><br>= $${{{\alpha ^{{3 \over 8}}}} \over {{\beta ^{{5 \over 8}}}}} + {{{\beta ^{{3 \o... | mcq | jee-main-2020-online-6th-september-morning-slot | 7,879 |
2VBumVZIDcyB2i8HAGjgy2xukg0cjhh6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ are the roots of the equation
<br/>2x(2x + 1) = 1, then $$\beta $$ is equal to :
| [{"identifier": "A", "content": "$$ - 2\\alpha \\left( {\\alpha + 1} \\right)$$"}, {"identifier": "B", "content": "$$ 2\\alpha \\left( {\\alpha + 1} \\right)$$"}, {"identifier": "C", "content": "$$2{\\alpha ^2}$$"}, {"identifier": "D", "content": "$$ 2\\alpha \\left( {\\alpha - 1} \\right)$$"}] | ["A"] | null | $$\alpha $$ and
$$\beta $$ are the roots of the equation
<br>4x<sup>2</sup> + 2x – 1 = 0.
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$ - {1 \over 2}$$
<br><br>$$ \Rightarrow $$ -1 = 2$$\alpha $$ + 2$$\beta $$
<br><br>and 4$$\alpha $$<sup>2</sup> + 2$$\alpha $$ - 1 = 0
<br><br>$$ \Rightarrow $$ 4$$\alpha $$<su... | mcq | jee-main-2020-online-6th-september-evening-slot | 7,880 |
ADtDSoqLJjuj5nPt5Zjgy2xukfqcls6x | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ are the roots of the equation,
<br/>7x<sup>2</sup> – 3x – 2 = 0, then the value of
<br/>$${\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {24}}$$"}, {"identifier": "B", "content": "$${{27} \\over {32}}$$"}, {"identifier": "C", "content": "$${{27} \\over {16}}$$"}, {"identifier": "D", "content": "$${3 \\over 8}$$"}] | ["C"] | null | Given, 7x<sup>2</sup> – 3x – 2 = 0
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $${3 \over 7}$$
<br><br>$$\alpha $$$$\beta $$ = - $${2 \over 7}$$
<br><br>$${\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}$$
<br><br>= $${{\alpha + \beta - \alpha \beta \left( {\alpha + \beta } \right)} \over ... | mcq | jee-main-2020-online-5th-september-evening-slot | 7,881 |
PAZAL0qE1kEzLNFYmvjgy2xukfak9s9b | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\lambda \ne 0$$ be in R. If $$\alpha $$ and $$\beta $$ are the roots of the <br/>equation, x<sup>2</sup> - x + 2$$\lambda $$ = 0 and $$\alpha $$ and $$\gamma $$ are the roots of <br/>the equation, $$3{x^2} - 10x + 27\lambda = 0$$, then $${{\beta \gamma } \over \lambda }$$ is equal to:
| [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "27"}, {"identifier": "D", "content": "18"}] | ["D"] | null | $$\alpha $$ and $$\beta $$ are the roots of the <br>equation x<sup>2</sup> - x + 2$$\lambda $$ = 0 .....(1)
<br><br>$$ \therefore $$ $$\alpha + \beta = 1,\,\alpha \beta = 2\lambda $$
<br><br>$$\alpha $$ and $$\gamma $$ are the roots of <br>the equation, $$3{x^2} - 10x + 27\lambda = 0$$ ......(2)
<br><br>$$ \therefo... | mcq | jee-main-2020-online-4th-september-evening-slot | 7,882 |
0EWeqQcicU7tlKL9gijgy2xukf8z77mb | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be the roots of x<sup>2</sup> - 3x + p=0 and $$\gamma $$ and $$\delta $$ be the roots of x<sup>2</sup> - 6x + q = 0. If $$\alpha, \beta, \gamma, \delta $$
form a geometric progression.Then ratio (2q + p) : (2q - p) is: | [{"identifier": "A", "content": "9 : 7"}, {"identifier": "B", "content": "5 : 3\n"}, {"identifier": "C", "content": "3 : 1 "}, {"identifier": "D", "content": "33 :31 "}] | ["A"] | null | $$\alpha $$ and $$\beta $$ are the roots of x<sup>2</sup> $$-$$ 3x + p = 0<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = 3 and $$\alpha \beta $$ = p<br><br>$$\gamma $$ and $$\delta $$ are the roots of x<sup>2</sup> $$-$$ 6x + q = 0<br><br>$$ \therefore $$ $$\gamma $$ + $$\delta $$ = 6 and $$\gamma \delta $$ = q<br... | mcq | jee-main-2020-online-4th-september-morning-slot | 7,883 |
vQfn0SPLSnaGH8uZOkjgy2xukewro3a6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let
$$\alpha $$ and
$$\beta $$ be the roots of the equation
<br/>5x<sup>2</sup> + 6x – 2 = 0. If S<sub>n</sub>
=
$$\alpha $$<sup>n</sup> +
$$\beta $$<sup>n</sup>, n = 1, 2, 3....,
then : | [{"identifier": "A", "content": "5S<sub>6</sub>\n + 6S<sub>5</sub>\n = 2S<sub>4</sub>"}, {"identifier": "B", "content": "5S<sub>6</sub>\n + 6S<sub>5</sub>\n + 2S<sub>4</sub> = 0"}, {"identifier": "C", "content": "6S<sub>6</sub>\n + 5S<sub>5</sub>\n + 2S<sub>4</sub> = 0"}, {"identifier": "D", "content": "6S<sub>6</sub>\... | ["A"] | null | $$\alpha $$ and
$$\beta $$ be the roots of the equation
<br>5x<sup>2</sup> + 6x – 2 = 0.
<br><br>$$ \Rightarrow $$ 5$$\alpha $$<sup>2</sup> + 6$$\alpha $$ - 2 = 0
<br><br>$$ \Rightarrow $$ 5$$\alpha $$<sup>n + 2</sup> + 6$$\alpha $$<sup>n + 2</sup> - 2$$\alpha $$<sup>n</sup> = 0 ......(1)
<br><br>(By multiplying $$\al... | mcq | jee-main-2020-online-2nd-september-morning-slot | 7,884 |
8XBpeSuAbU3N3i9RH5jgy2xukf0x1fao | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ are the roots of the equation
<br/>x<sup>2</sup>
+ px + 2 = 0 and $${1 \over \alpha }$$ and $${1 \over \beta }$$ are the<br/> roots of
the equation 2x<sup>2</sup>
+ 2qx + 1 = 0, then
<br/>$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alp... | [{"identifier": "A", "content": "$${9 \\over 4}\\left( {9 - {q^2}} \\right)$$"}, {"identifier": "B", "content": "$${9 \\over 4}\\left( {9 + {q^2}} \\right)$$"}, {"identifier": "C", "content": "$${9 \\over 4}\\left( {9 - {p^2}} \\right)$$"}, {"identifier": "D", "content": "$${9 \\over 4}\\left( {9 + {p^2}} \\right)$$"}] | ["C"] | null | $$\alpha $$ and $$\beta $$ are the roots of the <br><br>equation
x<sup>2</sup>
+ px + 2 = 0
<br><br>$$ \therefore $$ $$\alpha + \beta = - p,\,\alpha \beta = 2$$
<br><br>$${1 \over \alpha }$$ and $${1 \over \beta }$$ are the roots of
the <br><br>equation 2x<sup>2</sup>
+ 2qx + 1 = 0
<br><br>$$ \therefore $$ $${1 \... | mcq | jee-main-2020-online-3rd-september-morning-slot | 7,885 |
B9WR7JawEEY279wmzL7k9k2k5e2h7g2 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be two real roots of the equation <br/>(k + 1)tan<sup>2</sup>x - $$\sqrt 2 $$ . $$\lambda $$tanx = (1 - k), where k($$ \ne $$ - 1)
and $$\lambda $$ are real numbers. if tan<sup>2</sup> ($$\alpha $$ + $$\beta $$) = 50, then a value of $$\lambda $$ is:
| [{"identifier": "A", "content": "5$$\\sqrt 2 $$"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "10$$\\sqrt 2 $$"}] | ["B"] | null | Let tan$$\alpha $$ and tan$$\beta $$ are the roots of
<br><br>(k + 1)tan<sup>2</sup>x - $$\sqrt 2 $$ . $$\lambda $$tanx - (1 - k) = 0
<br><br>$$ \therefore $$ tan$$\alpha $$ + tan$$\beta $$ = $${{\sqrt 2 \lambda } \over {k + 1}}$$
<br><br>and an$$\alpha $$.tan$$\beta $$ = $${{k - 1} \over {k + 1}}$$
<br><br>Now tan($$\... | mcq | jee-main-2020-online-7th-january-morning-slot | 7,886 |
oXLFdyeUzYjCQopsqF7k9k2k5hk5f09 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$. <br/>If $$a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $$ and<br/> $$b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}} $$, then a and b are the roots of the quadratic equation : | [{"identifier": "A", "content": "x<sup>2</sup> + 101x + 100 = 0"}, {"identifier": "B", "content": "x<sup>2</sup> + 102x + 101 = 0"}, {"identifier": "C", "content": "x<sup>2</sup> \u2013 102x + 101 = 0"}, {"identifier": "D", "content": "x<sup>2</sup> \u2013 101x + 100 = 0"}] | ["C"] | null | $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$ = $$\omega $$
<br><br>$$a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $$
<br><br>= $$\left( {1 + \omega } \right)\sum\limits_{k = 0}^{100} {{\omega ^{2k}}} $$
<br><br>= $$\left( {1 + \omega } \right){{1\left( {1 - {{\left( {{\omega ^2}} \right)}^{1... | mcq | jee-main-2020-online-8th-january-evening-slot | 7,887 |
jDnLrPSQv7zEMajD7o7k9k2k5flw7xw | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be the roots of the equation x<sup>2</sup>
- x - 1 = 0. <br/>If p<sub>k</sub>
= $${\left( \alpha \right)^k} + {\left( \beta \right)^k}$$
, k $$ \ge $$ 1, then which one
of the following statements is not true? | [{"identifier": "A", "content": "(p<sub>1</sub> + p<sub>2</sub> + p<sub>3</sub> + p<sub>4</sub> + p<sub>5</sub>) = 26"}, {"identifier": "B", "content": "p<sub>5</sub> = 11"}, {"identifier": "C", "content": "p<sub>3</sub> = p<sub>5</sub> \u2013 p<sub>4</sub>"}, {"identifier": "D", "content": "p<sub>5</sub> = p<sub>2</su... | ["D"] | null | x<sup>2</sup>
- x - 1 = 0
<br><br>$$ \therefore $$ $$\alpha $$<sup>2</sup>
- $$\alpha $$ - 1 = 0
<br><br>$$ \Rightarrow $$ $$\alpha $$<sup>2</sup> = $$\alpha $$ + 1
<br><br>$$ \therefore $$ $$\alpha $$<sup>3</sup> = $$\alpha $$<sup>2</sup> + $$\alpha $$
<br><br>= $$\alpha $$ + 1 + $$\alpha $$
<br><br>= 2$$\alpha $$ +... | mcq | jee-main-2020-online-7th-january-evening-slot | 7,888 |
qlRl2Oq6OSlxdJ5RT61klt7pumo | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha$$ and $$\beta$$ be the roots of x<sup>2</sup> $$-$$ 6x $$-$$ 2 = 0. If a<sub>n</sub> = $$\alpha$$<sup>n</sup> $$-$$ $$\beta$$<sup>n</sup> for n $$ \ge $$ 1, then the value of $${{{a_{10}} - 2{a_8}} \over {3{a_9}}}$$ is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "1"}] | ["B"] | null | Given, $$\alpha$$ and $$\beta$$ be the roots of $${x^2} - 6x - 2 = 0$$<br><br>$$\matrix{
{\alpha + \beta = 6} \cr
{\alpha \beta = - 2} \cr
} $$<br><br>and $${\alpha ^2} - 6\alpha - 2 = 0 \Rightarrow {\alpha ^2} - 2 = 6\alpha $$<br><br>$${\beta ^2} - 6\beta - 2 = 0 \Rightarrow {\beta ^2} - 2 = 6\beta $... | mcq | jee-main-2021-online-25th-february-evening-slot | 7,889 |
APggZAnowmhcpBCqH71kluy92o5 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha$$ and $$\beta$$ be two real numbers such that $$\alpha$$ + $$\beta$$ = 1 and $$\alpha$$$$\beta$$ = $$-$$1. Let p<sub>n</sub> = ($$\alpha$$)<sup>n</sup> + ($$\beta$$)<sup>n</sup>, p<sub>n$$-$$1</sub> = 11 and p<sub>n+1</sub> = 29 for some integer n $$ \ge $$ 1. Then, the value of p$$_n^2$$ is ___________. | [] | null | 324 | Given, $$\alpha$$ + $$\beta$$ = 1, $$\alpha$$$$\beta$$ = $$-$$ 1<br><br>$$ \therefore $$ Quadratic equation with roots $$\alpha$$, $$\beta$$ is x<sup>2</sup> $$-$$ x $$-$$ 1 = 0<br><br>$$ \Rightarrow $$ $$\alpha$$<sup>2</sup> = $$\alpha$$ + 1<br><br>Multiplying both sides by $$\alpha$$<sup>n$$-$$1</sup><br><br>$$\alpha... | integer | jee-main-2021-online-26th-february-evening-slot | 7,890 |
5gwO0y8VqkWaKc4sGK1kmjavirx | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$4 + {1 \over {5 + {1 \over {4 + {1 \over {5 + {1 \over {4 + ......\infty }}}}}}}}$$ is : | [{"identifier": "A", "content": "2 + $${2 \\over 5}\\sqrt {30} $$"}, {"identifier": "B", "content": "2 + $${4 \\over {\\sqrt 5 }}\\sqrt {30} $$"}, {"identifier": "C", "content": "5 + $${2 \\over 5}\\sqrt {30} $$"}, {"identifier": "D", "content": "4 + $${4 \\over {\\sqrt 5 }}\\sqrt {30} $$"}] | ["A"] | null | $$y = 4 + {1 \over {5 + {1 \over y}}}$$<br><br>$$ \Rightarrow y = 4 + {y \over {5y + 1}}$$<br><br>$$ \Rightarrow 5{y^2} - 20y - 4 = 0$$<br><br>$$ \Rightarrow y = {{20 \pm \sqrt {400 + 80} } \over {10}}$$<br><br>$$ \Rightarrow y = {{20 \pm 4\sqrt {30} } \over {10}},y > 0$$<br><br>$$ \therefore $$ $$y = {{10 + 2\sqrt ... | mcq | jee-main-2021-online-17th-march-morning-shift | 7,891 |
Y7qXRAQT5BMwMqf4Iw1kmlj8n1j | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}$$ is equal to | [{"identifier": "A", "content": "1.5 + $$\\sqrt 3 $$"}, {"identifier": "B", "content": "2 + $$\\sqrt 3 $$"}, {"identifier": "C", "content": "3 + 2$$\\sqrt 3 $$"}, {"identifier": "D", "content": "4 + $$\\sqrt 3 $$"}] | ["A"] | null | Let $$x = 3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}$$<br><br>So, $$x = 3 + {1 \over {4 + {1 \over x}}} = 3 + {1 \over {{{4x + 1} \over x}}}$$<br><br>$$ \Rightarrow (x - 3) = {x \over {(4x + 1)}}$$<br><br>$$ \Rightarrow (4x + 1)(x - 3) = x$$<br><br>$$ \Rightarrow 4{x^2} - 12x + x - ... | mcq | jee-main-2021-online-18th-march-morning-shift | 7,892 |
1krw3bqey | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\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)$$ is eq... | [] | null | 1 | $${x^2} + 5\sqrt 2 x + 10 = 0$$<br><br>& $${P_n} = {\alpha ^n} - {\beta ^n}$$ (Given)<br><br>Now, $${{{{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}^{})}}}$$<br><br>$${{{P_{17}}... | integer | jee-main-2021-online-25th-july-morning-shift | 7,894 |
1l55h39ss | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let f(x) be a quadratic polynomial such that f($$-$$2) + f(3) = 0. If one of the roots of f(x) = 0 is $$-$$1, then the sum of the roots of f(x) = 0 is equal to :</p> | [{"identifier": "A", "content": "$${{11} \\over 3}$$"}, {"identifier": "B", "content": "$${{7} \\over 3}$$"}, {"identifier": "C", "content": "$${{13} \\over 3}$$"}, {"identifier": "D", "content": "$${{14} \\over 3}$$"}] | ["A"] | null | <p>$$\because$$ x = $$-$$1 be the roots of f(x) = 0</p>
<p>$$\therefore$$ Let $$f(x) = A(x + 1)(x - 1)$$ ...... (i)</p>
<p>Now, $$f( - 2) + f(3) = 0$$</p>
<p>$$ \Rightarrow A[ - 1( - 2 - b) + 4(3 - b)] = 0$$</p>
<p>$$b = {{14} \over 3}$$</p>
<p>$$\therefore$$ Second root of f(x) = 0 will be $${{14} \over 3}$$</p>
<p>$$... | mcq | jee-main-2022-online-28th-june-evening-shift | 7,897 |
1l5c0zzpy | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If the sum of the squares of the reciprocals of the roots $$\alpha$$ and $$\beta$$ of <br/><br/>the equation 3x<sup>2</sup> + $$\lambda$$x $$-$$ 1 = 0 is 15, then 6($$\alpha$$<sup>3</sup> + $$\beta$$<sup>3</sup>)<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "96"}] | ["B"] | null | <p>$$3{x^2} + \lambda x - 1 = 0$$</p>
<p>Given, two roots are $$\alpha$$ and $$\beta$$.</p>
<p>$$\therefore$$ Sum of roots $$ = \alpha + \beta = {-\lambda \over 3}$$</p>
<p>And product of roots $$ = \alpha \beta = {-1 \over 3}$$</p>
<p>Given that,</p>
<p>Sum of square of reciprocal of roots $$\alpha$$ and $$\beta$... | mcq | jee-main-2022-online-24th-june-morning-shift | 7,900 |
1l6hxh4o8 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The minimum value of the sum of the squares of the roots of $$x^{2}+(3-a) x+1=2 a$$ is:</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nny14a/71f56c77-eb10-4874-b074-072a3d7a554d/cce478a0-2c7e-11ed-a18d-5933e4fde865/file-1l7nny14b.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nny14a/71f56c77-eb10-4874-b074-072a3d7a554d/cce478a0-2c7e-11ed-a18d-5933e4fde865... | mcq | jee-main-2022-online-26th-july-evening-shift | 7,901 |
1l6kihzs6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If $$\alpha, \beta$$ are the roots of the equation</p>
<p>$$
x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0
$$,</p>
<p>then the equation, whose roots are $$\alpha+\frac{1}{\beta}$$ and $$\beta... | [{"identifier": "A", "content": "$$3 x^{2}-20 x-12=0$$"}, {"identifier": "B", "content": "$$3 x^{2}-10 x-4=0$$"}, {"identifier": "C", "content": "$$3 x^{2}-10 x+2=0$$"}, {"identifier": "D", "content": "$$3 x^{2}-20 x+16=0$$"}] | ["B"] | null | <p>$${3^{\sqrt {{{\log }_3}5} }} - {5^{\sqrt {{{\log }_5}3} }} = {3^{\sqrt {{{\log }_3}5} }} - {\left( {{3^{{{\log }_3}5}}} \right)^{\sqrt {{{\log }_5}3} }}$$</p>
<p>$${3^{{{\left( {{{\log }_3}5} \right)}^{{1 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = {5^{{{\left( {{{\log }_5}3} \right)}^{{2... | mcq | jee-main-2022-online-27th-july-evening-shift | 7,902 |
1l6nl5cn4 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha$$, $$\beta$$ be the roots of the equation $$x^{2}-\sqrt{2} x+\sqrt{6}=0$$ and $$\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$$ be the roots of the equation $$x^{2}+a x+b=0$$. Then the roots of the equation $$x^{2}-(a+b-2) x+(a+b+2)=0$$ are :</p> | [{"identifier": "A", "content": "non-real complex numbers"}, {"identifier": "B", "content": "real and both negative"}, {"identifier": "C", "content": "real and both positive"}, {"identifier": "D", "content": "real and exactly one of them is positive"}] | ["B"] | null | <p>$$\alpha + \beta = \sqrt 2 $$, $$\alpha \beta = \sqrt 6 $$</p>
<p>$${1 \over {{\alpha ^2}}} + 1 + {1 \over {{\beta ^2}}} + 1 = 2 + {{{\alpha ^2} + {\beta ^2}} \over 6}$$</p>
<p>$$ = 2 + {{2 - 2\sqrt 6 } \over 6} = - a$$</p>
<p>$$\left( {{1 \over {{\alpha ^2}}} + 1} \right)\left( {{1 \over {{\beta ^2}}} + 1} \rig... | mcq | jee-main-2022-online-28th-july-evening-shift | 7,903 |
1l6p0or7p | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If $$\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$$, then the maximum value of $$\mathrm{a}$$ is :</p> | [{"identifier": "A", "content": "198"}, {"identifier": "B", "content": "202"}, {"identifier": "C", "content": "212"}, {"identifier": "D", "content": "218"}] | ["C"] | null | <p>$${1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {40 - a}} + {1 \over {40 - a}} - {1 \over {60 - a}}\, + \,....\, + \,{1 \over {180 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$$</p>
<p>$$ \Rightarrow {1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$$</p>
<p>$$... | mcq | jee-main-2022-online-29th-july-morning-shift | 7,904 |
1l6rfi0r3 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta(\alpha>\beta)$$ be the roots of the quadratic equation $$x^{2}-x-4=0 .$$ If $$P_{n}=\alpha^{n}-\beta^{n}$$, $$n \in \mathrm{N}$$, then $$\frac{P_{15} P_{16}-P_{14} P_{16}-P_{15}^{2}+P_{14} P_{15}}{P_{13} P_{14}}$$ is equal to __________.</p> | [] | null | 16 | <p>$$\alpha$$ and $$\beta$$ are the roots of the quadratic equation $${x^2} - x - 4 = 0$$.</p>
<p>$$\therefore$$ $$\alpha$$ and $$\beta$$ are satisfy the given equation.</p>
<p>$${\alpha ^2} - \alpha - 4 = 0$$</p>
<p>$$ \Rightarrow {\alpha ^{n + 1}} - {\alpha ^n} - 4{\alpha ^{n - 1}} = 0$$ ...... (1)</p>
<p>and $${\be... | integer | jee-main-2022-online-29th-july-evening-shift | 7,905 |
1ldomtn4p | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$S = \left\{ {x:x \in \mathbb{R}\,\mathrm{and}\,{{(\sqrt 3 + \sqrt 2 )}^{{x^2} - 4}} + {{(\sqrt 3 - \sqrt 2 )}^{{x^2} - 4}} = 10} \right\}$$. Then $$n(S)$$ is equal to</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2"}] | ["B"] | null | Let $(\sqrt{3}+\sqrt{2})^{x^{2}-4}=t$
<br/><br/>$$
\begin{aligned}
& t+\frac{1}{t}=10 \\\\
\Rightarrow & t^{2}-10 t+1=0 \\\\
\Rightarrow & t=\frac{10 \pm \sqrt{100-4}}{2}=5 \pm 2 \sqrt{6}
\end{aligned}
$$
<br/><br/><b>Case-I</b>
<br/><br/>$$
\begin{aligned}
& t=5+2 \sqrt{6} = (\sqrt{3}+\sqrt{2})^{2} \\\\
\Rightarrow... | mcq | jee-main-2023-online-1st-february-morning-shift | 7,906 |
1ldsgawzn | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha_1,\alpha_2,....,\alpha_7$$ be the roots of the equation $${x^7} + 3{x^5} - 13{x^3} - 15x = 0$$ and $$|{\alpha _1}| \ge |{\alpha _2}| \ge \,...\, \ge \,|{\alpha _7}|$$. Then $$\alpha_1\alpha_2-\alpha_3\alpha_4+\alpha_5\alpha_6$$ is equal to _________.</p> | [] | null | 9 | <p>$${x^7} + 3{x^5} - 13{x^3} - 15x = 0$$</p>
<p>$$x({x^6} + 3{x^4} - 13{x^2} - 15) = 0$$</p>
<p>$$x = 0 = {\alpha _7}$$</p>
<p>Let $${x^2} = t$$</p>
<p>$${t^3} + 3{t^2} - 13t - 15 = 0$$</p>
<p>$$(t + 1)(t + 5)(t - 3) = 0$$</p>
<p>$$t = {x^2} = - 1, - 5,3$$</p>
<p>$$x\, = \, \pm \,i, \pm \,\sqrt 5 i, \pm \,\sqrt 3 $$<... | integer | jee-main-2023-online-29th-january-evening-shift | 7,907 |
1ldsu3cx2 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\lambda \ne 0$$ be a real number. Let $$\alpha,\beta$$ be the roots of the equation $$14{x^2} - 31x + 3\lambda = 0$$ and $$\alpha,\gamma$$ be the roots of the equation $$35{x^2} - 53x + 4\lambda = 0$$. Then $${{3\alpha } \over \beta }$$ and $${{4\alpha } \over \gamma }$$ are the roots of the equation</p> | [{"identifier": "A", "content": "$$7{x^2} - 245x + 250 = 0$$"}, {"identifier": "B", "content": "$$49{x^2} - 245x + 250 = 0$$"}, {"identifier": "C", "content": "$$49{x^2} + 245x + 250 = 0$$"}, {"identifier": "D", "content": "$$7{x^2} + 245x - 250 = 0$$"}] | ["B"] | null | $14 x^{2}-31 x+3 \lambda=0$
<br/><br/>
$$
\begin{aligned}
& \alpha+\beta=\frac{31}{14} \ldots .(1) \text { and } \alpha \beta=\frac{3 \lambda}{14}\quad...(2) \\\\
& 35 x^{2}-53 x+4 \lambda=0 \\\\
& \alpha+\gamma=\frac{53}{35} \ldots(3) \text { and } \alpha \gamma=\frac{4 \lambda}{35} \quad\ldots(4) \\\\
& \frac{(2)}{(4... | mcq | jee-main-2023-online-29th-january-morning-shift | 7,908 |
1ldu6c9z7 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha \in\mathbb{R}$$ and let $$\alpha,\beta$$ be the roots of the equation $${x^2} + {60^{{1 \over 4}}}x + a = 0$$. If $${\alpha ^4} + {\beta ^4} = - 30$$, then the product of all possible values of $$a$$ is ____________.</p> | [] | null | 45 | $x^{2}+60^{\frac{1}{4}} x+a=0$
<br/><br/>
$$
\therefore \alpha+\beta=-60^{\frac{1}{4}}, \alpha \beta=a
$$
<br/><br/>
Now $\alpha^{4}+\beta^{4}=-30$
<br/><br/>
$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}-2 a^{2}=-30$
<br/><br/>
$\Rightarrow\left[(\alpha+\beta)^{2}-2 a\right]^{2}-2 a^{2}=-30$
<br/><br/>
$\Rightarro... | integer | jee-main-2023-online-25th-january-evening-shift | 7,909 |
1lgowfari | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta$$ be the roots of the equation $$x^{2}-\sqrt{2} x+2=0$$. Then $$\alpha^{14}+\beta^{14}$$ is equal to</p> | [{"identifier": "A", "content": "$$-64$$"}, {"identifier": "B", "content": "$$-64 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$-128 \\sqrt{2}$$"}, {"identifier": "D", "content": "$$-128$$"}] | ["D"] | null |
<ol>
<li><p><strong>Find the roots of the quadratic equation:</strong></p>
<p>The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. For the quadratic equation $$x^2 - \sqrt{2}x + 2 = 0$$, we have $$a = 1, b = -\sqrt{2}, c = 2$$. Plugging these into the quadratic formula gives:
<br/><br/>$$x = \frac{\sqr... | mcq | jee-main-2023-online-13th-april-evening-shift | 7,910 |
1lgswd6s1 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The number of points, where the curve $$f(x)=\mathrm{e}^{8 x}-\mathrm{e}^{6 x}-3 \mathrm{e}^{4 x}-\mathrm{e}^{2 x}+1, x \in \mathbb{R}$$ cuts $$x$$-axis, is equal to _________.</p> | [] | null | 2 | Firstly, we know that the given function <br/><br/>$f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1$ intersects the x-axis <br/><br/>where $f(x) = 0$. Setting $f(x)$ equal to zero gives us :
<br/><br/>$e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0.$
<br/><br/>Let $t = e^{2x}$. The equation now becomes :
<br/><br/>$t^4 - t^3 - 3t^2 - t + 1 =... | integer | jee-main-2023-online-11th-april-evening-shift | 7,912 |
1lguwygfx | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If $$a$$ and $$b$$ are the roots of the equation $$x^{2}-7 x-1=0$$, then the value of $$\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$$ is equal to _____________.</p> | [] | null | 51 | We have, $a$ and $b$ are the roots of the equation
<br/><br/>$$
\begin{aligned}
& x^2-7 x-1=0 \\\\
& \Rightarrow a^2-7 a-1=0 \Rightarrow a^2-1=7 a .........(i)
\end{aligned}
$$
<br/><br/>On squaring both sides, we get $a^4+1=51 a^2$
<br/><br/>Similarly, $b^4+1=51 b^2$ ...........(ii)
<br/><br/>$$
\text { Now, } \frac{a... | integer | jee-main-2023-online-11th-april-morning-shift | 7,913 |
1lgzzo8pe | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta, \gamma$$ be the three roots of the equation $$x^{3}+b x+c=0$$. If $$\beta \gamma=1=-\alpha$$, then $$b^{3}+2 c^{3}-3 \alpha^{3}-6 \beta^{3}-8 \gamma^{3}$$ is equal to :</p> | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "19"}, {"identifier": "C", "content": "$$\\frac{169}{8}$$"}, {"identifier": "D", "content": "$$\\frac{155}{8}$$"}] | ["B"] | null | Given cubic equation is :
<br/><br/>$$
x^3+b x+c=0
$$
<br/><br/>$\because \alpha, \beta, \gamma$ are the roots of above equation.
<br/><br/>And $\beta \gamma=1=-\alpha$
<br/><br/>$$
\begin{aligned}
& \text { So, product of roots }=-c \\\\
& \Rightarrow \alpha \beta \gamma=-c \\\\
& \Rightarrow(-1)(1)=-c \\\\
& \Rightar... | mcq | jee-main-2023-online-8th-april-morning-shift | 7,914 |
1lh23bm1u | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The sum of all the roots of the equation $$\left|x^{2}-8 x+15\right|-2 x+7=0$$ is :</p> | [{"identifier": "A", "content": "$$11+\\sqrt{3}$$"}, {"identifier": "B", "content": "$$9+\\sqrt{3}$$"}, {"identifier": "C", "content": "$$9-\\sqrt{3}$$"}, {"identifier": "D", "content": "$$11-\\sqrt{3}$$"}] | ["B"] | null | $$
\begin{aligned}
& \text { We have, }\left|x^2-8 x+15\right|-2 x+7=0 \\\\
& \Rightarrow |(x-3)(x-5)|-2 x+7=0
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkn5295/e47cf7ce-efd4-4752-9bbb-34637d3af9f9/c36b1990-6798-11ee-97fe-41fa1903ca4f/file-6y3zli1lnkn5296.p... | mcq | jee-main-2023-online-6th-april-morning-shift | 7,915 |
lsamarl9 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $\alpha$ and $\beta$ be the roots of the equation $p x^2+q x-r=0$, where $p \neq 0$. If $p, q$ and $r$ be the consecutive terms of a non constant G.P. and $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$, then the value of $(\alpha-\beta)^2$ is : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "$\\frac{20}{3}$"}, {"identifier": "D", "content": "$\\frac{80}{9}$"}] | ["D"] | null | Given : $p x^2+q x-r=0$
<br/><br/>Let $p=\frac{a}{r_1}, q=a, r=a r_1$
<br/><br/>$\begin{aligned} & \text { and } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{-\frac{q}{p}}{-\frac{r}{p}}=\frac{3}{4} \\\\ & \Rightarrow \frac{q}{r}=\... | mcq | jee-main-2024-online-1st-february-evening-shift | 7,916 |
lsaolop2 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $\mathbf{S}=\left\{x \in \mathbf{R}:(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}$. Then the number of elements in $\mathrm{S}$ is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <p>Notice that $(\sqrt{3} + \sqrt{2})$ and $(\sqrt{3} - \sqrt{2})$ are reciprocals of each other because :</p>
<ul>
<li>$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$</li>
</ul>
<br/><strong>Using the Reciprocal Property :</strong>
<br/><ul>
<br/><li>This means $(\sqrt{3} - \sqrt{2})^x = \frac{1}{(\sqrt{3} ... | mcq | jee-main-2024-online-1st-february-morning-shift | 7,917 |
1lsgcitu2 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta \in \mathbf{N}$$ be roots of the equation $$x^2-70 x+\lambda=0$$, where $$\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbf{N}$$. If $$\lambda$$ assumes the minimum possible value, then $$\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$$ is equal to :</p> | [] | null | 60 | <p>$$\begin{aligned}
& x^2-70 x+\lambda=0 \\
& \alpha+\beta=70 \\
& \alpha \beta=\lambda \\
& \therefore \alpha(70-\alpha)=\lambda
\end{aligned}$$</p>
<p>Since, 2 and 3 does not divide $$\lambda$$</p>
<p>$$\therefore \alpha=5, \beta=65, \lambda=325$$</p>
<p>By putting value of $$\alpha, \beta, \lambda$$ we get the requ... | integer | jee-main-2024-online-30th-january-morning-shift | 7,918 |
luxwdj41 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta ; \alpha>\beta$$, be the roots of the equation $$x^2-\sqrt{2} x-\sqrt{3}=0$$. Let $$\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{N}$$. Then $$(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$$ is equal to</p> | [{"identifier": "A", "content": "$$10 \\sqrt{3} \\mathrm{P}_9$$\n"}, {"identifier": "B", "content": "$$11 \\sqrt{3} \\mathrm{P}_9$$\n"}, {"identifier": "C", "content": "$$11 \\sqrt{2} \\mathrm{P}_9$$\n"}, {"identifier": "D", "content": "$$10 \\sqrt{2} \\mathrm{P}_9$$"}] | ["A"] | null | <p>$$\begin{aligned}
& x^2-\sqrt{2} x-\sqrt{3}=0 \\
& P_n=\alpha^n-\beta^n
\end{aligned}$$</p>
<p>$$\alpha$$ and $$\beta$$ are the roots of the equation</p>
<p>Using Newton's theorem</p>
<p>$$\begin{aligned}
& P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0 \\
& \text { Put } n=10 \\
& P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0
... | mcq | jee-main-2024-online-9th-april-evening-shift | 7,919 |
luy6z57u | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta$$ be the roots of the equation $$x^2+2 \sqrt{2} x-1=0$$. The quadratic equation, whose roots are $$\alpha^4+\beta^4$$ and $$\frac{1}{10}(\alpha^6+\beta^6)$$, is:</p> | [{"identifier": "A", "content": "$$x^2-180 x+9506=0$$\n"}, {"identifier": "B", "content": "$$x^2-195 x+9506=0$$\n"}, {"identifier": "C", "content": "$$x^2-190 x+9466=0$$\n"}, {"identifier": "D", "content": "$$x^2-195 x+9466=0$$"}] | ["B"] | null | <p>$$\begin{aligned}
& x^2+2 \sqrt{2 x}-1=0 \\
& \alpha+\beta=-2 \sqrt{2} \text { and } \alpha \beta=-1 \\
& \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\
& =8+2=10 \\
& \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha \beta)^2 \\
& =100-2=98 \\
& \alpha^6+\beta^6=\left(\alpha^2+\beta^2\right)^3-3 \alpha... | mcq | jee-main-2024-online-9th-april-morning-shift | 7,920 |
lv0vxd39 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If 2 and 6 are the roots of the equation $$a x^2+b x+1=0$$, then the quadratic equation, whose roots are $$\frac{1}{2 a+b}$$ and $$\frac{1}{6 a+b}$$, is :</p> | [{"identifier": "A", "content": "$$x^2+8 x+12=0$$\n"}, {"identifier": "B", "content": "$$2 x^2+11 x+12=0$$\n"}, {"identifier": "C", "content": "$$4 x^2+14 x+12=0$$\n"}, {"identifier": "D", "content": "$$x^2+10 x+16=0$$"}] | ["A"] | null | <p>Given that the roots of the quadratic equation are $2$ and $6$, we can use Vieta's formulas which relate the coefficients of the polynomial to sums and products of its roots.</p>
<p>The given quadratic equation is:</p>
<p>$$a x^2 + b x + 1 = 0$$</p>
<p>By Vieta's formulas, the sum of the roots is:</p>
<p>$$2 + 6... | mcq | jee-main-2024-online-4th-april-morning-shift | 7,921 |
lv5gst25 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The sum of all the solutions of the equation $$(8)^{2 x}-16 \cdot(8)^x+48=0$$ is :</p> | [{"identifier": "A", "content": "$$1+\\log _8(6)$$\n"}, {"identifier": "B", "content": "$$1+\\log _6(8)$$\n"}, {"identifier": "C", "content": "$$\\log _8(6)$$\n"}, {"identifier": "D", "content": "$$\\log _8(4)$$"}] | ["A"] | null | <p>First, let's start by substituting $$y = (8)^x$$ in the given equation. By substituting, the equation $$8^{2x} - 16 \cdot 8^x + 48 = 0$$ will be transformed into</p>
<p>$$ y^2 - 16y + 48 = 0 $$</p>
<p>Now, we have a quadratic equation in $$y$$. To find the roots of this quadratic equation, we can use the quadratic... | mcq | jee-main-2024-online-8th-april-morning-shift | 7,922 |
lvb2952b | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta$$ be roots of $$x^2+\sqrt{2} x-8=0$$. If $$\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$$, then $$\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}$$ is equal to ________.</p> | [] | null | 4 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwapcytl/545aa1a6-edf5-4e15-9137-baf3d0f62d19/8d6a2390-144f-11ef-860c-d121cbcdd1fc/file-1lwapcytm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwapcytl/545aa1a6-edf5-4e15-9137-baf3d0f62d19/8d6a2390-144f-11ef-860c-d121cbcdd1fc... | integer | jee-main-2024-online-6th-april-evening-shift | 7,923 |
lvc57aw6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta$$ be the distinct roots of the equation $$x^2-\left(t^2-5 t+6\right) x+1=0, t \in \mathbb{R}$$ and $$a_n=\alpha^n+\beta^n$$. Then the minimum value of $$\frac{a_{2023}+a_{2025}}{a_{2024}}$$ is</p> | [{"identifier": "A", "content": "$$-1 / 2$$\n"}, {"identifier": "B", "content": "$$-1 / 4$$\n"}, {"identifier": "C", "content": "$$1 / 4$$\n"}, {"identifier": "D", "content": "$$1 / 2$$"}] | ["B"] | null | <p>$$\begin{aligned}
& x^2-\left(t^2-5 t+6\right) x+1=0 \\
& \therefore a_{2025}-\left(t^2-5 t+6\right) a_{2024}+a_{2023}=0 \\
& \Rightarrow \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \\
& =\left(t+\frac{5}{2}\right)^2+\left(\frac{-1}{4}\right) \\
& \text { Minimum value }=\frac{-1}{4}
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift | 7,924 |
YBEbaVOfFKiWxDON | maths | sequences-and-series | am,-gm-and-hm | If m is the A.M. of two distinct real numbers l and n $$(l,n > 1)$$ and $${G_1},{G_2}$$ and $${G_3}$$ are three geometric means between $$l$$ and n, then $$G_1^4\, + 2G_2^4\, + G_3^4$$ equals: | [{"identifier": "A", "content": "$$4\\,lm{n^2}$$ "}, {"identifier": "B", "content": "$$4\\,{l^2}{m^2}{n^2}$$ "}, {"identifier": "C", "content": "$$4\\,{l^2}m\\,n$$ "}, {"identifier": "D", "content": "$$4\\,l\\,{m^2}n$$ "}] | ["D"] | null | $$m = {{l + n} \over 2}$$ and common ratio of
<br><br>$$G.P.$$ $$ = r = {\left( {{n \over l}} \right)^{{1 \over 4}}}$$
<br><br>$$\therefore$$ $${G_1} = {l^{3/4}}\,{n^{1/4}},$$ $${G_2} = {l^{1/2}}{n^{1/2}},\,$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}$$
<br><br>$$G_1^4 + 2G_2^4 + G_3^4$$
<br><br... | mcq | jee-main-2015-offline | 7,925 |
UvIf51uHbb6DvoeXkhC7K | maths | sequences-and-series | am,-gm-and-hm | Let x, y, z be positive real numbers such that x + y + z = 12 and x<sup>3</sup>y<sup>4</sup>z<sup>5</sup> = (0.1) (600)<sup>3</sup>. Then x<sup>3</sup> + y<sup>3</sup> + z<sup>3</sup>is equal to : | [{"identifier": "A", "content": "270"}, {"identifier": "B", "content": "258"}, {"identifier": "C", "content": "342"}, {"identifier": "D", "content": "216"}] | ["D"] | null | As we know
<br><br>AM $$ \ge $$ GM
<br><br>$$ \Rightarrow $$ $${{3\left( {{x \over 3}} \right) + 4\left( {{y \over 4}} \right) + 5\left( {{z \over 5}} \right)} \over {12}}$$ $$ \ge $$ $${\left[ {{{\left( {{x \over 3}} \right)}^3}{{\left( {{y \over 4}} \right... | mcq | jee-main-2016-online-9th-april-morning-slot | 7,926 |
DhoQKCuInbj9DsKERyHLW | maths | sequences-and-series | am,-gm-and-hm | If A > 0, B > 0 and A + B = $${\pi \over 6}$$, <br/><br/>then the minimum value of tanA + tanB is : | [{"identifier": "A", "content": "$$\\sqrt 3 - \\sqrt 2 $$ "}, {"identifier": "B", "content": "$$2 - \\sqrt 3 $$"}, {"identifier": "C", "content": "$$4 - 2\\sqrt 3 $$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$ "}] | ["C"] | null | Given,
<br><br>A + B = $${\pi \over 6}$$
<br><br>$$ \therefore $$ tan(A + B) = tan$$\left( {{\pi \over 6}} \right)$$ = $${1 \over {\sqrt 3 }}$$
<br><br>We know,
<br><br>tan(A + B) = $${{\tan A + \tan B} \over {1 - \tan A\tan B}}$$
<br><br>$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}$$ = $${y \... | mcq | jee-main-2016-online-10th-april-morning-slot | 7,927 |
wuYGlr4ZcEF2WUJlcQDCt | maths | sequences-and-series | am,-gm-and-hm | If the arithmetic mean of two numbers a and b, a > b > 0, is five times their geometric mean, then $${{a + b} \over {a - b}}$$ is equal to : | [{"identifier": "A", "content": "$${{\\sqrt 6 } \\over 2}$$ "}, {"identifier": "B", "content": "$${{3\\sqrt 2 } \\over 4}$$"}, {"identifier": "C", "content": "$${{7\\sqrt 3 } \\over {12}}$$"}, {"identifier": "D", "content": "$${{5\\sqrt 6 } \\over {12}}$$ "}] | ["D"] | null | A.T.Q.,
<br><br>A.M. = 5G.M.
<br><br>$${{a + b} \over 2} = 5\sqrt {ab} $$
<br><br>$${{a + b} \over {\sqrt {ab} }}$$ $$ = 10$$
<br><br>$$ \therefore $$ $${a \over b} = {{10 + \sqrt {96} } \over {10 - \sqrt {96} }} = {{10 + 4\sqrt 6 } \over {10 - 4\sqrt 6 }}$$
<br><br>Use componendo and Dividendo
<br><br... | mcq | jee-main-2017-online-8th-april-morning-slot | 7,928 |
GNsf4xF6cvZXlPv0FTnFO | maths | sequences-and-series | am,-gm-and-hm | Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression $${{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}}$$ is :
| [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 4}$$"}, {"identifier": "C", "content": "$${{m + n} \\over {6mn}}$$"}, {"identifier": "D", "content": "1"}] | ["B"] | null | $${{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}} = {1 \over {\left( {{x^m} + {1 \over {{x^m}}}} \right)\left( {{y^n} + {1 \over {{y^n}}}} \right)}} \le {1 \over 4}$$
<br><br>using AM $$ \ge $$ GM | mcq | jee-main-2019-online-11th-january-evening-slot | 7,929 |
TQWG3sOpx0jCpXORb97nm | maths | sequences-and-series | am,-gm-and-hm | If sin<sup>4</sup>$$\alpha $$ + 4 cos<sup>4</sup>$$\beta $$ + 2 = 4$$\sqrt 2 $$ sin $$\alpha $$ cos $$\beta $$; $$\alpha $$, $$\beta $$ $$ \in $$ [0, $$\pi $$],
<br/>then cos($$\alpha $$ + $$\beta $$) $$-$$ cos($$\alpha $$ $$-$$ $$\beta $$) is equal to : | [{"identifier": "A", "content": "$$ - \\sqrt 2 $$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$-$$ 1"}, {"identifier": "D", "content": "$$\\sqrt 2 $$ "}] | ["A"] | null | A.M. $$ \ge $$ G.M.
<br><br>$${{{{\sin }^4}\alpha + 4{{\cos }^4}\beta + 1 + 1} \over 4} \ge {\left( {{{\sin }^4}\alpha .4{{\cos }^4}\beta .1.1} \right)^{{1 \over 4}}}$$
<br><br>sin<sup>4</sup><sup></sup>$$\alpha $$ + 4 cos<sup>2</sup>$$\beta $$ + 2 $$ \ge $$ 4 $$\sqrt 2 $$ sin $$\alpha $$ cos $$\beta $$
<br><br>Give... | mcq | jee-main-2019-online-12th-january-evening-slot | 7,930 |
PS85QIo4S7Avsvyt1bjgy2xukf49qm01 | maths | sequences-and-series | am,-gm-and-hm | If m arithmetic means (A.Ms) and three
geometric means (G.Ms) are inserted between
3 and 243 such that 4<sup>th</sup> A.M. is equal to 2<sup>nd</sup>
G.M., then m is equal to _________ . | [] | null | 39 | Given m arithmetic means (A.Ms) present between 3 and 243<br><br>$$ \therefore $$ Common difference, $$d = {{b - a} \over {m + 1}} = {{240} \over {m + 1}}$$<br><br>$$ \therefore $$ 4th A.M. = a + 4d<br><br>= 3 + 4 $$ \times $$ $${{240} \over {m + 1}}$$<br><br>Also there are 3 G.M between 3 and 243<br><br>$$ \therefore ... | integer | jee-main-2020-online-3rd-september-evening-slot | 7,931 |
FLfk7pqpuOUDRypvuS1klt80kz2 | maths | sequences-and-series | am,-gm-and-hm | The minimum value of $$f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}$$, where a, $$x \in R$$ and a > 0, is equal to : | [{"identifier": "A", "content": "$$a + {1 \\over a}$$"}, {"identifier": "B", "content": "2a"}, {"identifier": "C", "content": "a + 1"}, {"identifier": "D", "content": "$$2\\sqrt a $$"}] | ["D"] | null | We know, $$AM \ge GM$$<br><br>$$ \therefore $$ $${{{a^{a^x}} + {a \over {{a^{a^x}}}}} \over 2} \ge {\left( {{a^{a^x}}\,.\,{a \over {{a^{a^x}}}}} \right)^{1/2}} $$
<br><br>$$\Rightarrow {a^{a^x}} + {a^{1 - a^x}} \ge 2\sqrt a $$ | mcq | jee-main-2021-online-25th-february-evening-slot | 7,933 |
MYzMY4FrIZlEIa93ir1kluynti6 | maths | sequences-and-series | am,-gm-and-hm | If the arithmetic mean and geometric mean of the p<sup>th</sup> and q<sup>th</sup> terms of the <br/>sequence $$-$$16, 8, $$-$$4, 2, ...... satisfy the equation<br/> 4x<sup>2</sup> $$-$$ 9x + 5 = 0, then p + q is equal to __________. | [] | null | 10 | Given, $$4{x^2} - 9x + 5 = 0$$<br><br>$$ \Rightarrow (x - 1)(4x - 5) = 0$$<br><br>$$ \Rightarrow $$ A. M. $$ = {5 \over 4}$$, G. M. = 1 (As A. M. $$ \ge $$ G. M)<br><br>Again, for the series<br><br>$$-$$16, 8, $$-$$4, 2 ..........<br><br>$${p^{th}}$$ term $${t_p} = - 16{\left( {{{ - 1} \over 2}} \right)^{p - 1}}$$<br>... | integer | jee-main-2021-online-26th-february-evening-slot | 7,934 |
1l5b7vzyn | maths | sequences-and-series | am,-gm-and-hm | <p>Let x, y > 0. If x<sup>3</sup>y<sup>2</sup> = 2<sup>15</sup>, then the least value of 3x + 2y is</p> | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "40"}] | ["D"] | null | <p>x, y > 0 and x<sup>3</sup>y<sup>2</sup> = 2<sup>15</sup></p>
<p>Now, 3x + 2y = (x + x + x) + (y + y)</p>
<p>So, by A.M $$\ge$$ G.M inequality</p>
<p>$${{3x + 2y} \over 5} \ge \root 5 \of {{x^3}\,.\,{y^2}} $$</p>
<p>$$\therefore$$ $$3x + 2y \ge 5\root 5 \of {{2^{15}}} \ge 40$$</p>
<p>$$\therefore$$ Least value of $$... | mcq | jee-main-2022-online-24th-june-evening-shift | 7,936 |
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