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A. दोनों पक्षों में (7) जोड़ना/adding (7) to both sides
Step 1
Concept
Adding the same number to both sides does not change inequality direction. The sign changes only when multiplying or dividing by a negative number.
Step 2
Why this answer is correct
The correct answer is A. दोनों पक्षों में (7) जोड़ना / adding (7) to both sides. Adding the same number to both sides does not change inequality direction. The sign changes only when multiplying or dividing by a negative number.
Step 3
Exam Tip
दोनों पक्षों में समान संख्या जोड़ने से असमानता की दिशा नहीं बदलती। चिन्ह केवल ऋणात्मक गुणा या भाग में बदलता है।
We get \(-4x\ge 12\), and dividing by (-4) reverses the sign to \(x\le -3\). A negative coefficient is the most common trap.
Step 2
Why this answer is correct
The correct answer is B. \(x\le -3\). We get \(-4x\ge 12\), and dividing by (-4) reverses the sign to \(x\le -3\). A negative coefficient is the most common trap.
Step 3
Exam Tip
\(-4x\ge 12\) और (-4) से भाग देने पर चिन्ह उलटकर \(x\le -3\) होता है। ऋणात्मक गुणांक सबसे सामान्य गलती है।
The same (x) cannot be less than (-1) and greater than (3) at once. For compound inequalities, find the common part.
Step 2
Why this answer is correct
The correct answer is C. (x<-1) और (x>3) / (x<-1) and (x>3). The same (x) cannot be less than (-1) and greater than (3) at once. For compound inequalities, find the common part.
Step 3
Exam Tip
एक ही (x) साथ में (-1) से छोटा और (3) से बड़ा नहीं हो सकता। संयुक्त असमानता में साझा भाग खोजें।
The inequality \(x\le -2\) includes (-2) and all smaller real values. Infinity always uses an open parenthesis.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-2]\). The inequality \(x\le -2\) includes (-2) and all smaller real values. Infinity always uses an open parenthesis.
Step 3
Exam Tip
\(x\le -2\) में (-2) शामिल है और सभी छोटे वास्तविक मान आते हैं। \(\infty\) के साथ हमेशा खुला कोष्ठक लगता है।
For every real (x), (x+1) is (1) more than (x). Subtracting the same term gives (0<1).
Step 2
Why this answer is correct
The correct answer is A. सभी \(x\in\mathbb{R}\) / all \(x\in\mathbb{R}\). For every real (x), (x+1) is (1) more than (x). Subtracting the same term gives (0<1).
Step 3
Exam Tip
हर वास्तविक (x) के लिए (x+1), (x) से (1) अधिक है। समान पद हटाने पर (0<1) मिलता है।
D. कोई \(x\in\mathbb{R}\) नहीं/no \(x\in\mathbb{R}\)
Step 1
Concept
Subtracting the same (x) gives (3<-2), which is false. When the variable cancels, decide the solution from the remaining truth value.
Step 2
Why this answer is correct
The correct answer is D. कोई \(x\in\mathbb{R}\) नहीं / no \(x\in\mathbb{R}\). Subtracting the same (x) gives (3<-2), which is false. When the variable cancels, decide the solution from the remaining truth value.
Step 3
Exam Tip
समान (x) हटाने पर (3<-2) मिलता है, जो असत्य है। जब चर हट जाए तो शेष सत्यता से हल तय करें।
Multiplying \(x\le -1\) by (-5) gives \(-5x\ge 5\), then adding (2) gives \(-5x+2\ge 7\). Negative multiplication reverses the sign.
Step 2
Why this answer is correct
The correct answer is B. \(-5x+2\ge 7\). Multiplying \(x\le -1\) by (-5) gives \(-5x\ge 5\), then adding (2) gives \(-5x+2\ge 7\). Negative multiplication reverses the sign.
Step 3
Exam Tip
\(x\le -1\) को (-5) से गुणा करने पर \(-5x\ge 5\), फिर (2) जोड़ने पर \(-5x+2\ge 7\) मिलता है। ऋणात्मक गुणा चिन्ह उलटता है।
Multiplying both sides by positive (2) gives \(x-3\ge 8\), so \(x\ge 11\). A positive denominator does not change the sign.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 11\). Multiplying both sides by positive (2) gives \(x-3\ge 8\), so \(x\ge 11\). A positive denominator does not change the sign.
Step 3
Exam Tip
दोनों पक्षों को धनात्मक (2) से गुणा करने पर \(x-3\ge 8\), अतः \(x\ge 11\) मिलता है। धनात्मक हर चिन्ह नहीं बदलता।
Multiplying by (-3) reverses the sign, so (5-2x>-21), then (-2x>-26), and dividing by (-2) gives (x<13). The listed choices do not match this result.
Step 2
Why this answer is correct
The correct answer is B. (x> -8). Multiplying by (-3) reverses the sign, so (5-2x>-21), then (-2x>-26), and dividing by (-2) gives (x<13). The listed choices do not match this result.
Step 3
Exam Tip
(-3) से गुणा करने पर चिन्ह उलटता है, इसलिए (5-2x>-21), फिर (-2x>-26) और (x<13) नहीं, सही रूप (2x<26) से (x<13) होगा। इस पंक्ति में सही विकल्प उपलब्ध नहीं है।
Multiplying by (-2) gives \(7-3x\ge -10\), then \(-3x\ge -17\), so \(x\le \frac{17}{3}\). The correct solution is not among the options, so this question is invalid.
Step 2
Why this answer is correct
The correct answer is B. \(x\ge -1\). Multiplying by (-2) gives \(7-3x\ge -10\), then \(-3x\ge -17\), so \(x\le \frac{17}{3}\). The correct solution is not among the options, so this question is invalid.
Step 3
Exam Tip
(-2) से गुणा करने पर \(7-3x\ge -10\), फिर \(-3x\ge -17\) और \(x\le \frac{17}{3}\) मिलता है। सही हल विकल्पों में नहीं है, इसलिए प्रश्न अमान्य है।
D. कोई निश्चित निष्कर्ष नहीं/no definite conclusion
Step 1
Concept
Inequalities in opposite directions do not give a fixed order when added directly. The directions should match before adding.
Step 2
Why this answer is correct
The correct answer is D. कोई निश्चित निष्कर्ष नहीं / no definite conclusion. Inequalities in opposite directions do not give a fixed order when added directly. The directions should match before adding.
Step 3
Exam Tip
विपरीत दिशाओं की असमानताओं को सीधे जोड़कर निश्चित क्रम नहीं मिलता। जोड़ने से पहले दिशा समान होनी चाहिए।
Subtracting (2) from the whole interval subtracts (2) from its endpoints. Open and closed endpoint types remain the same.
Step 2
Why this answer is correct
The correct answer is A. ([0,5)). Subtracting (2) from the whole interval subtracts (2) from its endpoints. Open and closed endpoint types remain the same.
Step 3
Exam Tip
पूरे अंतराल से (2) घटाने पर सिरों से भी (2) घटता है। बंद और खुले सिरों की प्रकृति वही रहती है।
The value (-1) is excluded and (6) is included. Hence the left endpoint is open and the right endpoint is closed.
Step 2
Why this answer is correct
The correct answer is A. (x\in(-1,6]). The value (-1) is excluded and (6) is included. Hence the left endpoint is open and the right endpoint is closed.
Step 3
Exam Tip
(-1) शामिल नहीं है और (6) शामिल है। इसलिए बायां सिरा खुला और दायां सिरा बंद होगा।
The inequality \(|x|\ge 3\) means (x) is at least (3) units away from (0). Thus \(x\le -3\) or \(x\ge 3\).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-3]\cup[3,\infty\)). The inequality \(|x|\ge 3\) means (x) is at least (3) units away from (0). Thus \(x\le -3\) or \(x\ge 3\).
Step 3
Exam Tip
\(|x|\ge 3\) का अर्थ है (x), (0) से कम से कम (3) इकाई दूर है। इसलिए \(x\le -3\) या \(x\ge 3\) होगा।
When (x>1), (x) is positive and greater than (1), so its square is also greater than (1). Check the sign condition before squaring.
Step 2
Why this answer is correct
The correct answer is A. \(x^2>1\). When (x>1), (x) is positive and greater than (1), so its square is also greater than (1). Check the sign condition before squaring.
Step 3
Exam Tip
(x>1) होने पर (x) धनात्मक है और (1) से बड़ा है, इसलिए वर्ग भी (1) से बड़ा होगा। वर्ग करते समय चिन्ह की शर्त देखें।
From \(x\le y\) and (y<z), we get (x<z) because one link is strict. In a mixed chain, a strict sign can make the result strict.
Step 2
Why this answer is correct
The correct answer is A. (x<z). From \(x\le y\) and (y<z), we get (x<z) because one link is strict. In a mixed chain, a strict sign can make the result strict.
Step 3
Exam Tip
\(x\le y\) और (y<z) से (x<z) मिलता है क्योंकि अंत में एक कठोर असमानता है। मिश्रित क्रम में कठोर चिन्ह परिणाम को कठोर बना सकता है।
Multiplying by a positive number does not change inequality direction. Check the sign of the multiplier before writing the conclusion.
Step 2
Why this answer is correct
The correct answer is A. (ka<kb). Multiplying by a positive number does not change inequality direction. Check the sign of the multiplier before writing the conclusion.
Step 3
Exam Tip
धनात्मक संख्या से गुणा करने पर असमानता की दिशा नहीं बदलती। पहले गुणक का चिन्ह देखें, फिर निष्कर्ष लिखें।
Multiplying (a<b) by (-1) gives (-a>-b). The sign must reverse when multiplying by a negative number.
Step 2
Why this answer is correct
The correct answer is B. यदि (a<b), तो (-a<-b) / If (a<b), then (-a<-b). Multiplying (a<b) by (-1) gives (-a>-b). The sign must reverse when multiplying by a negative number.
Step 3
Exam Tip
(a<b) को (-1) से गुणा करने पर (-a>-b) होता है। ऋणात्मक से गुणा करते समय चिन्ह उलटना चाहिए।
Multiplying \(-3\le x<4\) by (-3) gives \(-12<-3x\le 9\), then adding (2) gives \(-10<2-3x\le 11\). Negative multiplication reverses the order.
Step 2
Why this answer is correct
The correct answer is A. \(-10<2-3x\le 11\). Multiplying \(-3\le x<4\) by (-3) gives \(-12<-3x\le 9\), then adding (2) gives \(-10<2-3x\le 11\). Negative multiplication reverses the order.
Step 3
Exam Tip
\(-3\le x<4\) को (-3) से गुणा करने पर \(-12<-3x\le 9\), फिर (2) जोड़ने पर \(-10<2-3x\le 11\) मिलता है। ऋणात्मक गुणा क्रम उलटता है।
At \(x=\frac{1}{2}\), \(x^2=\frac{1}{4}\), which is less than \(\frac{1}{2}\). One counterexample is enough to disprove a universal statement.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{2}\). At \(x=\frac{1}{2}\), \(x^2=\frac{1}{4}\), which is less than \(\frac{1}{2}\). One counterexample is enough to disprove a universal statement.
Step 3
Exam Tip
\(x=\frac{1}{2}\) पर \(x^2=\frac{1}{4}\), जो \(\frac{1}{2}\) से छोटा है। सार्वत्रिक कथन को गलत करने के लिए एक प्रतिउदाहरण काफी है।
A. यह सभी \(x\in\mathbb{R}\) के लिए सत्य है/true for all \(x\in\mathbb{R}\)
Step 1
Concept
Since \(x^2\ge 0\), we have \(x^2+1\ge 1>0\). Identify the minimum value using the square and constant term.
Step 2
Why this answer is correct
The correct answer is A. यह सभी \(x\in\mathbb{R}\) के लिए सत्य है / true for all \(x\in\mathbb{R}\). Since \(x^2\ge 0\), we have \(x^2+1\ge 1>0\). Identify the minimum value using the square and constant term.
Step 3
Exam Tip
क्योंकि \(x^2\ge 0\), इसलिए \(x^2+1\ge 1>0\)। वर्ग और स्थिरांक से न्यूनतम मान पहचानें।
A. अभिकथन और कारण दोनों सही हैं, और कारण सही व्याख्या है/both assertion and reason are true, and the reason explains it
Step 1
Concept
Adding (c) to both sides keeps the direction of (a<b) unchanged. The reason directly explains the assertion.
Step 2
Why this answer is correct
The correct answer is A. अभिकथन और कारण दोनों सही हैं, और कारण सही व्याख्या है / both assertion and reason are true, and the reason explains it. Adding (c) to both sides keeps the direction of (a<b) unchanged. The reason directly explains the assertion.
Step 3
Exam Tip
दोनों पक्षों में (c) जोड़ने से (a<b) की दिशा वही रहती है। कारण सीधे अभिकथन को सिद्ध करता है।
Multiplying (x>3) by (2) gives (2x>6), and subtracting (5) gives (2x-5>1). Positive multiplication does not reverse the sign.
Step 2
Why this answer is correct
The correct answer is A. (2x-5>1). Multiplying (x>3) by (2) gives (2x>6), and subtracting (5) gives (2x-5>1). Positive multiplication does not reverse the sign.
Step 3
Exam Tip
(x>3) को (2) से गुणा कर (2x>6) और (5) घटाकर (2x-5>1) मिलता है। धनात्मक गुणा में चिन्ह नहीं बदलता।
From \(4-3x\le -8\), we get \(-3x\le -12\), and dividing by (-3) gives \(x\ge 4\). Division by a negative reverses the sign.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 4\). From \(4-3x\le -8\), we get \(-3x\le -12\), and dividing by (-3) gives \(x\ge 4\). Division by a negative reverses the sign.
Step 3
Exam Tip
\(4-3x\le -8\) से \(-3x\le -12\) और (-3) से भाग देने पर \(x\ge 4\) मिलता है। ऋणात्मक भाग में चिन्ह उलटता है।
The common part of both conditions starts at (-1) and goes up to before (4). The value (-1) is included and (4) is excluded.
Step 2
Why this answer is correct
The correct answer is A. ([-1,4)). The common part of both conditions starts at (-1) and goes up to before (4). The value (-1) is included and (4) is excluded.
Step 3
Exam Tip
दोनों शर्तों का साझा भाग (-1) से शुरू होकर (4) से पहले तक है। (-1) शामिल है और (4) शामिल नहीं है।
Multiplying \(x\le 2\) by (-4) gives \(-4x\ge -8\), and adding (7) gives \(7-4x\ge -1\). Negative multiplication reverses the sign.
Step 2
Why this answer is correct
The correct answer is A. \(7-4x\ge -1\). Multiplying \(x\le 2\) by (-4) gives \(-4x\ge -8\), and adding (7) gives \(7-4x\ge -1\). Negative multiplication reverses the sign.
Step 3
Exam Tip
\(x\le 2\) को (-4) से गुणा करने पर \(-4x\ge -8\), फिर (7) जोड़ने पर \(7-4x\ge -1\) मिलता है। ऋणात्मक गुणा चिन्ह उलटता है।
From (1<x<4), we get (0<x-1<3), then dividing by positive (3) gives \(0<\frac{x-1}{3}<1\). Division by a positive number does not change the sign.
Step 2
Why this answer is correct
The correct answer is A. ((0,1)). From (1<x<4), we get (0<x-1<3), then dividing by positive (3) gives \(0<\frac{x-1}{3}<1\). Division by a positive number does not change the sign.
Step 3
Exam Tip
(1<x<4) से (0<x-1<3), फिर धनात्मक (3) से भाग देने पर \(0<\frac{x-1}{3}<1\) मिलता है। धनात्मक भाग में चिन्ह नहीं बदलता।
A. कोई \(x\in\mathbb{R}\) नहीं/no \(x\in\mathbb{R}\)
Step 1
Concept
No real number can be less than or equal to (6) and greater than (6) at the same time. The common part is empty.
Step 2
Why this answer is correct
The correct answer is A. कोई \(x\in\mathbb{R}\) नहीं / no \(x\in\mathbb{R}\). No real number can be less than or equal to (6) and greater than (6) at the same time. The common part is empty.
Step 3
Exam Tip
कोई वास्तविक संख्या (6) से कम या बराबर और (6) से बड़ी एक साथ नहीं हो सकती। साझा भाग खाली है।
Multiplying \(x\ge -2\) by \(-\frac{1}{2}\) gives \(-\frac{x}{2}\le 1\), then adding (5) gives \(-\frac{x}{2}+5\le 6\). A negative multiplier reverses the sign.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{x}{2}+5\le 6\). Multiplying \(x\ge -2\) by \(-\frac{1}{2}\) gives \(-\frac{x}{2}\le 1\), then adding (5) gives \(-\frac{x}{2}+5\le 6\). A negative multiplier reverses the sign.
Step 3
Exam Tip
\(x\ge -2\) को \(-\frac{1}{2}\) से गुणा करने पर \(-\frac{x}{2}\le 1\), फिर (5) जोड़ने पर \(-\frac{x}{2}+5\le 6\) मिलता है। ऋणात्मक गुणक चिन्ह उलटता है।
Subtracting (3) first gives \(-10<2x\le 6\). Then dividing by positive (2) gives the correct solution \(-5<x\le 3\).
Step 2
Why this answer is correct
The correct answer is A. \(-5<x\le 3\). Subtracting (3) first gives \(-10<2x\le 6\). Then dividing by positive (2) gives the correct solution \(-5<x\le 3\).
Step 3
Exam Tip
पहले (3) घटाने पर \(-10<2x\le 6\) मिलता है। फिर धनात्मक (2) से भाग देने पर \(-5<x\le 3\) सही हल है।
From (6-3x<x+5), we get (1<4x), so \(x>\frac{1}{4}\). While moving variable terms to one side, keep the sign and order carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x>\frac{1}{4}\). From (6-3x<x+5), we get (1<4x), so \(x>\frac{1}{4}\). While moving variable terms to one side, keep the sign and order carefully.
Step 3
Exam Tip
(6-3x<x+5) से (1<4x), इसलिए \(x>\frac{1}{4}\) मिलता है। चर पदों को एक ओर लाते समय चिन्ह और क्रम सावधानी से रखें।
From \(7-2x\le x+1\), we get \(6\le 3x\), so \(x\ge 2\). Apply the same operation to both sides while simplifying.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 2\). From \(7-2x\le x+1\), we get \(6\le 3x\), so \(x\ge 2\). Apply the same operation to both sides while simplifying.
Step 3
Exam Tip
\(7-2x\le x+1\) से \(6\le 3x\), इसलिए \(x\ge 2\) है। असमानता को सरल करते समय दोनों पक्षों पर समान क्रिया करें।
Multiplying by positive (6) gives (2(2x-1)\ge 3(x+5)), hence \(x\ge 17\). A positive multiplier used to clear denominators does not change the sign.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 17\). Multiplying by positive (6) gives (2(2x-1)\ge 3(x+5)), hence \(x\ge 17\). A positive multiplier used to clear denominators does not change the sign.
Step 3
Exam Tip
धनात्मक (6) से गुणा करने पर (2(2x-1)\ge 3(x+5)), इसलिए \(x\ge 17\) है। हर हटाते समय धनात्मक गुणक चिन्ह नहीं बदलता।
Subtracting (6) gives \(-10\le 2x<4\), then dividing by (2) gives \(-5\le x<2\). Check equality at each endpoint separately.
Step 2
Why this answer is correct
The correct answer is A. \(-5\le x<2\). Subtracting (6) gives \(-10\le 2x<4\), then dividing by (2) gives \(-5\le x<2\). Check equality at each endpoint separately.
Step 3
Exam Tip
पहले (6) घटाने पर \(-10\le 2x<4\) और फिर (2) से भाग देने पर \(-5\le x<2\) मिलता है। दोनों सिरों की बराबरी अलग-अलग देखें।
Inequalities in opposite directions do not give a fixed comparison between (a+c) and (b+d). Same direction is the safe rule for adding inequalities.
Step 2
Why this answer is correct
The correct answer is A. (a+c<b+d). Inequalities in opposite directions do not give a fixed comparison between (a+c) and (b+d). Same direction is the safe rule for adding inequalities.
Step 3
Exam Tip
विपरीत दिशा की असमानताओं को जोड़कर (a+c) और (b+d) की निश्चित तुलना नहीं मिलती। जोड़ने के लिए दिशा समान होना सुरक्षित नियम है।
The values (2) and (8) are excluded, so the natural values are from (3) to (7). Pay close attention to strict signs.
Step 2
Why this answer is correct
The correct answer is A. \(x\in{3,4,5,6,7}\). The values (2) and (8) are excluded, so the natural values are from (3) to (7). Pay close attention to strict signs.
Step 3
Exam Tip
(2) और (8) शामिल नहीं हैं, इसलिए प्राकृतिक मान (3) से (7) तक मिलते हैं। कठोर चिन्हों पर खास ध्यान दें।
For positive (x), equality in \(x+\frac{1}{x}\ge 2\) occurs at (x=1). Remember the equality condition in standard forms.
Step 2
Why this answer is correct
The correct answer is A. (x=1). For positive (x), equality in \(x+\frac{1}{x}\ge 2\) occurs at (x=1). Remember the equality condition in standard forms.
Step 3
Exam Tip
धनात्मक (x) के लिए \(x+\frac{1}{x}\ge 2\) में समानता (x=1) पर होती है। ऐसे मानक रूपों में समानता की शर्त याद रखें।
A. \(x\le -5\) या \(x\ge 3\)/\(x\le -5\) or \(x\ge 3\)
Step 1
Concept
For \(|x+1|\ge 4\), we have \(x+1\le -4\) or \(x+1\ge 4\). Thus \(x\le -5\) or \(x\ge 3\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le -5\) या \(x\ge 3\) / \(x\le -5\) or \(x\ge 3\). For \(|x+1|\ge 4\), we have \(x+1\le -4\) or \(x+1\ge 4\). Thus \(x\le -5\) or \(x\ge 3\).
Step 3
Exam Tip
\(|x+1|\ge 4\) में \(x+1\le -4\) या \(x+1\ge 4\) होता है। इसलिए \(x\le -5\) या \(x\ge 3\) मिलता है।
A. \(x\le -4\) या \(x\ge 4\)/\(x\le -4\) or \(x\ge 4\)
Step 1
Concept
The inequality \(x^2\ge 16\) means \(|x|\ge 4\). Hence (x) lies outside: \(x\le -4\) or \(x\ge 4\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le -4\) या \(x\ge 4\) / \(x\le -4\) or \(x\ge 4\). The inequality \(x^2\ge 16\) means \(|x|\ge 4\). Hence (x) lies outside: \(x\le -4\) or \(x\ge 4\).
Step 3
Exam Tip
\(x^2\ge 16\) का अर्थ \(|x|\ge 4\) है। इसलिए (x) बाहर की ओर \(x\le -4\) या \(x\ge 4\) होगा।
From (x>4), (x-4>0), so its reciprocal is also positive. The reciprocal keeps the same sign as the original quantity.
Step 2
Why this answer is correct
The correct answer is A. दोनों धनात्मक हैं / both are positive. From (x>4), (x-4>0), so its reciprocal is also positive. The reciprocal keeps the same sign as the original quantity.
Step 3
Exam Tip
(x>4) से (x-4>0), इसलिए उसका व्युत्क्रम भी धनात्मक होगा। व्युत्क्रम का चिन्ह मूल राशि जैसा ही रहता है।
A. यह सभी \(x\in\mathbb{R}\) के लिए सत्य है/it is true for all \(x\in\mathbb{R}\)
Step 1
Concept
Since \(x^2\ge 0\), \(x^2+4\ge 4\) is always true. Identify the minimum value of the square term.
Step 2
Why this answer is correct
The correct answer is A. यह सभी \(x\in\mathbb{R}\) के लिए सत्य है / it is true for all \(x\in\mathbb{R}\). Since \(x^2\ge 0\), \(x^2+4\ge 4\) is always true. Identify the minimum value of the square term.
Step 3
Exam Tip
क्योंकि \(x^2\ge 0\), इसलिए \(x^2+4\ge 4\) हमेशा सत्य है। वर्ग वाले पद का न्यूनतम मान पहचानें।
Let the number be (x); three times it is (3x), and less than (17) means (<17). Watch the order while translating words to symbols.
Step 2
Why this answer is correct
The correct answer is A. (3x<17). Let the number be (x); three times it is (3x), and less than (17) means (<17). Watch the order while translating words to symbols.
Step 3
Exam Tip
संख्या को (x) मानने पर उसका (3) गुना (3x) होगा और (17) से कम का अर्थ (<17) है। भाषा को प्रतीक में बदलते समय क्रम पर ध्यान दें।
Subtracting (4x) from both sides gives (1<9), which is always true. When the variable cancels, decide the solution from the truth of the remaining statement.
Step 2
Why this answer is correct
The correct answer is A. सभी \(x\in\mathbb{R}\) / all \(x\in\mathbb{R}\). Subtracting (4x) from both sides gives (1<9), which is always true. When the variable cancels, decide the solution from the truth of the remaining statement.
Step 3
Exam Tip
दोनों पक्षों से (4x) घटाने पर (1<9) मिलता है, जो हमेशा सत्य है। चर हटने पर शेष कथन की सत्यता से हल तय करें।
A. कोई \(x\in\mathbb{R}\) नहीं/no \(x\in\mathbb{R}\)
Step 1
Concept
Subtracting (6x) from both sides gives (-2>1), which is false. A false constant statement has no solution.
Step 2
Why this answer is correct
The correct answer is A. कोई \(x\in\mathbb{R}\) नहीं / no \(x\in\mathbb{R}\). Subtracting (6x) from both sides gives (-2>1), which is false. A false constant statement has no solution.
Step 3
Exam Tip
दोनों पक्षों से (6x) घटाने पर (-2>1) मिलता है, जो असत्य है। असत्य स्थिर कथन का कोई हल नहीं होता।
The expression (2x+4) is increasing, so the endpoint values are (2) and (10). Closed endpoints give a closed interval.
Step 2
Why this answer is correct
The correct answer is A. \(y\in[2,10]\). The expression (2x+4) is increasing, so the endpoint values are (2) and (10). Closed endpoints give a closed interval.
Step 3
Exam Tip
(2x+4) बढ़ने वाला रैखिक रूप है, इसलिए सिरों पर मान (2) और (10) मिलते हैं। बंद सिरों से बंद अंतराल मिलता है।
On this interval, (|x|) lies between (1) and (3), so \(1<x^2<9\). For squares on negative intervals, look at magnitude.
Step 2
Why this answer is correct
The correct answer is A. \(1<x^2<9\). On this interval, (|x|) lies between (1) and (3), so \(1<x^2<9\). For squares on negative intervals, look at magnitude.
Step 3
Exam Tip
इस अंतराल में (|x|) (1) और (3) के बीच है, इसलिए \(1<x^2<9\)। ऋणात्मक अंतराल में वर्ग के लिए परिमाण देखें।
When (x>1), (x), \(x^2\), and \(x^3\) are all greater than (1). Powers preserve the order for positive values above (1).
Step 2
Why this answer is correct
The correct answer is A. \(x^3>1\). When (x>1), (x), \(x^2\), and \(x^3\) are all greater than (1). Powers preserve the order for positive values above (1).
Step 3
Exam Tip
(x>1) होने पर (x), \(x^2\) और \(x^3\) सभी (1) से बड़े हैं। धनात्मक बड़े मानों में घात क्रम बनाए रखती है।
Multiplying by (-2) should reverse the sign, so the correct form is (-2a>-2b). This is the biggest error in negative multiplication.
Step 2
Why this answer is correct
The correct answer is A. \(a<b\Rightarrow -2a<-2b\). Multiplying by (-2) should reverse the sign, so the correct form is (-2a>-2b). This is the biggest error in negative multiplication.
Step 3
Exam Tip
(-2) से गुणा करने पर चिन्ह उलटना चाहिए, इसलिए सही रूप (-2a>-2b) है। ऋणात्मक गुणन में यही सबसे बड़ी गलती होती है।
Multiplying by positive (12) gives (3x+6<6x-2), so \(x>\frac{8}{3}\). A positive LCM does not reverse the sign.
Step 2
Why this answer is correct
The correct answer is A. \(x>\frac{8}{3}\). Multiplying by positive (12) gives (3x+6<6x-2), so \(x>\frac{8}{3}\). A positive LCM does not reverse the sign.
Step 3
Exam Tip
धनात्मक (12) से गुणा करने पर (3x+6<6x-2), इसलिए \(x>\frac{8}{3}\) है। धनात्मक लघुत्तम समापवर्त्य से चिन्ह नहीं बदलता।
Adding inequalities in the same direction gives (u+r<v+s). Subtraction, multiplication, and division need extra sign conditions.
Step 2
Why this answer is correct
The correct answer is A. (u+r<v+s). Adding inequalities in the same direction gives (u+r<v+s). Subtraction, multiplication, and division need extra sign conditions.
Step 3
Exam Tip
एक ही दिशा की असमानताओं को जोड़ने से (u+r<v+s) मिलता है। घटाव, गुणा और भाग के लिए अतिरिक्त चिन्ह शर्तें चाहिए।
The common part of both conditions starts at (3) and goes up to before (5). The value (3) is included and (5) is excluded.
Step 2
Why this answer is correct
The correct answer is A. \(3\le x<5\). The common part of both conditions starts at (3) and goes up to before (5). The value (3) is included and (5) is excluded.
Step 3
Exam Tip
दोनों शर्तों का साझा भाग (3) से शुरू होकर (5) से पहले तक है। (3) शामिल है और (5) शामिल नहीं है।
From (-6x+3>12), we get (-6x>9), and dividing by (-6) gives \(x<-\frac{3}{2}\). Division by a negative reverses the sign.
Step 2
Why this answer is correct
The correct answer is A. \(x<-\frac{3}{2}\). From (-6x+3>12), we get (-6x>9), and dividing by (-6) gives \(x<-\frac{3}{2}\). Division by a negative reverses the sign.
Step 3
Exam Tip
(-6x+3>12) से (-6x>9), फिर (-6) से भाग देने पर \(x<-\frac{3}{2}\) मिलता है। ऋणात्मक भाग में चिन्ह उलटता है।
A. अभिकथन और कारण दोनों सही हैं, और कारण सही व्याख्या है/both assertion and reason are true, and the reason explains it
Step 1
Concept
From (a<b), we directly get (b-a>0). The reason correctly explains this difference.
Step 2
Why this answer is correct
The correct answer is A. अभिकथन और कारण दोनों सही हैं, और कारण सही व्याख्या है / both assertion and reason are true, and the reason explains it. From (a<b), we directly get (b-a>0). The reason correctly explains this difference.
Step 3
Exam Tip
(a<b) से (b-a>0) सीधे मिलता है। कारण इसी अंतर की सही व्याख्या करता है।
The linear expression (5-3x) is decreasing, so the order of endpoints reverses. Since (x=1) is included, (2) is included, and since (x=4) is excluded, (-7) is excluded.
Step 2
Why this answer is correct
The correct answer is A. (y\in(-7,2]). The linear expression (5-3x) is decreasing, so the order of endpoints reverses. Since (x=1) is included, (2) is included, and since (x=4) is excluded, (-7) is excluded.
Step 3
Exam Tip
रैखिक रूप (5-3x) घटता है, इसलिए सिरों का क्रम उलटता है। (x=1) शामिल है इसलिए (2) शामिल है और (x=4) शामिल नहीं है इसलिए (-7) शामिल नहीं है।
Multiplying by positive (3) gives \(4x-5\le 3x+6\), so \(x\le 11\). Removing a positive denominator does not change the sign.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 11\). Multiplying by positive (3) gives \(4x-5\le 3x+6\), so \(x\le 11\). Removing a positive denominator does not change the sign.
Step 3
Exam Tip
धनात्मक (3) से गुणा करने पर \(4x-5\le 3x+6\), इसलिए \(x\le 11\) मिलता है। धनात्मक हर हटाने पर चिन्ह नहीं बदलता।
A. दोनों पक्षों में (4) जोड़ने पर/adding (4) to both sides
Step 1
Concept
Adding the same number to both sides does not change the inequality sign. In exams the sign reverses only for multiplication or division by a negative number.
Step 2
Why this answer is correct
The correct answer is A. दोनों पक्षों में (4) जोड़ने पर / adding (4) to both sides. Adding the same number to both sides does not change the inequality sign. In exams the sign reverses only for multiplication or division by a negative number.
Step 3
Exam Tip
दोनों पक्षों में समान संख्या जोड़ने से असमानता का चिह्न नहीं बदलता। परीक्षा में चिह्न केवल ऋणात्मक गुणा या भाग में उलटता है।
We get \(-4x\ge 12\), and dividing by (-4) gives \(x\le -3\). In exams do not forget to reverse the sign on negative division.
Step 2
Why this answer is correct
The correct answer is B. \(x\le -3\). We get \(-4x\ge 12\), and dividing by (-4) gives \(x\le -3\). In exams do not forget to reverse the sign on negative division.
Step 3
Exam Tip
\(-4x\ge 12\) और (-4) से भाग देने पर \(x\le -3\) मिलता है। परीक्षा में ऋणात्मक भाग से चिह्न उलटना न भूलें।
A. (x>-2) पर खुला वृत्त/open circle at (x=-2) with right shading
Step 1
Concept
From (5-2x<9), (-2x<4), so (x>-2). In exams (>) means right shading and an open circle.
Step 2
Why this answer is correct
The correct answer is A. (x>-2) पर खुला वृत्त / open circle at (x=-2) with right shading. From (5-2x<9), (-2x<4), so (x>-2). In exams (>) means right shading and an open circle.
Step 3
Exam Tip
(5-2x<9) से (-2x<4), इसलिए (x>-2)। परीक्षा में (>) के लिए दाईं ओर छाया और खुला वृत्त होता है।
Subtracting (1) from all parts gives \(-3\le 3x<9\), so \(-1\le x<3\). In exams keep all three parts together in compound inequalities.
Step 2
Why this answer is correct
The correct answer is A. \(-1\le x<3\). Subtracting (1) from all parts gives \(-3\le 3x<9\), so \(-1\le x<3\). In exams keep all three parts together in compound inequalities.
Step 3
Exam Tip
सभी भागों से (1) घटाने पर \(-3\le 3x<9\), इसलिए \(-1\le x<3\)। परीक्षा में संयुक्त असमानता में तीनों भाग साथ रखें।
In (x<-1), (-1) is not included, and in \(x\ge5\), (5) is included. In exams or usually means \(\cup\).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-1\)\cup[5,\infty)). In (x<-1), (-1) is not included, and in \(x\ge5\), (5) is included. In exams or usually means \(\cup\).
Step 3
Exam Tip
(x<-1) में (-1) शामिल नहीं और \(x\ge5\) में (5) शामिल है। परीक्षा में या का अर्थ प्रायः \(\cup\) होता है।
Since (3) is positive, the sign does not change, and \(x-1\ge6\) gives \(x\ge7\). In exams first check the sign of the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 7\). Since (3) is positive, the sign does not change, and \(x-1\ge6\) gives \(x\ge7\). In exams first check the sign of the denominator.
Step 3
Exam Tip
(3) धनात्मक है, इसलिए चिह्न नहीं बदलेगा और \(x-1\ge6\) से \(x\ge7\)। परीक्षा में हर का चिह्न पहले देखें।
\(-3x\le -6\), and dividing by (-3) gives \(x\ge2\). In exams reverse the sign when dividing by a negative coefficient.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge2\). \(-3x\le -6\), and dividing by (-3) gives \(x\ge2\). In exams reverse the sign when dividing by a negative coefficient.
Step 3
Exam Tip
\(-3x\le -6\) और (-3) से भाग देने पर \(x\ge2\)। परीक्षा में ऋणात्मक गुणांक से भाग देते समय चिह्न उलटें।
A. (p+r<q+r) हर वास्तविक (r) के लिए/(p+r<q+r) for every real (r)
Step 1
Concept
Adding the same real number does not change the direction of inequality. In exams the sign of the multiplier or divisor matters.
Step 2
Why this answer is correct
The correct answer is A. (p+r<q+r) हर वास्तविक (r) के लिए / (p+r<q+r) for every real (r). Adding the same real number does not change the direction of inequality. In exams the sign of the multiplier or divisor matters.
Step 3
Exam Tip
समान वास्तविक संख्या जोड़ने से असमानता की दिशा नहीं बदलती। परीक्षा में गुणा या भाग में संख्या का चिह्न जरूरी होता है।
A. \(x\le -3\) या \(x\ge3\)/\(x\le -3\) or \(x\ge3\)
Step 1
Concept
In \(|x|\ge3\), the distance from (0) is at least (3). In exams a greater-than absolute value gives two outer parts.
Step 2
Why this answer is correct
The correct answer is A. \(x\le -3\) या \(x\ge3\) / \(x\le -3\) or \(x\ge3\). In \(|x|\ge3\), the distance from (0) is at least (3). In exams a greater-than absolute value gives two outer parts.
Step 3
Exam Tip
\(|x|\ge3\) में (0) से दूरी कम से कम (3) है। परीक्षा में बड़ा वाला निरपेक्ष मान दो बाहरी भाग देता है।
The square of a real number is never negative, so \(x^2\ge0\) is always true. In exams remember the basic property of squares.
Step 2
Why this answer is correct
The correct answer is A. सदैव सत्य / always true. The square of a real number is never negative, so \(x^2\ge0\) is always true. In exams remember the basic property of squares.
Step 3
Exam Tip
किसी वास्तविक संख्या का वर्ग ऋणात्मक नहीं होता, इसलिए \(x^2\ge0\) सदैव सत्य है। परीक्षा में वर्ग का मूल गुण याद रखें।
The product is positive when both factors have the same sign. In exams keep the zero points (-3) and (2) open.
Step 2
Why this answer is correct
The correct answer is A. (x<-3) या (x>2) / (x<-3) or (x>2). The product is positive when both factors have the same sign. In exams keep the zero points (-3) and (2) open.
Step 3
Exam Tip
गुणनफल धनात्मक तब है जब दोनों गुणनखंड समान चिह्न के हों। परीक्षा में शून्य बिंदु (-3) और (2) को खुला रखें।
The fraction is positive when numerator and denominator have the same sign. In exams (x=-1) makes the denominator zero, so it is excluded.
Step 2
Why this answer is correct
The correct answer is A. (x<-1) या (x>2) / (x<-1) or (x>2). The fraction is positive when numerator and denominator have the same sign. In exams (x=-1) makes the denominator zero, so it is excluded.
Step 3
Exam Tip
भिन्न धनात्मक है जब अंश और हर समान चिह्न के हों। परीक्षा में (x=-1) हर को शून्य करता है, इसलिए शामिल नहीं होगा।
At (x=3), the denominator becomes (0), so the fraction is undefined. In exams always exclude values that make the denominator zero.
Step 2
Why this answer is correct
The correct answer is A. (3). At (x=3), the denominator becomes (0), so the fraction is undefined. In exams always exclude values that make the denominator zero.
Step 3
Exam Tip
(x=3) पर हर (0) हो जाता है, इसलिए भिन्न अपरिभाषित है। परीक्षा में हर को शून्य करने वाले मान हमेशा हटाएं।
After cancelling (x), we get (2>-5), which is always true. In exams such questions may have all real numbers as the answer.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\). After cancelling (x), we get (2>-5), which is always true. In exams such questions may have all real numbers as the answer.
Step 3
Exam Tip
(x) हटाने पर (2>-5) मिलता है, जो सदैव सत्य है। परीक्षा में ऐसे प्रश्नों में सभी वास्तविक संख्याएं उत्तर हो सकती हैं।
After cancelling (x), we get \(-6\ge1\), which is false. In exams a false constant statement means no solution.
Step 2
Why this answer is correct
The correct answer is A. \(\varnothing). After cancelling (x), we get \(-6\ge1\), which is false. In exams a false constant statement means no solution.
Step 3
Exam Tip
(x) हटाने पर \(-6\ge1\) मिलता है, जो असत्य है। परीक्षा में असत्य स्थिर कथन का अर्थ कोई हल नहीं है।
Greater than or equal to is written as \(m\ge40\). In exams when the word equal appears, look for \(\ge\) or \(\le\).
Step 2
Why this answer is correct
The correct answer is A. \(m\ge40\). Greater than or equal to is written as \(m\ge40\). In exams when the word equal appears, look for \(\ge\) or \(\le\).
Step 3
Exam Tip
अधिक या बराबर का गणितीय रूप \(m\ge40\) है। परीक्षा में शब्द बराबर आए तो \(\ge\) या \(\le\) पर ध्यान दें।
From \(x+3\ge2x-1\), \(4\ge x\), that is \(x\le4\). In exams understand the direction when rewriting \(4\ge x\) as \(x\le4\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le4\). From \(x+3\ge2x-1\), \(4\ge x\), that is \(x\le4\). In exams understand the direction when rewriting \(4\ge x\) as \(x\le4\).
Step 3
Exam Tip
\(x+3\ge2x-1\) से \(4\ge x\), अर्थात \(x\le4\)। परीक्षा में \(4\ge x\) को \(x\le4\) में बदलते समय दिशा समझें।
Multiplying all parts by positive (2) gives \(-2\le x<8\). In exams positive multiplication does not change signs.
Step 2
Why this answer is correct
The correct answer is A. \(-2\le x<8\). Multiplying all parts by positive (2) gives \(-2\le x<8\). In exams positive multiplication does not change signs.
Step 3
Exam Tip
सभी भागों को धनात्मक (2) से गुणा करने पर \(-2\le x<8\)। परीक्षा में धनात्मक गुणा से चिह्न नहीं बदलते।
Dividing by (-2) reverses both signs and gives \(3>x\ge-5\), that is \(-5\le x<3\). In exams rewrite compound inequalities in increasing order.
Step 2
Why this answer is correct
The correct answer is A. \(-5\le x<3\). Dividing by (-2) reverses both signs and gives \(3>x\ge-5\), that is \(-5\le x<3\). In exams rewrite compound inequalities in increasing order.
Step 3
Exam Tip
(-2) से भाग देने पर दोनों चिह्न उलटते हैं और \(3>x\ge-5\), यानी \(-5\le x<3\)। परीक्षा में संयुक्त असमानता को क्रम में फिर से लिखें।
Multiplying by positive (3) gives \(-12<x+2\le6\), so \(-14<x\le4\). In exams preserve open and closed signs separately.
Step 2
Why this answer is correct
The correct answer is A. \(-14<x\le4\). Multiplying by positive (3) gives \(-12<x+2\le6\), so \(-14<x\le4\). In exams preserve open and closed signs separately.
Step 3
Exam Tip
धनात्मक (3) से गुणा करने पर \(-12<x+2\le6\), इसलिए \(-14<x\le4\)। परीक्षा में खुले और बंद चिह्न अलग-अलग बनाए रखें।
The solution is \(-1\le x<\frac{11}{3}\), so the integers are ({-1,0,1,2,3}). In exams write the final answer according to the domain.
Step 2
Why this answer is correct
The correct answer is A. ({-1,0,1,2,3}). The solution is \(-1\le x<\frac{11}{3}\), so the integers are ({-1,0,1,2,3}). In exams write the final answer according to the domain.
Step 3
Exam Tip
हल \(-1\le x<\frac{11}{3}\) है, इसलिए पूर्णांक ({-1,0,1,2,3}) मिलते हैं। परीक्षा में अंतिम उत्तर डोमेन के अनुसार लिखें।
Multiplying by positive (12) gives (3(2x-5)>4(x+1)), so (2x>19). In exams the sign does not change when the LCM is positive.
Step 2
Why this answer is correct
The correct answer is A. \(x>\frac{19}{2}\). Multiplying by positive (12) gives (3(2x-5)>4(x+1)), so (2x>19). In exams the sign does not change when the LCM is positive.
Step 3
Exam Tip
धनात्मक (12) से गुणा करने पर (3(2x-5)>4(x+1)), इसलिए (2x>19)। परीक्षा में एलसीएम धनात्मक हो तो चिह्न नहीं बदलता।
Multiplying by positive (d) keeps the direction of the inequality unchanged. In exams identify the sign of the multiplier first.
Step 2
Why this answer is correct
The correct answer is A. (ad<bd). Multiplying by positive (d) keeps the direction of the inequality unchanged. In exams identify the sign of the multiplier first.
Step 3
Exam Tip
धनात्मक (d) से गुणा करने पर असमानता की दिशा वही रहती है। परीक्षा में गुणक का चिह्न पहले पहचानें।
The fraction is negative when numerator and denominator have opposite signs. In exams (x=-2) makes the denominator zero and (x=4) gives (0), so both are excluded.
Step 2
Why this answer is correct
The correct answer is A. (-2<x<4). The fraction is negative when numerator and denominator have opposite signs. In exams (x=-2) makes the denominator zero and (x=4) gives (0), so both are excluded.
Step 3
Exam Tip
भिन्न ऋणात्मक तब होता है जब अंश और हर विपरीत चिह्न के हों। परीक्षा में (x=-2) हर को शून्य करता है और (x=4) पर मान (0) है, इसलिए दोनों शामिल नहीं हैं।
