A. शीर्ष ((2,3)) और न्यूनतम (3)/vertex ((2,3)) and minimum (3)
Step 1
Concept
The minimum of (|x-2|) is (0) at (x=2), so (y=3). In exams, set the inside of the modulus equal to (0).
Step 2
Why this answer is correct
The correct answer is A. शीर्ष ((2,3)) और न्यूनतम (3) / vertex ((2,3)) and minimum (3). The minimum of (|x-2|) is (0) at (x=2), so (y=3). In exams, set the inside of the modulus equal to (0).
Step 3
Exam Tip
(|x-2|) का न्यूनतम (0) (x=2) पर होता है इसलिए (y=3) है। परीक्षा में मापांक के अंदर को (0) रखें।
For the square root, \(7-2x\ge 0\), so \(x\le \frac{7}{2}\). In exams, keep the expression inside the square root non-negative.
Step 2
Why this answer is correct
The correct answer is A. (\left\(-\infty,\frac{7}{2}\right]\). For the square root, \(7-2x\ge 0\), so \(x\le \frac{7}{2}\). In exams, keep the expression inside the square root non-negative.
Step 3
Exam Tip
वर्गमूल के लिए \(7-2x\ge 0\) इसलिए \(x\le \frac{7}{2}\)। परीक्षा में वर्गमूल के अंदर की राशि अनऋणात्मक रखें।
The denominator (x-1) gives vertical asymptote (x=1), and the outside (-2) gives horizontal asymptote (y=-2). In exams, check both the denominator and vertical shift.
Step 2
Why this answer is correct
The correct answer is A. (x=1) और (y=-2) / (x=1) and (y=-2). The denominator (x-1) gives vertical asymptote (x=1), and the outside (-2) gives horizontal asymptote (y=-2). In exams, check both the denominator and vertical shift.
Step 3
Exam Tip
हर (x-1) शून्य होने पर (x=1) लंबवत आसमापी है और बाहरी (-2) से (y=-2) क्षैतिज आसमापी है। परीक्षा में हर और ऊर्ध्व विस्थापन दोनों देखें।
Since (x-2-6x+8=(x-2)(x-4)), the zeroes are (x=2) and (x=4). In exams, set (y=0) for (x)-intercepts.
Step 2
Why this answer is correct
The correct answer is A. ((2,0)) और ((4,0)) / ((2,0)) and ((4,0)). Since (x-2-6x+8=(x-2)(x-4)), the zeroes are (x=2) and (x=4). In exams, set (y=0) for (x)-intercepts.
Step 3
Exam Tip
(x-2-6x+8=(x-2)(x-4)) इसलिए शून्य (x=2) और (x=4) हैं। परीक्षा में (x)-अवरोध के लिए (y=0) रखें।
Since \(\sqrt{x+2}\ge0\), (f(x)\ge -4). In exams, take the square-root minimum (0) and add the vertical shift.
Step 2
Why this answer is correct
The correct answer is A. \([-4,\infty\)). Since \(\sqrt{x+2}\ge0\), (f(x)\ge -4). In exams, take the square-root minimum (0) and add the vertical shift.
Step 3
Exam Tip
\(\sqrt{x+2}\ge0\) इसलिए (f(x)\ge -4)। परीक्षा में वर्गमूल का न्यूनतम (0) लेकर ऊर्ध्व विस्थापन जोड़ें।
(x-2) shifts the graph (2) units right and (+1) shifts it (1) unit up. In exams, read (x-a) as a right shift.
Step 2
Why this answer is correct
The correct answer is B. (2) दाईं ओर और (1) ऊपर / (2) right and (1) up. (x-2) shifts the graph (2) units right and (+1) shifts it (1) unit up. In exams, read (x-a) as a right shift.
Step 3
Exam Tip
(x-2) दाईं ओर (2) इकाई और (+1) ऊपर (1) इकाई खिसकाता है। परीक्षा में (x-a) को दाईं ओर मानें।
Equating gives \(x^2=2x+3\), that is \(x^2-2x-3=0\). In exams, set the two (y)-values equal for intersections.
Step 2
Why this answer is correct
The correct answer is A. (x=-1) और (x=3) / (x=-1) and (x=3). Equating gives \(x^2=2x+3\), that is \(x^2-2x-3=0\). In exams, set the two (y)-values equal for intersections.
Step 3
Exam Tip
समान करने पर \(x^2=2x+3\) यानी \(x^2-2x-3=0\) मिलता है। परीक्षा में प्रतिच्छेद के लिए दोनों (y)-मान बराबर रखें।
The denominator (x+3) becomes zero at (x=-3). In exams, find the vertical asymptote of a reciprocal graph from the denominator.
Step 2
Why this answer is correct
The correct answer is A. (x=-3). The denominator (x+3) becomes zero at (x=-3). In exams, find the vertical asymptote of a reciprocal graph from the denominator.
Step 3
Exam Tip
हर (x+3) शून्य होने पर (x=-3) मिलता है। परीक्षा में पारस्परिक ग्राफ का लंबवत आसमापी हर से निकालें।
From (2x+6=0), (x=-3) and the minimum is (0). In exams, find the vertex by setting the inside of the modulus to (0).
Step 2
Why this answer is correct
The correct answer is A. ((-3,0)). From (2x+6=0), (x=-3) and the minimum is (0). In exams, find the vertex by setting the inside of the modulus to (0).
Step 3
Exam Tip
(2x+6=0) से (x=-3) और न्यूनतम (0) मिलता है। परीक्षा में मापांक के अंदर को (0) रखकर शीर्ष खोजें।
A. केंद्र ((0,0)) और त्रिज्या (3) वाला वृत्त/circle with centre ((0,0)) and radius (3)
Step 1
Concept
Squaring gives \(x^2+y^2=9\) with \(y\ge0\). In exams, identify \(\sqrt{r^2-x^2}\) as the upper semicircle.
Step 2
Why this answer is correct
The correct answer is A. केंद्र ((0,0)) और त्रिज्या (3) वाला वृत्त / circle with centre ((0,0)) and radius (3). Squaring gives \(x^2+y^2=9\) with \(y\ge0\). In exams, identify \(\sqrt{r^2-x^2}\) as the upper semicircle.
Step 3
Exam Tip
वर्ग करने पर \(x^2+y^2=9\) और \(y\ge0\) मिलता है। परीक्षा में \(\sqrt{r^2-x^2}\) को ऊपरी अर्धवृत्त पहचानें।
\(\frac{x}{|x|}=1\) for (x>0) and (-1) for (x<0). In exams, remember to exclude (x=0) from the domain.
Step 2
Why this answer is correct
The correct answer is B. (f(x)=\operatorname{sgn}(x)). \(\frac{x}{|x|}=1\) for (x>0) and (-1) for (x<0). In exams, remember to exclude (x=0) from the domain.
Step 3
Exam Tip
\(\frac{x}{|x|}=1\) जब (x>0) और (-1) जब (x<0) होता है। परीक्षा में (x=0) को प्रांत से हटाना न भूलें।
Since \(\sqrt{x-1}\ge0\) and its least value is (0) at (x=1), \(y\le3\). In exams, for a negative square-root graph, check the starting point for maximum.
Step 2
Why this answer is correct
The correct answer is C. (3). Since \(\sqrt{x-1}\ge0\) and its least value is (0) at (x=1), \(y\le3\). In exams, for a negative square-root graph, check the starting point for maximum.
Step 3
Exam Tip
\(\sqrt{x-1}\ge0\) और सबसे छोटा मान (0) (x=1) पर है इसलिए \(y\le3\)। परीक्षा में ऋणात्मक वर्गमूल वाले ग्राफ का अधिकतम आरंभिक बिंदु पर देखें।
The axis of symmetry is \(x=-\frac{b}{2a}=-\frac{4}{2}=-2\). In exams, use the formula for the axis of a quadratic graph.
Step 2
Why this answer is correct
The correct answer is A. (x=-2). The axis of symmetry is \(x=-\frac{b}{2a}=-\frac{4}{2}=-2\). In exams, use the formula for the axis of a quadratic graph.
Step 3
Exam Tip
सममिति अक्ष \(x=-\frac{b}{2a}=-\frac{4}{2}=-2\) है। परीक्षा में द्विघात ग्राफ की अक्ष सूत्र से निकालें।
The degrees of numerator and denominator are equal, so the ratio of leading coefficients is \(\frac{2}{1}=2\). In exams, use the ratio of leading coefficients for equal degrees.
Step 2
Why this answer is correct
The correct answer is B. (y=2). The degrees of numerator and denominator are equal, so the ratio of leading coefficients is \(\frac{2}{1}=2\). In exams, use the ratio of leading coefficients for equal degrees.
Step 3
Exam Tip
अंश और हर की घात समान है इसलिए अग्र गुणांकों का अनुपात \(\frac{2}{1}=2\) है। परीक्षा में समान घातों पर अग्र गुणांक का अनुपात लें।
\(\lfloor3.1\rfloor+1=4\), so ((3.1,3)) is not on the graph. In exams, first find the greatest integer of (x) while checking a point.
Step 2
Why this answer is correct
The correct answer is D. ((3.1,3)). \(\lfloor3.1\rfloor+1=4\), so ((3.1,3)) is not on the graph. In exams, first find the greatest integer of (x) while checking a point.
Step 3
Exam Tip
\(\lfloor3.1\rfloor+1=4\) इसलिए ((3.1,3)) ग्राफ पर नहीं है। परीक्षा में बिंदु जांचते समय पहले (x) का महत्तम पूर्णांक निकालें।
The distance between (4) and (-2) is (6), so the minimum sum is (6). In exams, remember the minimum of (|x-a|+|x-b|) is (|a-b|).
Step 2
Why this answer is correct
The correct answer is C. (6). The distance between (4) and (-2) is (6), so the minimum sum is (6). In exams, remember the minimum of (|x-a|+|x-b|) is (|a-b|).
Step 3
Exam Tip
दो बिंदुओं (4) और (-2) के बीच दूरी (6) है इसलिए योग का न्यूनतम (6) है। परीक्षा में (|x-a|+|x-b|) का न्यूनतम (|a-b|) याद रखें।
The minimum of (|x+1|) is (0) at (x=-1), so (y=6) is maximum. In exams, a modulus graph with a negative multiplier has its maximum at the vertex.
Step 2
Why this answer is correct
The correct answer is A. अधिकतम (6) जब (x=-1) / maximum (6) when (x=-1). The minimum of (|x+1|) is (0) at (x=-1), so (y=6) is maximum. In exams, a modulus graph with a negative multiplier has its maximum at the vertex.
Step 3
Exam Tip
(|x+1|) का न्यूनतम (0) (x=-1) पर है इसलिए (y=6) अधिकतम है। परीक्षा में ऋणात्मक गुणक वाले मापांक में शीर्ष अधिकतम होता है।
\(x^2+1\) is an even function and its minimum is (1), so it does not cut the (x)-axis. In exams, check both symmetry and minimum value.
Step 2
Why this answer is correct
The correct answer is A. \(y=x^2+1\). \(x^2+1\) is an even function and its minimum is (1), so it does not cut the (x)-axis. In exams, check both symmetry and minimum value.
Step 3
Exam Tip
\(x^2+1\) सम फलन है और इसका न्यूनतम (1) है इसलिए (x)-अक्ष नहीं कटता। परीक्षा में सममिति और न्यूनतम दोनों जांचें।
The denominator (|x|) becomes zero at (x=0), so (0) is excluded. In exams, remove values that make the denominator zero.
Step 2
Why this answer is correct
The correct answer is B. \(\mathbb{R}\setminus{0}\). The denominator (|x|) becomes zero at (x=0), so (0) is excluded. In exams, remove values that make the denominator zero.
Step 3
Exam Tip
हर (|x|) (x=0) पर शून्य हो जाता है इसलिए (0) हटता है। परीक्षा में हर को शून्य बनाने वाला मान प्रांत से निकालें।
\(\lfloor2.3\rfloor=2\) and \(\lfloor-2.3\rfloor=-3\), so the value is (5). In exams, avoid mistakes with greatest integers of negative numbers.
Step 2
Why this answer is correct
The correct answer is C. (5). \(\lfloor2.3\rfloor=2\) and \(\lfloor-2.3\rfloor=-3\), so the value is (5). In exams, avoid mistakes with greatest integers of negative numbers.
Step 3
Exam Tip
\(\lfloor2.3\rfloor=2\) और \(\lfloor-2.3\rfloor=-3\) इसलिए मान (5) है। परीक्षा में ऋणात्मक महत्तम पूर्णांक में गलती न करें।
For (x>2), both (x+2) and (x-2) are positive, so both signum values are (1). In exams, check the sign of each signum term separately.
Step 2
Why this answer is correct
The correct answer is D. (2). For (x>2), both (x+2) and (x-2) are positive, so both signum values are (1). In exams, check the sign of each signum term separately.
Step 3
Exam Tip
(x>2) पर दोनों (x+2) और (x-2) धनात्मक हैं इसलिए दोनों साइनम मान (1) हैं। परीक्षा में हर साइनम पद का चिह्न अलग जांचें।
The denominator (x-2-9=(x-3)(x+3)) is zero at \(x=\pm3\). In exams, find vertical asymptotes from the zeroes of the denominator.
Step 2
Why this answer is correct
The correct answer is A. (x=-3) और (x=3) / (x=-3) and (x=3). The denominator (x-2-9=(x-3)(x+3)) is zero at \(x=\pm3\). In exams, find vertical asymptotes from the zeroes of the denominator.
Step 3
Exam Tip
हर (x-2-9=(x-3)(x+3)) शून्य होने पर \(x=\pm3\) मिलता है। परीक्षा में हर के शून्यों से लंबवत आसमापी खोजें।
A. प्रांत \([-1,\infty\)) और परिसर (\(-\infty,0]\)/domain \([-1,\infty\)) and range (\(-\infty,0]\)
Step 1
Concept
From \(x+1\ge0\), the domain is \([-1,\infty\)), and the negative sign gives \(y\le0\). In exams, check both the square root and the outside negative sign.
Step 2
Why this answer is correct
The correct answer is A. प्रांत \([-1,\infty\)) और परिसर (\(-\infty,0]\) / domain \([-1,\infty\)) and range (\(-\infty,0]\). From \(x+1\ge0\), the domain is \([-1,\infty\)), and the negative sign gives \(y\le0\). In exams, check both the square root and the outside negative sign.
Step 3
Exam Tip
\(x+1\ge0\) से प्रांत \([-1,\infty\)) है और ऋण चिह्न से \(y\le0\)। परीक्षा में वर्गमूल और बाहरी ऋण दोनों का प्रभाव देखें।
\(|x^2-4|=0\) when \(x^2-4=0\), so \(x=\pm2\). In exams, modulus is zero only when the inside expression is zero.
Step 2
Why this answer is correct
The correct answer is A. ((-2,0)) और ((2,0)) / ((-2,0)) and ((2,0)). \(|x^2-4|=0\) when \(x^2-4=0\), so \(x=\pm2\). In exams, modulus is zero only when the inside expression is zero.
Step 3
Exam Tip
\(|x^2-4|=0\) तब \(x^2-4=0\) होता है इसलिए \(x=\pm2\)। परीक्षा में मापांक शून्य तभी होता है जब अंदर की राशि शून्य हो।
A. (x=0) और (x=1) और (x=-1)/(x=0) and (x=1) and (x=-1)
Step 1
Concept
The equation \(x^2=|x|\) gives \(|x|^2=|x|\), so (|x|=0) or (|x|=1). In exams, use \(x^2=|x|^2\).
Step 2
Why this answer is correct
The correct answer is A. (x=0) और (x=1) और (x=-1) / (x=0) and (x=1) and (x=-1). The equation \(x^2=|x|\) gives \(|x|^2=|x|\), so (|x|=0) or (|x|=1). In exams, use \(x^2=|x|^2\).
Step 3
Exam Tip
समीकरण \(x^2=|x|\) से \(|x|^2=|x|\) मिलता है इसलिए (|x|=0) या (|x|=1)। परीक्षा में \(x^2=|x|^2\) का उपयोग करें।
The denominator (x+2=0) gives (x=-2), and the ratio of leading coefficients is (1). In exams, find the two asymptotes using different rules.
Step 2
Why this answer is correct
The correct answer is A. (x=-2) और (y=1) / (x=-2) and (y=1). The denominator (x+2=0) gives (x=-2), and the ratio of leading coefficients is (1). In exams, find the two asymptotes using different rules.
Step 3
Exam Tip
हर (x+2=0) से (x=-2) और समान घातों के अग्र गुणांकों का अनुपात (1) है। परीक्षा में दोनों आसमापी अलग नियम से निकालें।
The modulus graph decreases up to the vertex (x=3) and then increases. In exams, remember the left arm of a (V)-graph is decreasing.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,3]\). The modulus graph decreases up to the vertex (x=3) and then increases. In exams, remember the left arm of a (V)-graph is decreasing.
Step 3
Exam Tip
मापांक ग्राफ शीर्ष (x=3) तक घटता और फिर बढ़ता है। परीक्षा में (V)-ग्राफ का बायां भाग घटता हुआ याद रखें।
The denominator \(x^2+1\) has minimum (1) at (x=0), so the fraction has maximum (1). In exams, a smaller positive denominator gives a larger fraction.
Step 2
Why this answer is correct
The correct answer is B. (1). The denominator \(x^2+1\) has minimum (1) at (x=0), so the fraction has maximum (1). In exams, a smaller positive denominator gives a larger fraction.
Step 3
Exam Tip
हर \(x^2+1\) का न्यूनतम (1) (x=0) पर है इसलिए भिन्न का अधिकतम (1) है। परीक्षा में हर को छोटा करने से भिन्न बड़ा होता है।
From (0=-|x-2|+4), (|x-2|=4), so (x=-2) or (x=6). In exams, add the distance in both directions after modulus.
Step 2
Why this answer is correct
The correct answer is A. ((-2,0)) और ((6,0)) / ((-2,0)) and ((6,0)). From (0=-|x-2|+4), (|x-2|=4), so (x=-2) or (x=6). In exams, add the distance in both directions after modulus.
Step 3
Exam Tip
(0=-|x-2|+4) से (|x-2|=4) इसलिए (x=-2) या (x=6)। परीक्षा में मापांक के बाद दोनों दिशाओं में दूरी जोड़ें।
Reflection in the (x)-axis gives \(y=-x^2\), and shifting (3) up gives \(y=-x^2+3\). In exams, track reflection and shift carefully.
Step 2
Why this answer is correct
The correct answer is B. \(y=-x^2+3\). Reflection in the (x)-axis gives \(y=-x^2\), and shifting (3) up gives \(y=-x^2+3\). In exams, track reflection and shift carefully.
Step 3
Exam Tip
(x)-अक्ष में परावर्तन से \(y=-x^2\) और (3) ऊपर से \(y=-x^2+3\) मिलता है। परीक्षा में परावर्तन और विस्थापन का क्रम ध्यान रखें।
\(\frac{1}{x^2}>0\) for \(x\ne0\), so the graph stays above and is undefined at (x=0). In exams, remember positive branches on both sides when \(x^2\) is in the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(y=\frac{1}{x^2}\). \(\frac{1}{x^2}>0\) for \(x\ne0\), so the graph stays above and is undefined at (x=0). In exams, remember positive branches on both sides when \(x^2\) is in the denominator.
Step 3
Exam Tip
\(\frac{1}{x^2}>0\) जब \(x\ne0\) इसलिए ग्राफ ऊपर रहता है और (x=0) पर परिभाषित नहीं। परीक्षा में हर में \(x^2\) होने पर दोनों ओर धनात्मक शाखाएं याद रखें।
Between (-1) and (3), the total distance from both points remains constant (4). In exams, interpret the sum of two moduli as distance.
Step 2
Why this answer is correct
The correct answer is A. ([-1,3]). Between (-1) and (3), the total distance from both points remains constant (4). In exams, interpret the sum of two moduli as distance.
Step 3
Exam Tip
(-1) और (3) के बीच दोनों बिंदुओं से कुल दूरी स्थिर (4) रहती है। परीक्षा में दो मापांकों के योग को दूरी की तरह समझें।
A. बाएं से मान (2) के पास और बिंदु (3) पर मान (3)/left value near (2) and value (3) at the point
Step 1
Concept
On ([2,3)), the value is (2), but \(\lfloor3\rfloor=3\). In exams, watch open and closed endpoints in a step graph.
Step 2
Why this answer is correct
The correct answer is A. बाएं से मान (2) के पास और बिंदु (3) पर मान (3) / left value near (2) and value (3) at the point. On ([2,3)), the value is (2), but \(\lfloor3\rfloor=3\). In exams, watch open and closed endpoints in a step graph.
Step 3
Exam Tip
([2,3)) पर मान (2) है लेकिन \(\lfloor3\rfloor=3\)। परीक्षा में सीढ़ीनुमा ग्राफ में बंद और खुले बिंदु ध्यान से देखें।
Since \(\sqrt{|{-x}|}=\sqrt{|x|}\), the function is even. In exams, identify (y)-axis symmetry from (f(-x)=f(x)).
Step 2
Why this answer is correct
The correct answer is B. (y)-अक्ष के प्रति / about the (y)-axis. Since \(\sqrt{|{-x}|}=\sqrt{|x|}\), the function is even. In exams, identify (y)-axis symmetry from (f(-x)=f(x)).
Step 3
Exam Tip
\(\sqrt{|{-x}|}=\sqrt{|x|}\) इसलिए फलन सम है। परीक्षा में (f(-x)=f(x)) से (y)-अक्ष सममिति पहचानें।
From (|x-1|+2=4), (|x-1|=2), so (x=-1) or (x=3). In exams, set the (y)-values equal for intersection with a horizontal line.
Step 2
Why this answer is correct
The correct answer is A. (x=-1) और (x=3) / (x=-1) and (x=3). From (|x-1|+2=4), (|x-1|=2), so (x=-1) or (x=3). In exams, set the (y)-values equal for intersection with a horizontal line.
Step 3
Exam Tip
(|x-1|+2=4) से (|x-1|=2) इसलिए (x=-1) या (x=3)। परीक्षा में क्षैतिज रेखा से प्रतिच्छेद के लिए (y)-मान बराबर करें।
Between (0) and (2), the total distance has minimum (2), and it increases outside. In exams, use the distance idea for sums of moduli.
Step 2
Why this answer is correct
The correct answer is B. \([2,\infty\)). Between (0) and (2), the total distance has minimum (2), and it increases outside. In exams, use the distance idea for sums of moduli.
Step 3
Exam Tip
(0) और (2) के बीच कुल दूरी न्यूनतम (2) है और बाहर बढ़ती है। परीक्षा में दूरी वाले मापांक योग से न्यूनतम दूरी निकालें।
