The negative sign makes the graph open downward, and the maximum value is (2). For range, check the vertex and opening direction.
Step 2
Why this answer is correct
The correct answer is A. \(y\le2\). The negative sign makes the graph open downward, and the maximum value is (2). For range, check the vertex and opening direction.
Step 3
Exam Tip
ऋण चिह्न के कारण ग्राफ नीचे खुलता है और अधिकतम मान (2) है। परिसर निकालते समय शीर्ष और खुलने की दिशा देखें।
A right shift replaces (x) by (x-2), and an upward shift adds (5). In transformations, the inside sign works oppositely.
Step 2
Why this answer is correct
The correct answer is A. (y=(x-2)3+5). A right shift replaces (x) by (x-2), and an upward shift adds (5). In transformations, the inside sign works oppositely.
Step 3
Exam Tip
दाएँ खिसकाव में (x) की जगह (x-2) और ऊपर खिसकाव में (+5) जुड़ता है। ग्राफ रूपांतरण में अंदर का चिन्ह उलटा होता है।
For the greatest integer function, the value is (2) on \(2\le x<3\). Its step graph has a closed left end and open right end.
Step 2
Why this answer is correct
The correct answer is A. (2). For the greatest integer function, the value is (2) on \(2\le x<3\). Its step graph has a closed left end and open right end.
Step 3
Exam Tip
महत्तम पूर्णांक फलन में \(2\le x<3\) पर मान (2) रहता है। सीढ़ीनुमा ग्राफ में बायाँ सिरा बंद और दायाँ खुला होता है।
The fractional part is always from (0) up to but not including (1). Remember that (1) is not included in the range.
Step 2
Why this answer is correct
The correct answer is A. \(0\le y<1\). The fractional part is always from (0) up to but not including (1). Remember that (1) is not included in the range.
Step 3
Exam Tip
भिन्नांश भाग हमेशा (0) से (1) से कम होता है। ध्यान रखें कि (1) परिसर में शामिल नहीं होता।
This expression is the sum of distances from (1) and (-1), whose minimum is (2). In distance-type modulus graphs, use the gap between fixed points.
Step 2
Why this answer is correct
The correct answer is A. (2). This expression is the sum of distances from (1) and (-1), whose minimum is (2). In distance-type modulus graphs, use the gap between fixed points.
Step 3
Exam Tip
यह अभिव्यक्ति (x) की (1) और (-1) से दूरियों का योग है, जिसका न्यूनतम (2) है। दूरी वाले मापांक में बिंदुओं के बीच का अंतर उपयोगी होता है।
Here (f(-x)=-f(x)), so the graph is symmetric about the origin. Always compute (f(-x)) to test symmetry.
Step 2
Why this answer is correct
The correct answer is A. विषम सममिति / Odd symmetry. Here (f(-x)=-f(x)), so the graph is symmetric about the origin. Always compute (f(-x)) to test symmetry.
Step 3
Exam Tip
(f(-x)=-f(x)), इसलिए ग्राफ मूलबिंदु के सापेक्ष सममित है। सममिति जाँचने के लिए (f(-x)) अवश्य निकालें।
Since (f(-x)=f(x)), it is an even function. The graph of an even function is symmetric about the (y)-axis.
Step 2
Why this answer is correct
The correct answer is A. (y)-अक्ष सममिति / (y)-axis symmetry. Since (f(-x)=f(x)), it is an even function. The graph of an even function is symmetric about the (y)-axis.
Step 3
Exam Tip
(f(-x)=f(x)), इसलिए यह सम फलन है। सम फलन का ग्राफ (y)-अक्ष के सापेक्ष सममित होता है।
From \(y=\sqrt{9-x^2}\), we get \(x^2+y^2=9\) with \(y\ge0\). The square root gives only the upper semicircle.
Step 2
Why this answer is correct
The correct answer is A. वृत्त \(x^2+y^2=9\) / Circle \(x^2+y^2=9\). From \(y=\sqrt{9-x^2}\), we get \(x^2+y^2=9\) with \(y\ge0\). The square root gives only the upper semicircle.
Step 3
Exam Tip
\(y=\sqrt{9-x^2}\) से \(y^2=9-x^2\), इसलिए \(x^2+y^2=9\) और \(y\ge0\)। वर्गमूल के कारण केवल ऊपरी अर्धवृत्त मिलता है।
For (x)-intercepts, \(|x^2-4|=0\), so \(x^2-4=0\). An absolute value is zero only when its inside is zero.
Step 2
Why this answer is correct
The correct answer is A. (x=-2) और (x=2) / (x=-2) and (x=2). For (x)-intercepts, \(|x^2-4|=0\), so \(x^2-4=0\). An absolute value is zero only when its inside is zero.
Step 3
Exam Tip
(x)-अवरोध के लिए \(|x^2-4|=0\), इसलिए \(x^2-4=0\)। मापांक शून्य तभी होता है जब अंदर का भाग शून्य हो।
Since (f(|x|)=|x|2=x-2), the graph does not change. If the original function is even, using (|x|) may keep the same graph.
Step 2
Why this answer is correct
The correct answer is A. दोनों समान हैं / Both are identical. Since (f(|x|)=|x|2=x-2), the graph does not change. If the original function is even, using (|x|) may keep the same graph.
Step 3
Exam Tip
क्योंकि (f(|x|)=|x|2=x-2), इसलिए ग्राफ नहीं बदलता। यदि मूल फलन सम हो, तो (|x|) लगाने से ग्राफ समान रह सकता है।
A. यह (y)-अक्ष के सापेक्ष सममित है/It is symmetric about the (y)-axis
Step 1
Concept
Because of (|x|), (f(-x)=f(x)), so the graph is symmetric about the (y)-axis. Modulus often creates even symmetry.
Step 2
Why this answer is correct
The correct answer is A. यह (y)-अक्ष के सापेक्ष सममित है / It is symmetric about the (y)-axis. Because of (|x|), (f(-x)=f(x)), so the graph is symmetric about the (y)-axis. Modulus often creates even symmetry.
Step 3
Exam Tip
(|x|) के कारण (f(-x)=f(x)), इसलिए ग्राफ (y)-अक्ष के सापेक्ष सममित है। मापांक अक्सर सममिति बनाता है।
As (x) goes left, \(2^{x-1}\) approaches (0), so the asymptote is (y=3). The outside (+3) shifts the exponential graph upward.
Step 2
Why this answer is correct
The correct answer is A. (y=3). As (x) goes left, \(2^{x-1}\) approaches (0), so the asymptote is (y=3). The outside (+3) shifts the exponential graph upward.
Step 3
Exam Tip
\(2^{x-1}\) बाएँ ओर (0) के निकट जाता है, इसलिए असिम्पटोट (y=3) है। बाहरी (+3) पूरे घातीय ग्राफ को ऊपर खिसकाता है।
The log graph has an asymptote where its input approaches (0), so (x-2=0). In log graphs, the boundary of the domain gives the asymptote.
Step 2
Why this answer is correct
The correct answer is A. (x=2). The log graph has an asymptote where its input approaches (0), so (x-2=0). In log graphs, the boundary of the domain gives the asymptote.
Step 3
Exam Tip
लॉग का अंदरूनी भाग (0) के पास आने पर असिम्पटोट बनती है, इसलिए (x-2=0)। लॉग ग्राफ में सीमा रेखा प्रांत की सीमा होती है।
The maximum of \(\sin x\) is (1), attained at \(x=\frac{\pi}{2}\). Remember key points of standard trigonometric graphs.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{\pi}{2}\). The maximum of \(\sin x\) is (1), attained at \(x=\frac{\pi}{2}\). Remember key points of standard trigonometric graphs.
Step 3
Exam Tip
\(\sin x\) का अधिकतम मान (1) होता है, जो \(x=\frac{\pi}{2}\) पर मिलता है। मानक त्रिकोणमितीय ग्राफ के मुख्य बिंदु याद रखें।
The minimum of \(\cos x\) is (-1), which occurs at \(x=\pi\). When reading graphs, identify maximum and minimum points separately.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pi\). The minimum of \(\cos x\) is (-1), which occurs at \(x=\pi\). When reading graphs, identify maximum and minimum points separately.
Step 3
Exam Tip
\(\cos x\) का न्यूनतम मान (-1) होता है, जो \(x=\pi\) पर है। ग्राफ पढ़ते समय अधिकतम और न्यूनतम बिंदु अलग पहचानें।
A. \(\frac{\pi}{4}\) दाएँ खिसका/Shifted \(\frac{\pi}{4}\) right
Step 1
Concept
Because \(x-\frac{\pi}{4}\) is inside, the graph shifts \(\frac{\pi}{4}\) to the right. Inner transformations move opposite to the sign.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{4}\) दाएँ खिसका / Shifted \(\frac{\pi}{4}\) right. Because \(x-\frac{\pi}{4}\) is inside, the graph shifts \(\frac{\pi}{4}\) to the right. Inner transformations move opposite to the sign.
Step 3
Exam Tip
\(x-\frac{\pi}{4}\) अंदर है, इसलिए ग्राफ \(\frac{\pi}{4}\) दाएँ खिसकता है। अंदर वाले रूपांतरण का दिशा-चिह्न उलटा समझें।
It is the sum of distances from (-1) and (3), which stays minimum between them. The sum of distances to two points is minimized between those points.
Step 2
Why this answer is correct
The correct answer is A. \(-1\le x\le3\). It is the sum of distances from (-1) and (3), which stays minimum between them. The sum of distances to two points is minimized between those points.
Step 3
Exam Tip
यह (-1) और (3) से दूरियों का योग है, जो इनके बीच स्थिर न्यूनतम रहता है। दो बिंदुओं के बीच मापांक दूरी का योग न्यूनतम होता है।
A. वह (x)-अक्ष के ऊपर परावर्तित होगा/It is reflected above the (x)-axis
Step 1
Concept
(|f(x)|) changes negative (y)-values into positive ones. Hence the part below the axis is reflected upward.
Step 2
Why this answer is correct
The correct answer is A. वह (x)-अक्ष के ऊपर परावर्तित होगा / It is reflected above the (x)-axis. (|f(x)|) changes negative (y)-values into positive ones. Hence the part below the axis is reflected upward.
Step 3
Exam Tip
(|f(x)|) ऋणात्मक (y)-मानों को धनात्मक बना देता है। इसलिए नीचे का भाग ऊपर परावर्तित हो जाता है।
A. (y)-अक्ष में परावर्तन/Reflection in the (y)-axis
Step 1
Concept
Replacing (x) by (-x) gives a horizontal reflection. This reflection is about the (y)-axis.
Step 2
Why this answer is correct
The correct answer is A. (y)-अक्ष में परावर्तन / Reflection in the (y)-axis. Replacing (x) by (-x) gives a horizontal reflection. This reflection is about the (y)-axis.
Step 3
Exam Tip
(x) की जगह (-x) रखने से क्षैतिज परावर्तन होता है। यह परावर्तन (y)-अक्ष के सापेक्ष होता है।
Asymptotes occur when the denominator \(x^2-4=0\), so \(x=\pm2\). Always check the zeros of the denominator.
Step 2
Why this answer is correct
The correct answer is A. (x=-2) और (x=2) / (x=-2) and (x=2). Asymptotes occur when the denominator \(x^2-4=0\), so \(x=\pm2\). Always check the zeros of the denominator.
Step 3
Exam Tip
हर \(x^2-4=0\) होने पर असिम्पटोट मिलती है, इसलिए \(x=\pm2\)। हर के शून्य को हमेशा जाँचें।
A. यह शीर्ष और न्यूनतम बिंदु है/It is the vertex and minimum point
Step 1
Concept
Since (|x+1|+2=|x-(-1)|+2), the vertex is ((-1,2)). In an upward-opening modulus graph, the vertex is the minimum point.
Step 2
Why this answer is correct
The correct answer is A. यह शीर्ष और न्यूनतम बिंदु है / It is the vertex and minimum point. Since (|x+1|+2=|x-(-1)|+2), the vertex is ((-1,2)). In an upward-opening modulus graph, the vertex is the minimum point.
Step 3
Exam Tip
(|x+1|+2=|x-(-1)|+2), इसलिए शीर्ष ((-1,2)) है। ऊपर खुलने वाले मापांक ग्राफ में शीर्ष न्यूनतम होता है।
The graph of an inverse function is obtained by reflection in the line (y=x). In exams, check by changing a point ((a,b)) into ((b,a)).
Step 2
Why this answer is correct
The correct answer is A. (y=x). The graph of an inverse function is obtained by reflection in the line (y=x). In exams, check by changing a point ((a,b)) into ((b,a)).
Step 3
Exam Tip
व्युत्क्रम फलन का ग्राफ (y=x) रेखा में परावर्तन से मिलता है। परीक्षा में ((a,b)) बिंदु को ((b,a)) बनाकर जाँचें।
It is the sum of distances of (x) from (2) and (6), so the minimum value is (6-2=4). For such questions, look at the distance between the two fixed points.
Step 2
Why this answer is correct
The correct answer is A. (4). It is the sum of distances of (x) from (2) and (6), so the minimum value is (6-2=4). For such questions, look at the distance between the two fixed points.
Step 3
Exam Tip
यह (x) की (2) और (6) से दूरियों का योग है, जिसका न्यूनतम मान (6-2=4) है। ऐसे प्रश्नों में दोनों स्थिर बिंदुओं के बीच की दूरी देखें।
