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lv2ergyt | maths | probability | probability-distribution-of-a-random-variable | <p>If the mean of the following probability distribution of a radam variable $$\mathrm{X}$$ :</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;wo... | [{"identifier": "A", "content": "$$\\frac{581}{81}$$\n"}, {"identifier": "B", "content": "$$\\frac{566}{81}$$\n"}, {"identifier": "C", "content": "$$\\frac{151}{27}$$\n"}, {"identifier": "D", "content": "$$\\frac{173}{27}$$"}] | ["B"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial,... | mcq | jee-main-2024-online-4th-april-evening-shift |
lv5gs180 | maths | probability | probability-distribution-of-a-random-variable | <p>Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables $$X$$ and $$Y$$ respectively denote the number of blue and yellow balls. If $$\bar{X}$$ and $$\bar{Y}$$ are the means of $$X$$ and $$Y$$ respectively, then $$7 \bar{X}+4 \bar{Y}$$ is equal to ___________.</p> | [] | null | 17 | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial,... | integer | jee-main-2024-online-8th-april-morning-shift |
lv7v3quo | maths | probability | probability-distribution-of-a-random-variable | <p>From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $$X$$ denote the number of defective items in the sample. If the variance of $$X$$ is $$\sigma^2$$, then $$96 \sigma^2$$ is equal to __________.</p> | [] | null | 56 | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial,... | integer | jee-main-2024-online-5th-april-morning-shift |
lvb2954k | maths | probability | probability-distribution-of-a-random-variable | <p>From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $$X$$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $$X$$ is $$\frac{m}{n}$$, where $$\operatorname{gcd}(m, n)=1$$, the... | [] | null | 71 | <p>Given a lot of 12 items, 3 are defective.</p>
<p>Good items, $$12-3=9$$</p>
<p>Let $$X$$ denote the number of defective items.</p>
<p>So, value of $$X=0,1,2,3$$</p>
<p>A sample of $$S$$ items is drawn.</p>
<p>$$P(X=0)=G G G G G$$</p>
<p>(here $$G$$ is good item and $$d$$ is defective)</p>
<p>$$\begin{aligned}
& \fra... | integer | jee-main-2024-online-6th-april-evening-shift |
NlmoqFBur2q2cMiy | maths | probability | total-probability-theorem | A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and
this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at
random from the bag, then the probability that this drawn ball is red, is : | [{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${3 \\over 10}$$"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$${1 \\over 5}$$"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264051/exam_images/ijwqqvgif3jlthy0b2tu.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Probability Question 169 English Explanation">
<br><br>If we follow path 1, then probability of getting 1st ball black $$ = {6 ... | mcq | jee-main-2018-offline |
lgnxp7gc | maths | probability | total-probability-theorem | A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is : | [{"identifier": "A", "content": "$\\frac{1}{4}$"}, {"identifier": "B", "content": "$\\frac{9}{50}$"}, {"identifier": "C", "content": "$\\frac{1}{5}$"}, {"identifier": "D", "content": "$\\frac{11}{50}$"}] | ["C"] | null | Let $X$ be the number rolled on the die, and let $W$ be the event that all balls drawn are white. We want to find the probability $P(W)$, which can be calculated using the law of total probability as follows :
<br/><br/>$$P(W) = \sum\limits_{x=1}^{6} P(W|X=x)P(X=x)$$
<br/><br/>The probability of rolling any number fr... | mcq | jee-main-2023-online-15th-april-morning-shift |
lsao86jd | maths | probability | total-probability-theorem | A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is : | [{"identifier": "A", "content": "$\\frac{2}{5}$"}, {"identifier": "B", "content": "$\\frac{2}{7}$"}, {"identifier": "C", "content": "$\\frac{1}{7}$"}, {"identifier": "D", "content": "$\\frac{1}{5}$"}] | ["B"] | null | $\begin{aligned} & \mathrm{P}(4 \mathrm{~W} 4 \mathrm{~B} / 2 \mathrm{~W} 2 \mathrm{~B})= \\\\ & \frac{P(4 W 4 B) \times P(2 W 2 B / 4 W 4 B)}{P(2 W 6 B) \times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \times P(2 W 2 B / 3 W 5 B)} \\ & +\ldots \ldots \ldots \ldots+P(6 W 2 B) \times P(2 W 2 B / 6 W 2 B)\end{aligned}$
<br/><br/>$... | mcq | jee-main-2024-online-1st-february-morning-shift |
q6vgte5MQsjwFa1v | maths | probability | venn-diagram-and-set-theory | A problem in mathematics is given to three students $$A,B,C$$ and their respective probability of solving the problem is $${1 \over 2},{1 \over 3}$$ and $${1 \over 4}.$$ Probability that the problem is solved is : | [{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$ "}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["A"] | null | Given $$P\left( A \right) = {1 \over 2}$$, $$P\left( B \right) = {1 \over 3}$$, $$P\left( C \right) = {1 \over 4}$$
<br><br>So, $$P\left( {\overline A } \right) = {1 \over 2}$$ (Probablity that the problem can't be solve by A)
<br>$$P\left( {\overline B } \right) = {2 \over 3}$$ (Probablity that the problem can't be so... | mcq | aieee-2002 |
Inc63WUwWyWr1GZE | maths | probability | venn-diagram-and-set-theory | $$A$$ and $$B$$ are events such that $$P\left( {A \cup B} \right) = 3/4$$,$$P\left( {A \cap B} \right) = 1/4,$$
<br/>$$P\left( {\overline A } \right) = 2/3$$ then $$P\left( {\overline A \cap B} \right)$$ is : | [{"identifier": "A", "content": "$$5/12$$"}, {"identifier": "B", "content": "$$3/8$$"}, {"identifier": "C", "content": "$$5/8$$ "}, {"identifier": "D", "content": "$$1/4$$ "}] | ["A"] | null | Given $$P\left( {A \cup B} \right) = 3/4$$,
<br>$$P\left( {A \cap B} \right) = 1/4,$$
<br>$$P\left( {\overline A } \right) = 2/3$$
<br><br>We know, $$P\left( A \right)$$ = 1 - $$P\left( {\overline A } \right)$$
<br>$$\therefore$$ $$P\left( A \right)$$ = 1 - $${2 \over 3}$$ = $${1 \over 3}$$
<br><br>We know $$P\left( {... | mcq | aieee-2002 |
m09Dh9Fy3uo4ADUI | maths | probability | venn-diagram-and-set-theory | Events $$A, B, C$$ are mutually exclusive events such that $$P\left( A \right) = {{3x + 1} \over 3},$$ $$P\left( B \right) = {{1 - x} \over 4}$$ and $$P\left( C \right) = {{1 - 2x} \over 2}$$ The set of possible values of $$x$$ are in the interval. | [{"identifier": "A", "content": "$$\\left[ {0,1} \\right]$$ "}, {"identifier": "B", "content": "$$\\left[ {{1 \\over 3},{1 \\over 2}} \\right]$$ "}, {"identifier": "C", "content": "$$\\left[ {{1 \\over 3},{2 \\over 3}} \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {{1 \\<br>3},{13 \\over 3}} \\right]$$"}] | ["B"] | null | Given $$P\left( A \right) = {{3x + 1} \over 3},$$ $$P\left( B \right) = {{1 - x} \over 4}$$ and $$P\left( C \right) = {{1 - 2x} \over 2}$$
<br><br>We know for any event X, $$0 \le P\left( X \right) \le 1$$
<br><br>$$\therefore$$ $$0 \le {{3x + 1} \over 3} \le 1$$
<br>$$ \Rightarrow - 1 \le 3x \le 2$$
<br>$$ \Rightarro... | mcq | aieee-2003 |
kJkLQp3LNsJbLPI9 | maths | probability | venn-diagram-and-set-theory | Let $$A$$ and $$B$$ two events such that $$P\left( {\overline {A \cup B} } \right) = {1 \over 6},$$ $$P\left( {A \cap B} \right) = {1 \over 4}$$ and $$P\left( {\overline A } \right) = {1 \over 4},$$ where $${\overline A }$$ stands for complement of event $$A$$. Then events $$A$$ and $$B$$ are : | [{"identifier": "A", "content": "equally likely and mutually exclusive"}, {"identifier": "B", "content": "equally likely but not independent "}, {"identifier": "C", "content": "independent but not equally likely"}, {"identifier": "D", "content": "mutually exclusive and independent"}] | ["C"] | null | <p>Given that,</p>
<p>$$P(\overline {A \cup B} ) = {1 \over 6}$$, $$P(A \cap B) = {1 \over 4}$$, $$P(\overline A ) = {1 \over 4}$$</p>
<p>$$\because$$ $$P(\overline {A \cup B} ) = {1 \over 6}$$</p>
<p>$$ \Rightarrow 1 - P(A \cup B) = {1 \over 6}$$</p>
<p>$$ \Rightarrow 1 - P(A) - P(B) + P(A \cap B) = {1 \over 6}$$</p>
... | mcq | aieee-2005 |
CvljnrLZQyp22oGR | maths | probability | venn-diagram-and-set-theory | For three events A, B and C, <br/><br/>P(Exactly one of A or B occurs) <br/>= P(Exactly one of B or C occurs) <br/>= P
(Exactly one of C or A occurs) = $${1 \over 4}$$
<br/>and P(All the three events occur simultaneously) = $${1 \over {16}}$$.
<br/><br/> Then the
probability that at least one of the events occurs, is ... | [{"identifier": "A", "content": "$${7 \\over {16}}$$"}, {"identifier": "B", "content": "$${7 \\over {64}}$$"}, {"identifier": "C", "content": "$${3 \\over {16}}$$"}, {"identifier": "D", "content": "$${7 \\over {32}}$$"}] | ["A"] | null | Given, P (A $$ \cap $$ B $$ \cap $$ C) = $${1 \over {16}}$$
<br><br>P (exactly one of A or B occurs)
<br><br>= P(A) + P (B) – 2P (A $$ \cap $$ B) = $${1 \over 4}$$ .....(1)
<br><br>P (Exactly one of B or C occurs)
<br><br>= P(B) + P (C) – 2P (B $$ \cap $$ C) = $${1 \over 4}$$ .....(2)
<br><br>P (Exactly one of C or A o... | mcq | jee-main-2017-offline |
dGx2Muc10197ojWmZcE9D | maths | probability | venn-diagram-and-set-theory | In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the students selected has opted neither for NCC
nor for NSS is : | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over 6}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${5 \\over 6}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264827/exam_images/atidu8b68l8usxxqwsp4.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Evening Slot Mathematics - Probability Question 146 English Explanation">
<br>A $$ ... | mcq | jee-main-2019-online-12th-january-evening-slot |
QkHi8fNAnBXDglkK9K7k9k2k5hjz0pp | maths | probability | venn-diagram-and-set-theory | Let A and B be two events such that the
probability that exactly one of them occurs is $${2 \over 5}$$ and the probability that A or B occurs is $${1 \over 2}$$ ,
then the probability of both of them occur
together is : | [{"identifier": "A", "content": "0.20"}, {"identifier": "B", "content": "0.02"}, {"identifier": "C", "content": "0.01"}, {"identifier": "D", "content": "0.10"}] | ["D"] | null | Probability that exactly one of them occurs
<br><br>P(A) + P(B) – 2P (A $$ \cap $$ B) = $${2 \over 5}$$ .....(1)
<br><br>Probability
that A or B occurs is
<br><br>P(A) + P(B) – P(A $$ \cap $$ B) = $${1 \over 2}$$ ......(2)
<br><br>Doing (2) - (1)
<br><br>P(A $$ \cap $$ B) = $${1 \over 2} - {2 \over 5}$$ = 0.10 | mcq | jee-main-2020-online-8th-january-evening-slot |
eziF1WqbTH9NrxKj6Hjgy2xukg0cfpa9 | maths | probability | venn-diagram-and-set-theory | The probabilities of three events A, B and C are
given by <br/>P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.
<br/>If P(A$$ \cup $$B) = 0.8, P(A$$ \cap $$C) = 0.3, P(A$$ \cap $$B$$ \cap $$C) = 0.2,
P(B$$ \cap $$C) = $$\beta $$<br/> and P(A$$ \cup $$B$$ \cup $$C) = $$\alpha $$, where
0.85 $$ \le \alpha \le $$ 0.95, then $$\beta... | [{"identifier": "A", "content": "[0.35, 0.36]\n"}, {"identifier": "B", "content": "[0.20, 0.25]"}, {"identifier": "C", "content": "[0.25, 0.35]"}, {"identifier": "D", "content": "[0.36, 0.40]"}] | ["C"] | null | P(A $$ \cup $$ B) = P(A) + P(B) – P(A $$ \cup $$ B)
<br><br>$$ \Rightarrow $$ 0.8 = 0.6 + 0.4 – P(A $$ \cap $$ B)
<br><br>$$ \Rightarrow $$ P(A $$ \cap $$ B) = 0.2
<br><br>P(A$$ \cup $$B$$ \cup $$C) = P(A) + P(B) + P(C) – P(A $$ \cap $$ B) – P(B $$ \cap $$ C) –P(C $$ \cap $$ A) + P(A $$ \cap $$ B $$ \cap $$ C)
<br><br>... | mcq | jee-main-2020-online-6th-september-evening-slot |
1krrr6wf1 | maths | probability | venn-diagram-and-set-theory | Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 $$-$$ k), the probability that exactly one of B and C occurs is (1 $$-$$ 2k), the probability that exactly one of C and A occurs is (1 $$-$$ k) and the probability of all A, B and C occur simultaneously is k<sup>2</sup>, w... | [{"identifier": "A", "content": "greater than $${1 \\over 8}$$ but less than $${1 \\over 4}$$"}, {"identifier": "B", "content": "greater than $${1 \\over 2}$$"}, {"identifier": "C", "content": "greater than $${1 \\over 4}$$ but less than $${1 \\over 2}$$"}, {"identifier": "D", "content": "exactly equal to $${1 \\over 2... | ["B"] | null | $$P(\overline A \cap B) + P(A \cap \overline B ) = 1 - k$$<br><br>$$P(\overline A \cap C) + P(A \cap \overline C ) = 1 - 2k$$<br><br>$$P(\overline B \cap C) + P(B \cap \overline C ) = 1 - k$$<br><br>$$P(A \cap B \cap C) = {k^2}$$<br><br>$$P(A) + P(B) - 2P(A \cap B) = 1 - k$$ .....(i)<br><br>$$P(B) + P(C) - 2P(B \cap... | mcq | jee-main-2021-online-20th-july-evening-shift |
1l54tbhne | maths | probability | venn-diagram-and-set-theory | <p>The probability that a relation R from {x, y} to {x, y} is both symmetric and transitive, is equal to :</p> | [{"identifier": "A", "content": "$${5 \\over {16}}$$"}, {"identifier": "B", "content": "$${9 \\over {16}}$$"}, {"identifier": "C", "content": "$${11 \\over {16}}$$"}, {"identifier": "D", "content": "$${13 \\over {16}}$$"}] | ["A"] | null | Total number of relations $=2^{2^{2}}=2^{4}=16$
<br/><br/>
Relations that are symmetric as well as transitive are
<br/><br/>
$\phi,\{(x, x)\},\{(y, y)\},\{(x, x),(x, y),(y, y),(y, x)\},\{(x, x),(y, y)\}$
<br/><br/>
$\therefore \quad$ favourable cases $=5$
<br/><br/>
$\therefore \quad P_{r}=\frac{5}{16}$ | mcq | jee-main-2022-online-29th-june-evening-shift |
1l6givpji | maths | probability | venn-diagram-and-set-theory | <p>Let $$\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3}$$ be three mutually exclusive events such that $$\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{2+3 \mathrm{p}}{6}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{2-\mathrm{p}}{8}$$ and $$\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{1-\mathrm{p}}{2}$$. If the maximum and... | [{"identifier": "A", "content": "$$\\frac{2}{3}$$"}, {"identifier": "B", "content": "$$\\frac{5}{3}$$"}, {"identifier": "C", "content": "$$\\frac{5}{4}$$"}, {"identifier": "D", "content": "1"}] | ["B"] | null | <p>$$0 \le {{2 + 3P} \over 6} \le 1 \Rightarrow P \in \left[ { - {2 \over 3},{4 \over 3}} \right]$$</p>
<p>$$0 \le {{2 - P} \over 8} \le 1 \Rightarrow P \in [ - 6,2]$$</p>
<p>$$0 \le {{1 - P} \over 2} \le 1 \Rightarrow P \in [ - 1,1]$$</p>
<p>$$0 < P({E_1}) + P({E_2}) + P({E_3}) \le 1$$</p>
<p>$$0 < {{13} \over {12}} -... | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6p2wf0h | maths | probability | venn-diagram-and-set-theory | <p>Let $$S=\{1,2,3, \ldots, 2022\}$$. Then the probability, that a randomly chosen number n from the set S such that $$\mathrm{HCF}\,(\mathrm{n}, 2022)=1$$, is :</p> | [{"identifier": "A", "content": "$$\\frac{128}{1011}$$"}, {"identifier": "B", "content": "$$\\frac{166}{1011}$$"}, {"identifier": "C", "content": "$$\\frac{127}{337}$$"}, {"identifier": "D", "content": "$$\\frac{112}{337}$$"}] | ["D"] | null | <p>S = {1, 2, 3, .......... 2022}</p>
<p>HCF (n, 2022) = 1</p>
<p>$$\Rightarrow$$ n and 2022 have no common factor</p>
<p>Total elements = 2022</p>
<p>2022 = 2 $$\times$$ 3 $$\times$$ 337</p>
<p>M : numbers divisible by 2.</p>
<p>{2, 4, 6, ........, 2022}$$\,\,\,\,$$ n(M) = 1011</p>
<p>N : numbers divisible by 3.</p>
<... | mcq | jee-main-2022-online-29th-july-morning-shift |
1ldo6jbm8 | maths | probability | venn-diagram-and-set-theory | <p>Two dice are thrown independently. Let $$\mathrm{A}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is less than the number appeared on the $$2^{\text {nd }}$$ die, $$\mathrm{B}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is even and that on the second die is odd, an... | [{"identifier": "A", "content": "A and B are mutually exclusive"}, {"identifier": "B", "content": "the number of favourable cases of the events A, B and C are 15, 6 and 6 respectively"}, {"identifier": "C", "content": "B and C are independent"}, {"identifier": "D", "content": "the number of favourable cases of the even... | ["D"] | null | $\begin{aligned} & A=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\} \\\\ & n(A)=15 \\\\ & B=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\} \\\\ & n(B)=9 \\\\ & C=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\} \\\\ & n(C)=9\end{aligned}$
<br/><br/>$$
(... | mcq | jee-main-2023-online-1st-february-evening-shift |
1lsgb3d4o | maths | probability | venn-diagram-and-set-theory | <p>A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics and Chemistry. It was found that all students passed in atleast one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, atmost 11 students passed in both Mathematics ... | [] | null | 10 | $\begin{aligned} & n(M)=20 \\\\ & n(P)=25 \\\\ & n(C)=16 \\\\ & n(M \cap P)=11 \\\\ & n(P \cap C)=15 \\\\ & n(M \cap C)=15\end{aligned}$
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnvub5/791e93e6-65e8-4ec4-acf1-6dac1a0eab27/74eb5a10-cde5-11ee-a0d3-7b75c4537559/f... | integer | jee-main-2024-online-30th-january-morning-shift |
PuWjyRwOQzS1fiJs | maths | properties-of-triangle | area-of-triangle | If in a $$\Delta ABC$$, the altitudes from the vertices $$A, B, C$$ on opposite sides are in H.P, then $$\sin A,\sin B,\sin C$$ are in : | [{"identifier": "A", "content": "G. P."}, {"identifier": "B", "content": "A. P."}, {"identifier": "C", "content": "A.P-G.P."}, {"identifier": "D", "content": "H. P"}] | ["B"] | null | $$\Delta = {1 \over 2}{p_1}a = {1 \over 2}{p_2}b = {1 \over 2}{p_3}b$$
<br><br>$${p_1},{p_2},{p_3},$$ are in $$H.P.$$
<br><br>$$ \Rightarrow {{2\Delta } \over a},{{2\Delta } \over b},{{2\Delta } \over c}$$ are in $$H.P.$$
<br><br>$$ \Rightarrow {1 \over a},{1 \over b},{1 \over c},$$ are in $$H.P.$$
<br><br>$$ \Right... | mcq | aieee-2005 |
uF5vBTaoRp0w8iRS | maths | properties-of-triangle | area-of-triangle | In a $$\Delta PQR,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} $$ If $$3{\mkern 1mu} \sin {\mkern 1mu} P + 4{\mkern 1mu} \cos {\mkern 1mu} Q = 6$$ and $$4\sin Q + 3\cos P = 1,$$ then the angle R is equal to : | [{"identifier": "A", "content": "$${{5\\pi } \\over 6}$$ "}, {"identifier": "B", "content": "$${{\\pi } \\over 6}$$"}, {"identifier": "C", "content": "$${{\\pi } \\over 4}$$"}, {"identifier": "D", "content": "$${{3\\pi } \\over 4}$$"}] | ["B"] | null | Given $$3$$ $$\sin \,P + 4\cos Q = 6$$ $$\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$4\sin Q + 3\cos P = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Squaring and adding $$(i)$$ & $$(ii)$$ we get
<br><br>$$9\,{\sin ^2}P + 16{\cos ^2}Q + 24\sin P\cos Q$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,... | mcq | aieee-2012 |
fdmnaeLrvX4aMdxOwkjgy2xukf7gxjfo | maths | properties-of-triangle | area-of-triangle | A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If $$\angle BAC = {90^o}$$ and area$$\left( {\Delta ABC} \right) = 5\sqrt 5 $$ s units, then the abscissa of the vertex C is : | [{"identifier": "A", "content": "$$1 + 2\\sqrt 5 $$"}, {"identifier": "B", "content": "$$ 2\\sqrt 5 - 1$$"}, {"identifier": "C", "content": "$$1 + \\sqrt 5 $$"}, {"identifier": "D", "content": "$$2 + \\sqrt 5 $$"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264973/exam_images/fxiejzh5aeuzbgfxdpro.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267653/exam_images/matb2vixbiqzmc3xsv1z.webp"><source media="(max-wid... | mcq | jee-main-2020-online-4th-september-morning-slot |
JPpOTLK1LgUoYqsvMt1kluxiym4 | maths | properties-of-triangle | area-of-triangle | The triangle of maximum area that can be inscribed in a given circle of radius 'r' is : | [{"identifier": "A", "content": "An equilateral triangle having each of its side of length $$\\sqrt 3 $$r."}, {"identifier": "B", "content": "An equilateral triangle of height $${{2r} \\over 3}$$."}, {"identifier": "C", "content": "A right angle triangle having two of its sides of length 2r and r."}, {"identifier": "D"... | ["A"] | null | Area of triangle ABC<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266363/exam_images/t4j7msbaucclnkye1pro.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Properties of Triangle Quest... | mcq | jee-main-2021-online-26th-february-evening-slot |
d8zoYrCZDGgB8OMG3I1kmhzdcqe | maths | properties-of-triangle | area-of-triangle | Let ABCD be a square of side of unit length. Let a circle C<sub>1</sub> centered at A with unit radius is drawn. Another circle C<sub>2</sub> which touches C<sub>1</sub> and the lines AD and AB are tangent to it, is also drawn. Let a tangent line from the point C to the circle C<sub>2</sub> meet the side AB at E. If th... | [] | null | 1 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266007/exam_images/gqh1muo7zwcf1cjiz3fa.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Mathematics - Properties of Triangle Question 16 English Explanation"><br... | integer | jee-main-2021-online-16th-march-morning-shift |
ZzEH5xuVZGeyr66uaK1kmiznmbw | maths | properties-of-triangle | area-of-triangle | In $$\Delta$$ABC, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of $$\Delta$$ABC is 30 cm<sup>2</sup> and R and r are respectively the radii of circumcircle and incircle of $$\Delta$$ABC, then the value of 2R + r (in cm) is equal to ___________. | [] | null | 15 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264566/exam_images/c0o3n3zizv5saixvrv1t.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Properties of Triangle Question 15 English Explanation 1"><... | integer | jee-main-2021-online-16th-march-evening-shift |
1krzr2ib8 | maths | properties-of-triangle | area-of-triangle | If a rectangle is inscribed in an equilateral triangle of side length $$2\sqrt 2 $$ as shown in the figure, then the square of the largest area of such a rectangle is _____________.<br/><br/><img src="data:image/png;base64,UklGRsIFAABXRUJQVlA4ILYFAABQLQCdASoEAc4APm02mUgkIyKhJVh5aIANiWlu4WueXRnZ12fpF/gO1v/IdI05u52DuCPov... | [] | null | 3 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266129/exam_images/ouqduiix1vmfx5nsutac.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263970/exam_images/qsqsvhbym8ueyy7jh4jv.webp"><source media="(max-wid... | integer | jee-main-2021-online-25th-july-evening-shift |
1l57o61t8 | maths | properties-of-triangle | area-of-triangle | <p>The lengths of the sides of a triangle are 10 + x<sup>2</sup>, 10 + x<sup>2</sup> and 20 $$-$$ 2x<sup>2</sup>. If for x = k, the area of the triangle is maximum, then 3k<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9umae/dcd93d6b-ea7f-4626-9b1a-432648434cd3/99642f60-0655-11ed-903e-c9687588b3f3/file-1l5q9umaf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5q9umae/dcd93d6b-ea7f-4626-9b1a-432648434cd3/99642f60-0655-11ed-903e-c9687588b3f3... | mcq | jee-main-2022-online-27th-june-morning-shift |
1ldr5mq8n | maths | properties-of-triangle | area-of-triangle | <p>A straight line cuts off the intercepts $$\mathrm{OA}=\mathrm{a}$$ and $$\mathrm{OB}=\mathrm{b}$$ on the positive directions of $$x$$-axis and $$y$$ axis respectively. If the perpendicular from origin $$O$$ to this line makes an angle of $$\frac{\pi}{6}$$ with positive direction of $$y$$-axis and the area of $$\tria... | [{"identifier": "A", "content": "$$\\frac{392}{3}$$"}, {"identifier": "B", "content": "98"}, {"identifier": "C", "content": "196"}, {"identifier": "D", "content": "$$\\frac{196}{3}$$"}] | ["A"] | null | <p>$${1 \over 2}ab = {{98\sqrt 3 } \over 3}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leq0ppnt/4e6ea50d-9a35-4fc8-8434-5621ca292f68/d132e190-b85f-11ed-8195-4f3c56fa1eb5/file-1leq0ppnu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leq0ppnt/4e6ea50d-9a35-4fc8-8434... | mcq | jee-main-2023-online-30th-january-morning-shift |
1lgvqyw2i | maths | properties-of-triangle | area-of-triangle | <p>In the figure, $$\theta_{1}+\theta_{2}=\frac{\pi}{2}$$ and $$\sqrt{3}(\mathrm{BE})=4(\mathrm{AB})$$. If the area of $$\triangle \mathrm{CAB}$$ is $$2 \sqrt{3}-3$$ unit $${ }^{2}$$, when $$\frac{\theta_{2}}{\theta_{1}}$$ is the largest, then the perimeter (in unit) of $$\triangle \mathrm{CED}$$ is equal to _________.... | [] | null | 6 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnk6r0mj/2e1e2dc2-1bc0-46e6-95cc-c6b581c795c0/aac42bb0-6758-11ee-a06a-699a057b80c4/file-6y3zli1lnk6r0mk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnk6r0mj/2e1e2dc2-1bc0-46e6-95cc-c6b581c795c0/aac42bb0-6758-11ee-a0... | integer | jee-main-2023-online-10th-april-evening-shift |
jaoe38c1lseyc1v8 | maths | properties-of-triangle | circumcenter,-incenter-and-orthocenter | <p>Let $$\left(5, \frac{a}{4}\right)$$ be the circumcenter of a triangle with vertices $$\mathrm{A}(a,-2), \mathrm{B}(a, 6)$$ and $$C\left(\frac{a}{4},-2\right)$$. Let $$\alpha$$ denote the circumradius, $$\beta$$ denote the area and $$\gamma$$ denote the perimeter of the triangle. Then $$\alpha+\beta+\gamma$$ is</p> | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "62"}, {"identifier": "C", "content": "53"}, {"identifier": "D", "content": "30"}] | ["C"] | null | <p>$$\begin{aligned}
& A(a,-2), B(a, 6), C\left(\frac{a}{4},-2\right), O\left(5, \frac{a}{4}\right) \\
& A O=B O \\
& (a-5)^2+\left(\frac{a}{4}+2\right)^2=(a-5)^2+\left(\frac{a}{4}-6\right)^2 \\
& a=8 \\
& A B=8, A C=6, B C=10 \\
& \alpha=5, \beta=24, \gamma=24
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
luxwe7pg | maths | properties-of-triangle | circumcenter,-incenter-and-orthocenter | <p>Two vertices of a triangle $$\mathrm{ABC}$$ are $$\mathrm{A}(3,-1)$$ and $$\mathrm{B}(-2,3)$$, and its orthocentre is $$\mathrm{P}(1,1)$$. If the coordinates of the point $$\mathrm{C}$$ are $$(\alpha, \beta)$$ and the centre of the of the circle circumscribing the triangle $$\mathrm{PAB}$$ is $$(\mathrm{h}, \mathrm{... | [{"identifier": "A", "content": "81"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "51"}, {"identifier": "D", "content": "5"}] | ["D"] | null | <p>$$\begin{aligned}
& m_{P A}=\frac{2}{-2}=-1 \\
& \therefore \quad m_{B C}=1
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1m11ba/909a9c7c-64f8-4f41-9021-c0d2d0a9d656/7865b860-0f4f-11ef-ad52-af909612c772/file-1lw1m11bb.png?format=png" data-orsrc="https://app-content... | mcq | jee-main-2024-online-9th-april-evening-shift |
lv2erz6q | maths | properties-of-triangle | cosine-rule | <p>Consider a triangle $$\mathrm{ABC}$$ having the vertices $$\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$$ and $$\mathrm{C}(\gamma, \delta)$$ and angles $$\angle A B C=\frac{\pi}{6}$$ and $$\angle B A C=\frac{2 \pi}{3}$$. If the points $$\mathrm{B}$$ and $$\mathrm{C}$$ lie on the line $$y=x+4$$, then $$\alpha^2+\gamma^... | [] | null | 14 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhhd8gk/a86d447c-ca28-4c88-a2d0-2170f95b6e56/c85f5d40-1809-11ef-b156-f754785ad3ce/file-1lwhhd8gl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhhd8gk/a86d447c-ca28-4c88-a2d0-2170f95b6e56/c85f5d40-1809-11ef-b156-f754785ad3ce... | integer | jee-main-2024-online-4th-april-evening-shift |
lvb294yu | maths | properties-of-triangle | cosine-rule | <p>In a triangle $$\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N}$$ and $$\cos \mathrm{A}=\frac{2}{3}$$. If $$49 \cos (3 \mathrm{C})+42=\frac{\mathrm{m}}{\mathrm{n}}$$, where $$\operatorname{gcd}(m, n)=1$$, then $$m+n$$ is equal to _________.</p> | [] | null | 39 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwap0izi/fd6cee3f-e437-4a65-830c-d11ed7b1c7a4/3378e3e0-144e-11ef-860c-d121cbcdd1fc/file-1lwap0izj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwap0izi/fd6cee3f-e437-4a65-830c-d11ed7b1c7a4/3378e3e0-144e-11ef-860c-d121cbcdd1fc... | integer | jee-main-2024-online-6th-april-evening-shift |
L8Rjt2UM5OY6mLNw | maths | properties-of-triangle | ex-circle | In a triangle with sides $$a, b, c,$$ $${r_1} > {r_2} > {r_3}$$ (which are the ex-radii) then : | [{"identifier": "A", "content": "$$a>b>c$$"}, {"identifier": "B", "content": "$$a < b < c$$ "}, {"identifier": "C", "content": "$$a > b$$ and $$b < c$$ "}, {"identifier": "D", "content": "$$a < b$$ and $$b > c$$ "}] | ["A"] | null | $${r_1} > {r_2} > {r_3}$$
<br><br>$$ \Rightarrow {\Delta \over {s - a}} > {\Delta \over {s - b}} > {\Delta \over {s - c}};$$
<br><br>$$ \Rightarrow s - a < s - b < s - c$$
<br><br>$$ \Rightarrow - a < - b < - c$$
<br><br>$$ \Rightarrow a > b > c$$ | mcq | aieee-2002 |
R4DzEtYltigsfO3b | maths | properties-of-triangle | half-angle-formulae | The sum of the radii of inscribed and circumscribed circles for an $$n$$ sided regular polygon of side $$a, $$ is : | [{"identifier": "A", "content": "$${a \\over 4}\\cot \\left( {{\\pi \\over {2n}}} \\right)$$ "}, {"identifier": "B", "content": "$$a\\cot \\left( {{\\pi \\over {n}}} \\right)$$"}, {"identifier": "C", "content": "$${a \\over 2}\\cot \\left( {{\\pi \\over {2n}}} \\right)$$"}, {"identifier": "D", "content": "$$a\\cot \... | ["C"] | null | $$\tan \left( {{\pi \over n}} \right) = {a \over {2r}};\,\,\sin \left( {{\pi \over n}} \right) = {a \over {2R}}$$
<br><br>$$r + R = {a \over 2}\left[ {\cot {\pi \over n} + \cos ec{\pi \over n}} \right]$$
<br><br>
<br>$$ = {a \over 2}\left[ {{{\cos {\pi \over n} + 1} \over {\sin {\pi \over n}}}} \right]$$
<br><br... | mcq | aieee-2003 |
9HMUIzA8ObB5e9Yg | maths | properties-of-triangle | half-angle-formulae | If in a $$\Delta ABC$$ $$a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2},$$ then the sides $$a, b$$ and $$c$$ : | [{"identifier": "A", "content": "satisfy $$a+b=c$$"}, {"identifier": "B", "content": "are in A.P"}, {"identifier": "C", "content": "are in G.P "}, {"identifier": "D", "content": "are in H.P"}] | ["B"] | null | If $$a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2}$$
<br><br>$$a\left[ {\cos C + 1} \right] + c\left[ {\cos A + 1} \right] = 3b$$
<br><br>$$\left( {a + c} \right) + \left( {a\cos C + c\cos \,B} \right) = 3b$$
<br><br>$$a + c + b = 3b$$ or $$a + c = 2b$$
<br><br>or ... | mcq | aieee-2003 |
Y0Ghl4TgTp0GRL5s | maths | properties-of-triangle | half-angle-formulae | For a regular polygon, let $$r$$ and $$R$$ be the radii of the inscribed and the circumscribed circles. A $$false$$ statement among the following is : | [{"identifier": "A", "content": "There is a regular polygon with $${r \\over R} = {1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "There is a regular polygon with $${r \\over R} = {2 \\over 3}$$ "}, {"identifier": "C", "content": "There is a regular polygon with $${r \\over R} = {{\\sqrt 3 } \\over 2}$$ "},... | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265911/exam_images/oijmrfmejq2eg8qdfmzd.webp" loading="lazy" alt="AIEEE 2010 Mathematics - Properties of Triangle Question 27 English Explanation">
<br><br>If $$O$$ is center of polygon and
<br><br>$$AB$$ is one of the side, then b... | mcq | aieee-2010 |
lgnyq5za | maths | properties-of-triangle | half-angle-formulae | If the line $x=y=z$ intersects the line
<br/><br/>$x \sin A+y \sin B+z \sin C-18=0=x \sin 2 A+y \sin 2 B+z \sin 2 C-9$,
<br/><br/>where $A, B, C$ are the angles of a triangle $A B C$, then $80\left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)$
<br/><br/>is equal to ______________. | [] | null | 5 | $$
\begin{aligned}
&x= y=z=k(\text { let }) \\\\
&\therefore k(\sin A+\sin B+\sin C)=18 \\\\
&\Rightarrow k\left(4 \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}\right)=18 \\\\
& k(\sin 2 A+\sin 2 B+\sin 2 C)=9 \\\\
&\Rightarrow k(4 \sin A \cdot \sin B \cdot \sin C)=9 \ldots \text { (ii) } \\\\
& \text ... | integer | jee-main-2023-online-15th-april-morning-shift |
6TY7IytUNsnzYJJy | maths | properties-of-triangle | mediun-and-angle-bisector | In a triangle $$ABC$$, medians $$AD$$ and $$BE$$ are drawn. If $$AD=4$$,
<br/>$$\angle DAB = {\pi \over 6}$$ and $$\angle ABE = {\pi \over 3}$$, then the area of the $$\angle \Delta ABC$$ is : | [{"identifier": "A", "content": "$${{64} \\over 3}$$ "}, {"identifier": "B", "content": "$${8 \\over 3}$$ "}, {"identifier": "C", "content": "$${{16} \\over 3}$$ "}, {"identifier": "D", "content": "$${{32} \\over {3\\sqrt 3 }}$$ "}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264957/exam_images/mahs6riio0a6znddsa2r.webp" loading="lazy" alt="AIEEE 2003 Mathematics - Properties of Triangle Question 32 English Explanation">
<br><br>$$AP = {2 \over 3}AD = {8 \over 3};\,\,PD = {4 \over 3};\,\,$$
<br><br>Let $... | mcq | aieee-2003 |
aZw5wVknZ2tAwD4dSv3rsa0w2w9jxadfk62 | maths | properties-of-triangle | mediun-and-angle-bisector | A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (–1, 1) and (2, 3). Then the centroid of this triangle is : | [{"identifier": "A", "content": "$$\\left( {{1 \\over 3},2} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{1 \\over 3},{5 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {1,{7 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 3},1} \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265818/exam_images/u883z0iudbpn9rsvcpfy.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Properties of Triangle Question 20 English Explanation"><b... | mcq | jee-main-2019-online-12th-april-evening-slot |
2Pi5cD2rl2m48jUoBJ1klrkm8wt | maths | properties-of-triangle | mediun-and-angle-bisector | Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be $$\left( {{{10} \over 3},{7 \over 3}} \right)$$. If $$\alpha$$, $$\beta$$ are the roots of the equation $$a{x^2} + bx + 1 = 0$$, then the value of $${\alpha ^2} + {\beta ^2} - \alpha \beta $$ is : | [{"identifier": "A", "content": "$${{69} \\over {256}}$$"}, {"identifier": "B", "content": "$${{71} \\over {256}}$$"}, {"identifier": "C", "content": "$$ - {{71} \\over {256}}$$"}, {"identifier": "D", "content": "$$ - {{69} \\over {256}}$$"}] | ["C"] | null | 2b = a + c<br><br>$${{2a + 2} \over 3} = {{10} \over 3}$$ and $${{2b + c} \over 3} = {7 \over 3}$$<br><br>a = 4, <br><br>$$\left\{ \matrix{
2b + c = 7 \hfill \cr
2b - c = 4 \hfill \cr} \right\}$$, solving<br><br>$$b = {{11} \over 4}$$<br><br>$$c = {3 \over 2}$$<br><br>$$ \therefore $$ Quadratic Equation is $$4{x^... | mcq | jee-main-2021-online-24th-february-evening-slot |
oNsrvtAu54sUcSHeU7ufP | maths | properties-of-triangle | sine-rule | In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x<sup>2</sup> – c<sup>2</sup> = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is : | [{"identifier": "A", "content": "$${y \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${c \\over 3}$$"}, {"identifier": "C", "content": "$${c \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$y"}] | ["C"] | null | Given a + b = x and ab = y
<br><br>If x<sup>2</sup> $$-$$ c<sup>2</sup> = y $$ \Rightarrow $$ (a + b)<sup>2</sup> $$-$$ c<sup>2</sup> = ab
<br><br>$$ \Rightarrow $$ a<sup>2</sup> + b<sup>2</sup> $$-$$ c<sup>2</sup> = $$-$$ ab
<br><br>$$ \Rightarrow $$ $${{{a^2} + {b^2} - {c^2}} \over {2ab}}... | mcq | jee-main-2019-online-11th-january-morning-slot |
T5cYEycLTxY97vpIuZ3rsa0w2w9jx22m6nv | maths | properties-of-triangle | sine-rule | The angles A, B and C of a triangle ABC are in A.P. and a : b = 1 : $$\sqrt 3 $$. If c = 4 cm, then the area (in sq. cm)
of this triangle is : | [{"identifier": "A", "content": "2$$\\sqrt 3 $$"}, {"identifier": "B", "content": "4$$\\sqrt 3 $$"}, {"identifier": "C", "content": "$${4 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$"}] | ["A"] | null | From the question 2B = A + C & A + B + C = $$\pi $$<br><br>
$$ \Rightarrow $$ 3B = $$\pi $$<br><br>
$$ \Rightarrow $$ B = $${\pi \over 3}$$<br><br>
$$ \therefore A + C = {{2\pi } \over 3}\sigma $$<br><br>
$${a \over b} = {1 \over {\sqrt 3 }}$$<br><br>
$${{2R\sin A} \over {2R\sin B}} = {1 \over {\sqrt 3 }}$$<br><b... | mcq | jee-main-2019-online-10th-april-evening-slot |
1krpur0db | maths | properties-of-triangle | sine-rule | If in a triangle ABC, AB = 5 units, $$\angle B = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$ and radius of circumcircle of $$\Delta$$ABC is 5 units, then the area (in sq. units) of $$\Delta$$ABC is : | [{"identifier": "A", "content": "$$10 + 6\\sqrt 2 $$"}, {"identifier": "B", "content": "$$8 + 2\\sqrt 2 $$"}, {"identifier": "C", "content": "$$6 + 8\\sqrt 3 $$"}, {"identifier": "D", "content": "$$4 + 2\\sqrt 3 $$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l14vgj3y/f007ae64-29a9-4ba7-a06b-961129209c7b/fea546d0-ab5f-11ec-ba67-db8ddd9d738c/file-1l14vgj3z.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l14vgj3y/f007ae64-29a9-4ba7-a06b-961129209c7b/fea546d0-ab5f-11ec-ba67-db8ddd9d738c/fi... | mcq | jee-main-2021-online-20th-july-morning-shift |
1ktemdhtu | maths | properties-of-triangle | sine-rule | Let $${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$$, where A, B, C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then : | [{"identifier": "A", "content": "b<sup>2</sup> $$-$$ a<sup>2</sup> = a<sup>2</sup> + c<sup>2</sup>"}, {"identifier": "B", "content": "b<sup>2</sup>, c<sup>2</sup>, a<sup>2</sup> are in A.P."}, {"identifier": "C", "content": "c<sup>2</sup>, a<sup>2</sup>, b<sup>2</sup> are in A.P."}, {"identifier": "D", "content": "a<su... | ["B"] | null | $${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$$<br><br>As A, B, C are angles of triangle.<br><br>A + B + C = $$\pi$$<br><br>A = $$\pi$$ $$-$$ (B + C) ...... (1)<br><br>Similarly sinB = sin(A + C) ..... (2)<br><br>From (1) and (2)<br><br>$${{\sin (B + C)} \over {\sin (A + C)}} = {{\sin (A - C)} \ov... | mcq | jee-main-2021-online-27th-august-morning-shift |
sz85xbGPliuKueJlPT7k9k2k5irbfkf | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The number of real roots of the equation,
<br/>e<sup>4x</sup> + e<sup>3x</sup> – 4e<sup>2x</sup> + e<sup>x</sup> + 1 = 0 is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["A"] | null | e<sup>4x</sup> + e<sup>3x</sup> – 4e<sup>2x</sup> + e<sup>x</sup> + 1 = 0
<br><br>Dividing by e<sup>2x</sup>, we get
<br><br>e<sup>2x</sup> + e<sup>x</sup> - 4 + $${1 \over {{e^x}}}$$ + $${1 \over {{e^{2x}}}}$$ = 0
<br><br>$$ \Rightarrow $$ $$\left( {{e^{2x}} + {1 \over {{e^{2x}}}}} \right) + \left( {{e^x} + {1 \over {... | mcq | jee-main-2020-online-9th-january-morning-slot |
2aAWMYAQBGnzJe5QFG1kluhn0w9 | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The sum of 162<sup>th</sup> power of the roots of the equation x<sup>3</sup> $$-$$ 2x<sup>2</sup> + 2x $$-$$ 1 = 0 is ________. | [] | null | 3 | x<sup>3</sup> $$-$$ 2x<sup>2</sup> + 2x $$-$$ 1 = 0<br><br>x = 1 satisfying the equation<br><br>$$ \therefore $$ x $$-$$ 1 is factor of <br><br>x<sup>3</sup> $$-$$ 2x<sup>2</sup> + 2x $$-$$ 1<br><br>= (x $$-$$ 1) (x<sup>2</sup> $$-$$ x + 1) = 0<br><br>x = 1, $${{1 + i\sqrt 3 } \over 2},{{1 - i\sqrt 3 } \over 2}$$<br><b... | integer | jee-main-2021-online-26th-february-morning-slot |
1krw18wj6 | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The number of real roots of the equation $${e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0$$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $${e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0$$<br><br>$$ \Rightarrow {\left( {{e^{3x}} - 1} \right)^2} - {e^x}\left( {{e^{3x}} - 1} \right) = 12{e^{2x}}$$<br><br>$${\left( {{e^{3x}} - 1} \right)^2}\left( {{e^x} - {e^{ - x}} - {e^{ - 2x}}} \right) = 12$$<br><br>$$ \Rightarrow \underbrace {{e^x} - {e^{ ... | mcq | jee-main-2021-online-25th-july-morning-shift |
1krygctud | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The number of real roots of the equation e<sup>4x</sup> $$-$$ e<sup>3x</sup> $$-$$ 4e<sup>2x</sup> $$-$$ e<sup>x</sup> + 1 = 0 is equal to ______________. | [] | null | 2 | t<sup>4</sup> $$-$$ t<sup>3</sup> $$-$$ 4t<sup>2</sup> $$-$$ t + 1 = 0, e<sup>x</sup> = t > 0<br><br>$$ \Rightarrow {t^2} - t - 4 - {1 \over t} + {1 \over {{t^2}}} = 0$$<br><br>$$ \Rightarrow {\alpha ^2} - \alpha - 6 = 0,\alpha = t + {1 \over t} \ge 2$$<br><br>$$ \Rightarrow \alpha = 3, - 2$$ (reject)<br><br>$$ \... | integer | jee-main-2021-online-27th-july-evening-shift |
1ks09h31j | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | Let $$\alpha$$, $$\beta$$ be two roots of the <br/><br/>equation x<sup>2</sup> + (20)<sup>1/4</sup>x + (5)<sup>1/2</sup> = 0. Then $$\alpha$$<sup>8</sup> + $$\beta$$<sup>8</sup> is equal to | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "100"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "160"}] | ["C"] | null | x<sup>2</sup> + (20)<sup>1/4</sup>x + (5)<sup>1/2</sup> = 0
<br/><br/>$$ \Rightarrow $$ x<sup>2</sup> + $$\sqrt 5$$ = - (20)<sup>1/4</sup>x
<br/><br/>Squaring both sides, we get
<br/><br/>$${\left( {{x^2} + \sqrt 5 } \right)^2} = \sqrt {20} {x^2}$$<br><br>$$ \Rightarrow $$ x<sup>4</sup> = $$-$$5 $$\Rightarrow$$ x<sup>8... | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktk93082 | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The sum of the roots of the equation <br/><br/>$$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$$, is : | [{"identifier": "A", "content": "log<sub>2</sub> 14"}, {"identifier": "B", "content": "log<sub>2</sub> 11"}, {"identifier": "C", "content": "log<sub>2</sub> 12"}, {"identifier": "D", "content": "log<sub>2</sub> 13"}] | ["B"] | null | $$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$$<br><br>$${\log _2}({2^{x + 1}}) - {\log _2}{(3 + {2^x})^2} + {\log _2}(10 - {2^{ - x}}) = 0$$<br><br>$$lo{g_2}\left( {{{{2^{x + 1}}.(10 - {2^{ - x}})} \over {{{(3 + {2^x})}^2}}}} \right) = 0$$<br><br>$${{2({{10.2}^{ - x}} - 1)} \over {{{(3 + {2^x})}^2}... | mcq | jee-main-2021-online-31st-august-evening-shift |
1ktob9klc | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | Let f(x) be a polynomial of degree 3 such that<br/> $$f(k) = - {2 \over k}$$ for k = 2, 3, 4, 5. Then the value of 52 $$-$$ 10f(10) is equal to : | [] | null | 26 | $$k\,f(k) + 2 = \lambda (x - 2)(x - 3)(x - 4)(x - 5)$$ .... (1)<br><br>put x = 0<br><br>we get $$\lambda = {1 \over {60}}$$<br><br>Now, put $$\lambda$$ in equation (1)<br><br>$$ \Rightarrow kf(k) + 2 = {1 \over {60}}(x - 2)(x - 3)(x - 4)(x - 5)$$<br><br>Put x = 10<br><br>$$ \Rightarrow 10f(10) + 2 = {1 \over {60}}(8)(... | integer | jee-main-2021-online-1st-september-evening-shift |
1l54aj2cv | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>Let $$\alpha$$ be a root of the equation 1 + x<sup>2</sup> + x<sup>4</sup> = 0. Then, the value of $$\alpha$$<sup>1011</sup> + $$\alpha$$<sup>2022</sup> $$-$$ $$\alpha$$<sup>3033</sup> is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$\\alpha$$"}, {"identifier": "C", "content": "1 + $$\\alpha$$"}, {"identifier": "D", "content": "1 + 2$$\\alpha$$"}] | ["A"] | null | <p>Given, $$\alpha$$ is a root of the equation 1 + x<sup>2</sup> + x<sup>4</sup> = 0</p>
<p>$$\therefore$$ $$\alpha$$ will satisfy the equation.</p>
<p>$$\therefore$$ 1 + $$\alpha$$<sup>2</sup> + $$\alpha$$<sup>4</sup> = 0</p>
<p>$${\alpha ^2} = {{ - 1 \pm \sqrt {1 - 4} } \over 2}$$</p>
<p>$$ = {{ - 1 \pm \sqrt 3 i} \o... | mcq | jee-main-2022-online-29th-june-evening-shift |
1l57oztwx | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>If the sum of all the roots of the equation <br/><br/>$${e^{2x}} - 11{e^x} - 45{e^{ - x}} + {{81} \over 2} = 0$$ is $${\log _e}p$$, then p is equal to ____________.</p> | [] | null | 45 | Given that
<br/><br/>$$e^{2 x}-11 e^x-45 e^{-x}+\frac{81}{2}=0 $$
<br/><br/>$$\Rightarrow 2 e^{3 x}-22 e^{2 x}-90+81 e^x=0 $$
<br/><br/>$$\Rightarrow 2\left(e^x\right)^3-22\left(e^x\right)^2+81 e^x-90=0$$
<br/><br/>Let $ e^x=y$
<br/><br/>$$
\Rightarrow 2 y^3-22 y^2+81 y-90=0
$$
<br/><br/>Product of roots $\left(y_1, ... | integer | jee-main-2022-online-27th-june-morning-shift |
1l6duvpmp | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>If $$\alpha, \beta, \gamma, \delta$$ are the roots of the equation $$x^{4}+x^{3}+x^{2}+x+1=0$$, then $$\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$4"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}] | ["B"] | null | <p>When, $${x^5} = 1$$</p>
<p>then $${x^5} - 1 = 0$$</p>
<p>$$ \Rightarrow (x - 1)({x^4} + {x^3} + {x^2} + x + 1) = 0$$</p>
<p>Given, $${x^4} + {x^3} + {x^2} + x + 1 = 0$$ has roots $$\alpha$$, $$\beta$$, $$\gamma$$ and 8.</p>
<p>$$\therefore$$ Roots of $${x^5} - 1 = 0$$ are 1, $$\alpha$$, $$\beta$$, $$\gamma$$ and 8.<... | mcq | jee-main-2022-online-25th-july-morning-shift |
jaoe38c1lscn5p78 | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>If $$\alpha, \beta$$ are the roots of the equation, $$x^2-x-1=0$$ and $$S_n=2023 \alpha^n+2024 \beta^n$$, then :</p> | [{"identifier": "A", "content": "$$2 S_{12}=S_{11}+S_{10}$$\n"}, {"identifier": "B", "content": "$$S_{12}=S_{11}+S_{10}$$\n"}, {"identifier": "C", "content": "$$S_{11}=S_{10}+S_{12}$$\n"}, {"identifier": "D", "content": "$$2 S_{11}=S_{12}+S_{10}$$"}] | ["B"] | null | <p>$$\begin{aligned}
& x^2-x-1=0 \\
& S_n=2023 \alpha^n+2024 \beta^n \\
& S_{n-1}+S_{n-2}=2023 \alpha^{n-1}+2024 \beta^{n-1}+2023 \alpha^{n-2}+2024 \beta^{n-2} \\
& =2023 \alpha^{n-2}[1+\alpha]+2024 \beta^{n-2}[1+\beta] \\
& =2023 \alpha^{n-2}\left[\alpha^2\right]+2024 \beta^{n-2}\left[\beta^2\right] \\
& =2023 \alpha^... | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsfl1j5e | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>Let the set $$C=\left\{(x, y) \mid x^2-2^y=2023, x, y \in \mathbb{N}\right\}$$. Then $$\sum_\limits{(x, y) \in C}(x+y)$$ is equal to _________.</p> | [] | null | 46 | <p>First, let's consider the equation $$x^2 - 2^y = 2023$$ where $$x$$ and $$y$$ are natural numbers. Our goal is to find all the pairs $$(x, y)$$ that satisfy this equation and then sum the values of $$x+y$$ for each pair in set $$C$$.
<p>Since $$2023$$ is an odd number, and $$x^2$$, the square of any natural number,... | integer | jee-main-2024-online-29th-january-evening-shift |
lvc57nwy | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>Let $$x_1, x_2, x_3, x_4$$ be the solution of the equation $$4 x^4+8 x^3-17 x^2-12 x+9=0$$ and $$\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$$. Then the value of $$m$$ is _________.</p> | [] | null | 221 | <p>$$\begin{aligned}
& 4 x^4+8 x^3-17 x^2-12 x+9=0 \\
& (x+1)\left(4 x^3+4 x^2-21 x+9\right)=0 \\
& (x+1)(x+3)\left(4 x^2-8 x+3\right)=0 \\
& (x+1)(x+3)\left(4 x^2-6 x-2 x+3\right)=0 \\
& (x+1)(x+3)(2 x(2 x-3)-1(2 x-3))=0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@wid... | integer | jee-main-2024-online-6th-april-morning-shift |
i4NFcZin92dKvntr | maths | quadratic-equation-and-inequalities | common-roots | If one root of the equation $${x^2} + px + 12 = 0$$ is 4, while the equation $${x^2} + px + q = 0$$ has equal roots,
<br/>then the value of $$'q'$$ is | [{"identifier": "A", "content": "4 "}, {"identifier": "B", "content": "12 "}, {"identifier": "C", "content": "3 "}, {"identifier": "D", "content": "$${{49} \\over 4}$$ "}] | ["D"] | null | $$4$$ is a root of $${x^2} + px + 12 = 0$$
<br><br>$$ \Rightarrow 16 + 4p + 12 = 0$$
<br><br>$$ \Rightarrow p = - 7$$
<br><br>Now, the equation $${x^2} + px + q = 0$$
<br><br>has equal roots.
<br><br>$$\therefore$$ $${p^2} - 4q = 0$$ $$ \Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$ | mcq | aieee-2004 |
Y1ywUGSBBkETN6q8 | maths | quadratic-equation-and-inequalities | common-roots | The quadratic equations $${x^2} - 6x + a = 0$$ and $${x^2} - cx + 6 = 0$$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is | [{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "4 "}, {"identifier": "C", "content": "3 "}, {"identifier": "D", "content": "2"}] | ["D"] | null | Let the roots of equation $${x^2} - 6x + a = 0$$ be $$\alpha $$
<br><br>and $$4$$ $$\beta $$ and that of the equation
<br><br>$${x^2} - cx + 6 = 0$$ be $$\alpha $$ and $$3\beta .$$ Then
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a$$
<br><br>and $$\,\,\... | mcq | aieee-2008 |
hpSOJ5blAQVflwER | maths | quadratic-equation-and-inequalities | common-roots | If the equations $${x^2} + 2x + 3 = 0$$ and $$a{x^2} + bx + c = 0,$$ $$a,\,b,\,c\, \in \,R,$$ have a common root, then $$a\,:b\,:c\,$$ is | [{"identifier": "A", "content": "$$1:2:3$$ "}, {"identifier": "B", "content": "$$3:2:1$$"}, {"identifier": "C", "content": "$$1:3:2$$"}, {"identifier": "D", "content": "$$3:1:2$$"}] | ["A"] | null | Given equations are
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 2x + 3 = 0\,\,\,\,\,...\left( i \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,a{x^2} + bx + c = 0\,\,\,...\left( {ii} \right)$$
<br><br>Roots of equation $$(i)$$ are imaginary roots.
<br><br>According to the question $$(ii)$$ will also have both roots same a... | mcq | jee-main-2013-offline |
pHnNgwGeazDuA1PHdH1JN | maths | quadratic-equation-and-inequalities | common-roots | If the equations x<sup>2</sup> + bx−1 = 0 and x<sup>2</sup> + x + b = 0 have a common root different from −1, then $$\left| b \right|$$ is equal to : | [{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\sqrt 3 $$"}] | ["D"] | null | Given,
<br><br>x<sup>2</sup> + bx $$-$$ 1 = 0 . . . . .(1)
<br><br>and x<sup>2</sup> + x + b = 0 . . . . . (2)
<br><br>Performing (1) $$-$$ (2) we get,
<br><br>bx $$-$$ 1 $$-$$ x $$-$$ b = 0
<br><br>$$ \Rightarrow $$ x(b $$-$$ 1) = b + 1
<br><br>$$ \Rightarrow $$ x = $${{b + 1} \over... | mcq | jee-main-2016-online-9th-april-morning-slot |
2ZbFQGhj0vnf1RUxo53rsa0w2w9jxaoprqa | maths | quadratic-equation-and-inequalities | common-roots | If $$\alpha $$, $$\beta $$ and $$\gamma $$ are three consecutive terms of a non-constant G.P. such that the equations $$\alpha $$x
<sup>2</sup>
+ 2$$\beta $$x + $$\gamma $$ = 0 and
x<sup>2</sup>
+ x – 1 = 0 have a common root, then $$\alpha $$($$\beta $$ + $$\gamma $$) is equal to : | [{"identifier": "A", "content": "$$\\alpha $$$$\\gamma $$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$\\beta $$$$\\gamma $$"}, {"identifier": "D", "content": "$$\\alpha $$$$\\beta $$"}] | ["C"] | null | Let the common ratio of G.P is <b>r</b><br><br>
Then the equation of $$\alpha {x^2} + 2\beta x + \gamma = 0$$<br><br>
$$ \Rightarrow $$ $$\alpha {x^2} + 2\alpha rx + \alpha {r^2} = 0$$<br><br>
$$ \Rightarrow $$ x<sup>2</sup> + 2rx + r<sup>2</sup> = 0 ........<b>(i)</b><br>
Equation (i) and x<sup>2</sup> + x - 1 = 0 .... | mcq | jee-main-2019-online-12th-april-evening-slot |
zmYqIAqG6Q4t0K97bk7k9k2k5k7hec6 | maths | quadratic-equation-and-inequalities | common-roots | Let a, b $$ \in $$ R, a $$ \ne $$ 0 be such that the equation,
ax<sup>2</sup> – 2bx + 5 = 0 has a repeated root $$\alpha $$, which
is also a root of the equation, x<sup>2</sup> – 2bx – 10 = 0.
If $$\beta $$ is the other root of this equation, then
$$\alpha $$<sup>2</sup> + $$\beta $$<sup>2</sup> is equal to : | [{"identifier": "A", "content": "28"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "26"}, {"identifier": "D", "content": "25"}] | ["D"] | null | Roots of equation ax<sup>2</sup>
– 2bx + 5 = 0 are $$\alpha $$, $$\alpha $$.
<br><br>$$ \therefore $$ $$\alpha $$ + $$\alpha $$ = $${{2b} \over a}$$
<br><br>$$ \Rightarrow $$ 2$$\alpha $$ = $${{2b} \over a}$$
<br><br>$$ \Rightarrow $$ $$\alpha $$ = $${{b} \over a}$$ ....(1)
<br><br>and $$\alpha $$<sup>2</sup> = $${5 \o... | mcq | jee-main-2020-online-9th-january-evening-slot |
1ktd4mp2o | maths | quadratic-equation-and-inequalities | common-roots | Let $$\lambda$$ $$\ne$$ 0 be in R. If $$\alpha$$ and $$\beta$$ are the roots of the equation x<sup>2</sup> $$-$$ x + 2$$\lambda$$ = 0, and $$\alpha$$ and $$\gamma$$ are the roots of equation 3x<sup>2</sup> $$-$$ 10x + 27$$\lambda$$ = 0, then $${{\beta \gamma } \over \lambda }$$ is equal to ____________. | [] | null | 18 | 3$$\alpha$$<sup>2</sup> $$-$$ 10$$\alpha$$ + 27$$\lambda$$ = 0 ..... (1)<br><br>$$\alpha$$<sup>2</sup> $$-$$ $$\alpha$$ + 2$$\lambda$$ = 0 ...... (2)<br><br>(1) $$-$$ 3(2) gives<br><br>$$-$$7$$\alpha$$ + 21$$\lambda$$ = 0 $$\Rightarrow$$ $$\alpha$$ = 3$$\lambda$$<br><br>Put $$\alpha$$ = 3$$\lambda$$ in equation (1) we ... | integer | jee-main-2021-online-26th-august-evening-shift |
1l59jpipu | maths | quadratic-equation-and-inequalities | common-roots | <p>Let a, b $$\in$$ R be such that the equation $$a{x^2} - 2bx + 15 = 0$$ has a repeated root $$\alpha$$. If $$\alpha$$ and $$\beta$$ are the roots of the equation $${x^2} - 2bx + 21 = 0$$, then $${\alpha ^2} + {\beta ^2}$$ is equal to :</p> | [{"identifier": "A", "content": "37"}, {"identifier": "B", "content": "58"}, {"identifier": "C", "content": "68"}, {"identifier": "D", "content": "92"}] | ["B"] | null | <p>$$a{x^2} - 2bx + 15 = 0$$ has repeated root so $${b^2} = 15a$$ and $$\alpha = {{15} \over b}$$</p>
<p>$$\because$$ $$\alpha$$ is a root of $${x^2} - 2bx + 21 = 0$$</p>
<p>So $${{225} \over {{b^2}}} = 9 \Rightarrow {b^2} = 25$$</p>
<p>Now $${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta = 4{b^2} ... | mcq | jee-main-2022-online-25th-june-evening-shift |
1l6gj9sdd | maths | quadratic-equation-and-inequalities | common-roots | <p>If for some $$\mathrm{p}, \mathrm{q}, \mathrm{r} \in \mathbf{R}$$, not all have same sign, one of the roots of the equation $$\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right) x^{2}-2 \mathrm{q}(\mathrm{p}+\mathrm{r}) x+\mathrm{q}^{2}+\mathrm{r}^{2}=0$$ is also a root of the equation $$x^{2}+2 x-8=0$$, then $$\frac{\mathrm... | [] | null | 272 | <p>Let roots of</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbuiql/3cbb58ae-c17d-4fc8-9ae2-606226e918d5/7db780d0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbuiqm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbuiql/3cbb58ae-c17d-4fc8-9ae2-606226e918d5/7db780d0-2c4f-1... | integer | jee-main-2022-online-26th-july-morning-shift |
ldr04kaw | maths | quadratic-equation-and-inequalities | common-roots | If the value of real number $a>0$ for which $x^2-5 a x+1=0$ and $x^2-a x-5=0$
<br/><br/>have a common real root is $\frac{3}{\sqrt{2 \beta}}$ then $\beta$ is equal to ___________. | [] | null | 13 | <p>$${x^2} - 5\alpha x + 1 = 0$$ ..... (1)</p>
<p>$${x^2} - \alpha x - 5 = 0$$ ...... (2)</p>
<p>have a common root.</p>
<p>Subtracting (1) with (2) we'll get $$x = {6 \over {4\alpha }}$$</p>
<p>Substituting in (1)</p>
<p>$${{36} \over {16{\alpha ^2}}} - {{30} \over 4} + 1 = 0$$</p>
<p>$$ \Rightarrow {\alpha ^2} = {9 \... | integer | jee-main-2023-online-30th-january-evening-shift |
Jkj398nu3J6NteP2 | maths | quadratic-equation-and-inequalities | graph-and-sign-of-quadratic-expression | If both the roots of the quadratic equation $${x^2} - 2kx + {k^2} + k - 5 = 0$$ are less than 5, then $$k$$ lies in the interval | [{"identifier": "A", "content": "$$\\left( {5,6} \\right]$$ "}, {"identifier": "B", "content": "$$\\left( {6,\\,\\infty } \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { - \\infty ,\\,4} \\right)$$ "}, {"identifier": "D", "content": "$$\\left[ {4,\\,5} \\right]$$ "}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267782/exam_images/zwhawncrjdcscooeyd20.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Quadratic Equation and Inequalities Question 154 English Explanation">
<br><br>both roots are less than $$5,$$
<br><br>then $$(i)$$ Discrimi... | mcq | aieee-2005 |
krncAZPJvsT4aI3N | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | If $$a \in R$$ and the equation $$ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$$ (where [$$x$$] denotes the greater integer $$ \le x$$) has no integral solution, then all possible values of a lie in the interval : | [{"identifier": "A", "content": "$$\\left( { - 2, - 1} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - \\infty , - 2} \\right) \\cup \\left( {2,\\infty } \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { - 1,0} \\right) \\cup \\left( {0,1} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {... | ["C"] | null | Given, $$ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$$
<br>As we know, $$\left[ x \right] + \left\{ x \right\} = x$$
<br>where $$\left[ x \right]$$ is integral part and $$\left\{ x \right\}$$ is fractional part.
<br>$$\therefore$$$$\left\{ x \right\} = x - \left[ x... | mcq | jee-main-2014-offline |
t5TJfuWEpRBw5JFGnLjgy2xukf7fkxwm | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | Let [t] denote the greatest integer $$ \le $$ t. Then the equation in x,
<br/>[x]<sup>2</sup> + 2[x+2] - 7 = 0 has : | [{"identifier": "A", "content": "no integral solution."}, {"identifier": "B", "content": "exactly two solutions."}, {"identifier": "C", "content": "exactly four integral solutions."}, {"identifier": "D", "content": "infinitely many solutions.\n"}] | ["D"] | null | $${[x]^2} + 2[x + 2] - 7 = 0$$
<br><br>$$ \Rightarrow $$ $${[x]^2} + 2[x] + 4 - 7 = 0$$
<br><br>Using the property [x + n] = [x] + n ; n $$ \in $$ I
<br><br>$$ \Rightarrow $$ $${[x]^2} + 2[x] - 3 = 0$$<br><br>let [x] = y<br><br>$${y^2} + 3y - y - 3 = 0$$<br><br>$$ \Rightarrow $$ $$(y - 1)(y + 3) = 0$$<br><br>$$[x] = 1... | mcq | jee-main-2020-online-4th-september-morning-slot |
1kru42urc | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | 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 : | [{"identifier": "A", "content": "$$\\left[ {0,{1 \\over e}} \\right)$$"}, {"identifier": "B", "content": "[log<sub>e</sub>2, log<sub>e</sub>3)"}, {"identifier": "C", "content": "[1, e)"}, {"identifier": "D", "content": "[0, log<sub>e</sub>2)"}] | ["D"] | null | $${[{e^x}]^2} + [{e^x} + 1] - 3 = 0$$<br><br>$$ \Rightarrow {[{e^x}]^2} + [{e^x}] + 1 - 3 = 0$$<br><br>Let $$[{e^x}] = t$$<br><br>$$ \Rightarrow {t^2} + t - 2 = 0$$<br><br>$$ \Rightarrow t = - 2,1$$<br><br>$$[{e^x}] = - 2$$ (Not possible)<br><br>or $$[{e^x}] = 1$$ $$\therefore$$ $$1 \le {e^x} < 2$$<br><br>$$ \Righ... | mcq | jee-main-2021-online-22th-july-evening-shift |
1krzmnrse | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | If [x] be the greatest integer less than or equal to x, <br/><br/>then $$\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]} $$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$2"}, {"identifier": "D", "content": "2"}] | ["B"] | null | $$\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]} $$<br><br>= [4] + [-4.5] + [5] + [-5.5] + [6] +..... + [-49.5] + [50]
<br><br>= 4 - 5 + 5 - 6 + 6 ......-50 + 50
<br><br> = 4 | mcq | jee-main-2021-online-25th-july-evening-shift |
1ldybbre4 | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | <p>The equation $${x^2} - 4x + [x] + 3 = x[x]$$, where $$[x]$$ denotes the greatest integer function, has :</p> | [{"identifier": "A", "content": "exactly two solutions in ($$-\\infty,\\infty$$)"}, {"identifier": "B", "content": "no solution"}, {"identifier": "C", "content": "a unique solution in ($$-\\infty,\\infty$$)"}, {"identifier": "D", "content": "a unique solution in ($$-\\infty,1$$)"}] | ["C"] | null | <p>$${x^2} - 4x + [x] + 3 = x[x]$$</p>
<p>$$ \Rightarrow {x^2} - 4x + [x] + 3 - x[x] = 0$$</p>
<p>$$ \Rightarrow (x - 1)(x - 3) - [x](x - 1) = 0$$</p>
<p>$$ \Rightarrow (x - 1)(x - [x] - 3) = 0$$</p>
<p>$$\therefore$$ $$x = 1$$</p>
<p>or</p>
<p>$$x - [x] - 3 = 0$$</p>
<p>$$ \Rightarrow \{ x\} - 3 = 0$$ [As $$\{ x\} =... | mcq | jee-main-2023-online-24th-january-morning-shift |
1lgoy6y4d | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | <p>Let $$[\alpha]$$ denote the greatest integer $$\leq \alpha$$. Then $$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[\sqrt{120}]$$ is equal to __________</p> | [] | null | 825 | $$
\begin{aligned}
& {[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots .+[120]} \\\\
& E=1+1+1+2+2+2+2+2+3+3+3+3+3 \\\\
& +3+3+4+4+\ldots \\\\
& E=3 \times 1+5 \times 2+7 \times 3+\ldots .+19 \times 9+10 \times 21 \\\\
& =\sum_{r=1}^{10}(2 r+1) r=2\left[\frac{10 \times 11 \times 21}{6}\right]+\frac{10 \times 11}{2} \\\\
& =770+... | integer | jee-main-2023-online-13th-april-evening-shift |
1lh1zzike | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | <p>Let $$A = \{ x \in R:[x + 3] + [x + 4] \le 3\} ,$$
<br/><br/>$$B = \left\{ {x \in R:{3^x}{{\left( {\sum\limits_{r = 1}^\infty {{3 \over {{{10}^r}}}} } \right)}^{x - 3}} < {3^{ - 3x}}} \right\},$$ where [t] denotes greatest integer function. Then,</p> | [{"identifier": "A", "content": "$$B \\subset C,A \\ne B$$"}, {"identifier": "B", "content": "$$A \\subset B,A \\ne B$$"}, {"identifier": "C", "content": "$$A = B$$"}, {"identifier": "D", "content": "$$A \\cap B = \\phi $$"}] | ["C"] | null | We have,
<br/><br/>$$
\begin{aligned}
& A=\{x \in R:[x+3]+[x+4] \leq 3\} \\\\
& \text { Here, }[x+3]+[x+4] \leq 3 \\\\
& \Rightarrow [x]+3+[x]+4 \leq 3 \\\\
& (\because[x+n]=[x]+n, n \in I) \\\\
& \Rightarrow 2[x]+4 \leq 0 \Rightarrow[x] \leq-2 \\\\
& \Rightarrow x \in(-\infty,-1) \\\\
& A \equiv(-\infty,-1) ............ | mcq | jee-main-2023-online-6th-april-morning-shift |
GvxG3TohU0ETSsR9 | maths | quadratic-equation-and-inequalities | inequalities | If $$a,\,b,\,c$$ are distinct $$ + ve$$ real numbers and $${a^2} + {b^2} + {c^2} = 1$$ then $$ab + bc + ca$$ is | [{"identifier": "A", "content": "less than 1 "}, {"identifier": "B", "content": "equal to 1 "}, {"identifier": "C", "content": "greater than 1 "}, {"identifier": "D", "content": "any real no."}] | ["A"] | null | As $$\,\,\,\,\,{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} > 0$$
<br><br>$$ \Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0$$
<br><br>$$ \Rightarrow 2 > 2\left( {ab + bc + ca} \right)$$
<br><br>$$ \Rightarrow ab + bc + ca < 1$$ | mcq | aieee-2002 |
1hqxM4JgTHL47dSP | maths | quadratic-equation-and-inequalities | inequalities | <b>STATEMENT - 1 :</b> For every natural number $$n \ge 2,$$
$$${1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + ........ + {1 \over {\sqrt n }} > \sqrt n .$$$
<p><b>STATEMENT - 2 :</b> For every natural number $$n \ge 2,$$,
$$$\sqrt {n\left( {n + 1} \right)} < n + 1.$$$</p>
| [{"identifier": "A", "content": "Statement - 1 is false, Statement - 2 is true "}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for statement - 1"}, {"identifier": "C", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is not ... | ["B"] | null | Statements $$2$$ is $$\sqrt {n\left( {n + 1} \right)} < n + 1,n \ge 2$$
<br><br>$$ \Rightarrow \sqrt n < \sqrt {n + 1} ,n \ge 2$$ which is true
<br><br>$$ \Rightarrow \sqrt 2 < \sqrt 3 < \sqrt 4 < \sqrt 5 < - - - - - - \sqrt n $$
<br><br>Now $$\sqrt 2 < \sqrt n \Rightarrow {1 \over {... | mcq | aieee-2008 |
4ho1wNjg6e6O80jXth3rsa0w2w9jwxdf2pn | maths | quadratic-equation-and-inequalities | inequalities | All the pairs (x, y) that satisfy the inequality
<br/>$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$
<br/>also satisfy the equation | [{"identifier": "A", "content": "sin x = |sin y|"}, {"identifier": "B", "content": "sin x = 2sin y"}, {"identifier": "C", "content": "2 sin x = sin y"}, {"identifier": "D", "content": "2 |sin x | = 3 sin y"}] | ["A"] | null | $${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }} \le {2^{2{{\sin }^2}y}}$$<br><br>
$$ \Rightarrow $$ $$\sqrt {{{\sin }^2}x - 2\sin x + 5} \le 2{\sin ^2}y$$<br><br>
$$ \Rightarrow \sqrt {{{\left( {\sin x - 1} \right)}^2} + 4} \le 2{\sin ^2}y$$<br><br>
it is true when sinx = 1, |siny| = 1<br><br>
so sinx = |siny|
| mcq | jee-main-2019-online-10th-april-morning-slot |
VdoWy5AixI4ojhT7Ph1klrgdqnh | maths | quadratic-equation-and-inequalities | inequalities | Let p and q be two positive numbers such that p + q = 2 and p<sup>4</sup>+q<sup>4</sup> = 272. Then p and q are
roots of the equation : | [{"identifier": "A", "content": "x<sup>2</sup> \u2013 2x + 8 = 0"}, {"identifier": "B", "content": "x<sup>2</sup> - 2x + 136=0"}, {"identifier": "C", "content": "x<sup>2</sup> \u2013 2x + 16 = 0"}, {"identifier": "D", "content": "x<sup>2</sup> \u2013 2x + 2 = 0"}] | ["C"] | null | $${p^2} + {q^2} = {(p + q)^2} - 2pq$$<br><br>$$ = 4 - 2pq$$<br><br>Now, $${\left( {{p^2} + {q^2}} \right)^2} = {p^4} + {q^4} + 2{p^2}{q^2}$$<br><br>$$ \Rightarrow {\left( {4 - 2pq} \right)^2} = 272 + 2{p^2}{q^2}$$<br><br>$$ \Rightarrow 16 + 4{p^2}{q^2} - 16pq = 272 + 2{p^2}{q^2}$$<br><br>$$ \Rightarrow 2{p^2}{q^2} - 16... | mcq | jee-main-2021-online-24th-february-morning-slot |
ABe8PX8aWeMsgSLkwa1kls48put | maths | quadratic-equation-and-inequalities | inequalities | The integer 'k', for which the inequality x<sup>2</sup> $$-$$ 2(3k $$-$$ 1)x + 8k<sup>2</sup> $$-$$ 7 > 0 is valid for every x in R, is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "0"}] | ["C"] | null | $${x^2} - 2(3k - 1)x + 8{k^2} - 7 > 0$$<br><br>Now, D < 0<br><br>$$ \Rightarrow 4{(3k - 1)^2} - 4 \times 1 \times (8{k^2} - 7) < 0$$<br><br>$$ \Rightarrow 9{k^2} - 6k + 1 - 8{k^2} + 7 < 0$$<br><br>$$ \Rightarrow {k^2} - 6k + 8 < 0$$<br><br>$$ \Rightarrow (k - 4)(k - 2) < 0$$<br><br>2 < k < 4
<br... | mcq | jee-main-2021-online-25th-february-morning-slot |
1krzrgqrk | maths | quadratic-equation-and-inequalities | inequalities | If a + b + c = 1, ab + bc + ca = 2 and abc = 3, then the value of a<sup>4</sup> + b<sup>4</sup> + c<sup>4</sup> is equal to ______________. | [] | null | 13 | (a + b + c)<sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ a<sup>2</sup>
+ b<sup>2</sup>
+ c<sup>2</sup>
+ 2(ab + bc + ca) = 1
<br><br>$$ \Rightarrow $$ a<sup>2</sup> + b<sup>2</sup>
+ c<sup>2</sup> = – 3 ….(i)
<br><br>$$ \Rightarrow $$ ab + bc + ca = 2 ….(ii)
<br><br>Squaring of equation (ii),
<br><br>$$ \Rightarrow $... | integer | jee-main-2021-online-25th-july-evening-shift |
1l5vz0zg9 | maths | quadratic-equation-and-inequalities | inequalities | <p>Let $${S_1} = \left\{ {x \in R - \{ 1,2\} :{{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0} \right\}$$ and $${S_2} = \left\{ {x \in R:{3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0} \right\}$$. Then, $${S_1} \cup {S_2}$$ is equal to :</p> | [{"identifier": "A", "content": "$$( - \\infty , - 2] \\cup (1,2)$$"}, {"identifier": "B", "content": "$$( - \\infty , - 2] \\cup [1,2]$$"}, {"identifier": "C", "content": "$$( - 2,1] \\cup [2,\\infty )$$"}, {"identifier": "D", "content": "$$( - \\infty ,2]$$"}] | ["B"] | null | <p>Given,</p>
<p>$${{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0$$</p>
<p>$${x^2} + 3x + 5$$ is a quadratic equation</p>
<p>$$a = 1 > 0$$ and $$D = {( - 3)^2} - 4\,.\,1\,.\,5 = - 11 < 0$$</p>
<p>$$\therefore$$ $${x^2} + 3x + 5 > 0$$ (always)</p>
<p>So, we can ignore this quadratic term</p>
<p>$$... | mcq | jee-main-2022-online-30th-june-morning-shift |
jaoe38c1lse4uha6 | maths | quadratic-equation-and-inequalities | inequalities | <p>Let $$\mathrm{S}$$ be the set of positive integral values of $$a$$ for which $$\frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32} < 0, \forall x \in \mathbb{R}$$. Then, the number of elements in $$\mathrm{S}$$ is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$\\infty$$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $x^2-8 x+32>0 \forall x \in R$ as discriminant of this quadratic is $64-4 \times 32<0$
<br/><br/>$$
\Rightarrow a x^2+2(a+1) x+9 a+4<0 \forall x \in R
$$
<br/><br/>$\Rightarrow$ Only possible when $a<0$ and $D<0$
<br/><br/>$\Rightarrow$ Since $S$ is set of positive
values of $a \Rightarrow S$ is a null set
<br/><br/>$$... | mcq | jee-main-2024-online-31st-january-morning-shift |
pQ8kOFxyh1NO0kIK | maths | quadratic-equation-and-inequalities | location-of-roots | All the values of $$m$$ for which both roots of the equation $${x^2} - 2mx + {m^2} - 1 = 0$$ are greater than $$ - 2$$ but less then 4, lie in the interval | [{"identifier": "A", "content": "$$ - 2 < m < 0$$ "}, {"identifier": "B", "content": "$$m > 3$$ "}, {"identifier": "C", "content": "$$ - 1 < m < 3$$ "}, {"identifier": "D", "content": "$$1 < m < 4$$ "}] | ["C"] | null | Equation $${x^2} - 2mx + {m^2} - 1 = 0$$
<br><br>$${\left( {x - m} \right)^2} - 1 = 0$$
<br><br>or $$\left( {x - m + 1} \right)\left( {x - m - 1} \right) = 0$$
<br><br>$$x = m - 1,m + 1$$
<br><br>$$m - 1 > - 2$$ and $$m + 1 < 4$$
<br><br>$$ \Rightarrow m > - 1$$ and $$m<3$$
<br><br>or $$\,\,\, - 1 < m... | mcq | aieee-2006 |
pX6k4PzHt3u5DKh2xhfDB | maths | quadratic-equation-and-inequalities | location-of-roots | If both the roots of the quadratic equation x<sup>2</sup> $$-$$ mx + 4 = 0 are real and distinct and they lie in the interval [1, 5], then m lies in the interval : | [{"identifier": "A", "content": "($$-$$5, $$-$$4)"}, {"identifier": "B", "content": "(4, 5)"}, {"identifier": "C", "content": "(5, 6)"}, {"identifier": "D", "content": "(3, 4)"}] | ["B"] | null | x<sup>2</sup> $$-$$mx + 4 = 0
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264994/exam_images/bkaah5pafcac92p7jsty.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Mathematics - Quadratic Equation ... | mcq | jee-main-2019-online-9th-january-evening-slot |
mSNXXAHDAMqkcYrI59bPZ | maths | quadratic-equation-and-inequalities | location-of-roots | Consider the quadratic equation (c – 5)x<sup>2</sup> – 2cx + (c – 4) = 0, c $$ \ne $$ 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is - | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "11"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267172/exam_images/rziw7accnofwe87kl8ml.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Mathematics - Quadratic Equation and Inequalities Question 122 English... | mcq | jee-main-2019-online-10th-january-morning-slot |
NjnfwqqKfFebUKmJODjgy2xukf44aor4 | maths | quadratic-equation-and-inequalities | location-of-roots | The set of all real values of $$\lambda $$ for which the
quadratic equations, <br/>($$\lambda $$<sup>2</sup>
+ 1)x<sup>2</sup>
– 4$$\lambda $$x + 2 = 0
always have exactly one root in the interval
(0, 1) is : | [{"identifier": "A", "content": "(\u20133, \u20131)"}, {"identifier": "B", "content": "(2, 4]"}, {"identifier": "C", "content": "(0, 2)"}, {"identifier": "D", "content": "(1, 3]"}] | ["D"] | null | Given quadratic equation,<br><br>$$({\lambda ^2} + 1){x^2} - 4\lambda x + 2 = 0$$<br><br>Here coefficient of x<sup>2</sup> is ($$\lambda $$<sup>2</sup> + 1) which is always positive. So quadratic equation is upward parabola.<br><br><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy... | mcq | jee-main-2020-online-3rd-september-evening-slot |
1ldo68yjn | maths | quadratic-equation-and-inequalities | location-of-roots | <p>The number of integral values of k, for which one root of the equation $$2x^2-8x+k=0$$ lies in the interval (1, 2) and its other root lies in the interval (2, 3), is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5hv330/3bf12c29-312e-4d43-a5a3-c05d3e6a864c/6c2ba9c0-ad16-11ed-8a8c-4d67f5492755/file-1le5hv331.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5hv330/3bf12c29-312e-4d43-a5a3-c05d3e6a864c/6c2ba9c0-ad16-11ed-8a8c-4d67f5492755/fi... | mcq | jee-main-2023-online-1st-february-evening-shift |
J4uBIzApDyw2Lggl | maths | quadratic-equation-and-inequalities | modulus-function | Product of real roots of equation $${t^2}{x^2} + \left| x \right| + 9 = 0$$ | [{"identifier": "A", "content": "is always positive "}, {"identifier": "B", "content": "is always negative "}, {"identifier": "C", "content": "does not exist"}, {"identifier": "D", "content": "none of these "}] | ["A"] | null | Product of real roots $$ = {9 \over {{t^2}}} > 0,\forall \,t\, \in R$$
<br><br>$$\therefore$$ Product of real roots is always positive. | mcq | aieee-2002 |
YCu5If6M2eZmxcua | maths | quadratic-equation-and-inequalities | modulus-function | The number of real solutions of the equation $${x^2} - 3\left| x \right| + 2 = 0$$ is | [{"identifier": "A", "content": "3 "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "4 "}, {"identifier": "D", "content": "1 "}] | ["C"] | null | $${x^2} - 3\left| x \right| + 2 = 0$$
<br><br>$$ \Rightarrow {\left| x \right|^2} - 3\left| x \right| + 2 = 0$$
<br><br>$$\left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 1} \right) = 0$$
<br><br>$$\left| x \right| = 1,2$$ or $$x = \pm 1, \pm 2$$
<br><br>$$\therefore$$ No. of solution $$=4$$ | mcq | aieee-2003 |
8zfNJ8kRVBDzRMei | maths | quadratic-equation-and-inequalities | modulus-function | Let S = { $$x$$ $$ \in $$ R : $$x$$ $$ \ge $$ 0 and
<br/><br/>$$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$}. Then S | [{"identifier": "A", "content": "contains exactly four elements"}, {"identifier": "B", "content": "is an empty set"}, {"identifier": "C", "content": "contains exactly one element"}, {"identifier": "D", "content": "contains exactly two elements"}] | ["D"] | null | Given,
<br><br>$$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$
<br><br><b><u>Case 1</u> :</b>
<br><br>When $$\sqrt x - 3 \ge 0,$$ then equation becomes
<br><br>$$2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$
<br><br>$$ \Rightarrow \,\,\,\,2\sqrt... | mcq | jee-main-2018-offline |
GucnZdzvR3MFEr855XuFJ | maths | quadratic-equation-and-inequalities | modulus-function | The sum of the solutions of the equation <br/>
$$\left| {\sqrt x - 2} \right| + \sqrt x \left( {\sqrt x - 4} \right) + 2 = 0$$<br/>
(x > 0) is equal to: | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "10"}] | ["D"] | null | <b>Case 1 :</b> When $$\sqrt x \ge 2$$
<br><br>then $$\left| {\sqrt x - 2} \right| = \sqrt x - 2$$
<br><br>$$ \therefore $$ The given equation becomes,
<br><br>$$\left( {\sqrt x - 2} \right)$$ + $$\sqrt x \left( {\sqrt x - 4} \right) + 2$$ = 0
<br><br>$$ \Rightarrow $$ $$\left( {\sqrt x - 2} \right)$$ + $$x - 4\s... | mcq | jee-main-2019-online-8th-april-morning-slot |
ZaQAnwZJq0uaay0nM43rsa0w2w9jx1ywhvd | maths | quadratic-equation-and-inequalities | modulus-function | The number of real roots of the equation
<br/> 5 + |2<sup>x</sup>
– 1| = 2<sup>x</sup>
(2<sup>x</sup>
– 2) is | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["B"] | null | When 2<sup>x</sup> $$ \ge $$ 1 <br><br>
5 + 2<sup>x</sup> –1 = 2<sup>x</sup> (2<sup>x</sup> – 2) <br><br>
Let 2x = t <br><br>
$$ \Rightarrow $$5 + t – 1 = t (t – 2) <br><br>
$$ \Rightarrow $$ t = 4, – 1(rejected) <br><br>
$$ \Rightarrow $$ 2x = 4 <br><br>
$$ \Rightarrow $$ x = 2 <br><br>
Now when 2<sup>x</sup> < 1 <... | mcq | jee-main-2019-online-10th-april-evening-slot |
tLULgaIgrnpDTGKEEF7k9k2k5hk1kck | maths | quadratic-equation-and-inequalities | modulus-function | Let S be the set of all real roots of the equation,<br/>
3<sup>x</sup>(3<sup>x</sup> – 1) + 2 = |3<sup>x</sup> – 1| + |3<sup>x</sup> – 2|. Then S : | [{"identifier": "A", "content": "contains exactly two elements."}, {"identifier": "B", "content": "is an empty set."}, {"identifier": "C", "content": "is a singleton."}, {"identifier": "D", "content": "contains at least four elements."}] | ["C"] | null | Let 3<sup>x</sup> = t ; t $$>$$ 0
<br><br>t(t – 1) + 2 = |t – 1| + |t – 2|
<br><br>t<sup>2</sup> – t + 2 = |t – 1| + |t – 2|
<br><br><b>Case-I</b> : t $$<$$ 1
<br><br>t<sup>2</sup> – t + 2 = 1 – t + 2 – t
<br><br>$$ \Rightarrow $$ t<sup>2</sup> + 2 = 3 – t
<br><br>$$ \Rightarrow $$ t<sup>2</sup> + t – 1 = 0
<br><... | mcq | jee-main-2020-online-8th-january-evening-slot |
RcJaa2kkStUK6mm6FBjgy2xukfg6wcxg | maths | quadratic-equation-and-inequalities | modulus-function | The product of the roots of the <br/>equation 9x<sup>2</sup> - 18|x| + 5 = 0 is : | [{"identifier": "A", "content": "$${{5} \\over {9}}$$"}, {"identifier": "B", "content": "$${{5} \\over {27}}$$"}, {"identifier": "C", "content": "$${{25} \\over {81}}$$"}, {"identifier": "D", "content": "$${{25} \\over {9}}$$"}] | ["C"] | null | $$9{x^2} - 18\left| x \right| + 5 = 0$$<br><br>$$ \Rightarrow $$ $$9{x^2} - 15\left| x \right| - 3\left| x \right| + 5 = 0$$ ($$ \because $$ x<sup>2</sup> = $${\left| x \right|^2}$$)<br><br>$$ \Rightarrow $$ $$3\left| x \right|(3\left| x \right| - 5) - (3\left| x \right| - 5) = 0$$<br><br>$$ \Rightarrow $$$$\left| x \r... | mcq | jee-main-2020-online-5th-september-morning-slot |
YmhKJLqLfmHUC0KJZD1klrmwr0o | maths | quadratic-equation-and-inequalities | modulus-function | The number of the real roots of the equation $${(x + 1)^2} + |x - 5| = {{27} \over 4}$$ is ________. | [] | null | 2 | When $$x > 5$$<br><br>$${(x + 1)^2} + (x - 5) = {{27} \over 4}$$<br><br>$$ \Rightarrow {x^2} + 3x - 4 = {{27} \over 4}$$<br><br>$$ \Rightarrow {x^2} + 3x - {{43} \over 4} = 0$$<br><br>$$ \Rightarrow 4{x^2} + 12x - 43 = 0$$<br><br>$$x = {{ - 12 \pm \sqrt {144 + 688} } \over 8}$$<br><br>$$x = {{ - 12 \pm \sqrt {832} }... | integer | jee-main-2021-online-24th-february-evening-slot |
1krzn03g8 | maths | quadratic-equation-and-inequalities | modulus-function | The number of real solutions of the equation, x<sup>2</sup> $$-$$ |x| $$-$$ 12 = 0 is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}] | ["A"] | null | |x|<sup>2</sup> $$-$$ |x| $$-$$ 12 = 0<br><br>$$ \Rightarrow $$ (|x| + 3)(|x| $$-$$ 4) = 0<br><br>$$ \Rightarrow $$ |x| = 4
<br><br>$$\Rightarrow$$ x = $$\pm$$4 | mcq | jee-main-2021-online-25th-july-evening-shift |
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