The square root gives \(x\ge 3\) and the denominator gives \(x\ne 5\). In exams take the intersection of all restrictions.
Step 2
Why this answer is correct
The correct answer is A. \([3,\infty\)\setminus{5}). The square root gives \(x\ge 3\) and the denominator gives \(x\ne 5\). In exams take the intersection of all restrictions.
Step 3
Exam Tip
वर्गमूल से \(x\ge 3\) और हर से \(x\ne 5\) चाहिए। परीक्षा में सभी प्रतिबंधों का प्रतिच्छेद लें।
The square root needs \(2x-1\ge 0\), so \(x\ge \frac{1}{2}\). Since (x=-4) is not in this interval, it need not be removed separately.
Step 2
Why this answer is correct
The correct answer is A. \([\frac{1}{2},\infty\)). The square root needs \(2x-1\ge 0\), so \(x\ge \frac{1}{2}\). Since (x=-4) is not in this interval, it need not be removed separately.
Step 3
Exam Tip
वर्गमूल के लिए \(2x-1\ge 0\) से \(x\ge \frac{1}{2}\) मिलता है। (x=-4) इस अंतराल में नहीं है, इसलिए अलग से हटाने की जरूरत नहीं है।
The square root needs \(x^2-16\ge 0\), that is \(x^2\ge 16\). In exams \(x^2\ge a^2\) gives the outer intervals.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-4]\cup[4,\infty\)). The square root needs \(x^2-16\ge 0\), that is \(x^2\ge 16\). In exams \(x^2\ge a^2\) gives the outer intervals.
Step 3
Exam Tip
वर्गमूल के लिए \(x^2-16\ge 0\), यानी \(x^2\ge 16\)। परीक्षा में \(x^2\ge a^2\) से बाहरी अंतराल मिलते हैं।
The square root is in the denominator, so \(x^2-25>0\) is required. In exams do not include equality for a denominator square root.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-5\)\cup\(5,\infty\)). The square root is in the denominator, so \(x^2-25>0\) is required. In exams do not include equality for a denominator square root.
Step 3
Exam Tip
हर में वर्गमूल है इसलिए \(x^2-25>0\) चाहिए। परीक्षा में हर वाले वर्गमूल के लिए बराबरी शामिल न करें।
The original denominator is (x-2-4=(x-2)(x+2)), so \(x\ne -2,2\). In exams remove zeros of the original denominator before cancellation.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-2,2}\). The original denominator is (x-2-4=(x-2)(x+2)), so \(x\ne -2,2\). In exams remove zeros of the original denominator before cancellation.
Step 3
Exam Tip
मूल हर (x-2-4=(x-2)(x+2)) है, इसलिए \(x\ne -2,2\)। परीक्षा में काटने से पहले मूल हर के शून्य हटाएं।
The denominator (x-2+2x-3=(x+3)(x-1)) removes (x=-3) and (x=1). In exams factorize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{1,-3}\). The denominator (x-2+2x-3=(x+3)(x-1)) removes (x=-3) and (x=1). In exams factorize the denominator.
Step 3
Exam Tip
हर (x-2+2x-3=(x+3)(x-1)) से (x=-3) और (x=1) हटेंगे। परीक्षा में हर को गुणनखंड करें।
The conditions \(x-1\ge 0\) and \(7-x\ge 0\) together give \(1\le x\le 7\). In exams take the intersection for the combined domain.
Step 2
Why this answer is correct
The correct answer is A. ([1,7]). The conditions \(x-1\ge 0\) and \(7-x\ge 0\) together give \(1\le x\le 7\). In exams take the intersection for the combined domain.
Step 3
Exam Tip
शर्तें \(x-1\ge 0\) और \(7-x\ge 0\) मिलकर \(1\le x\le 7\) देती हैं। परीक्षा में संयुक्त डोमेन के लिए प्रतिच्छेद लें।
The square root needs \(12-3x\ge 0\), giving \(x\le 4\). In exams dividing by a negative coefficient reverses the inequality.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,4]\). The square root needs \(12-3x\ge 0\), giving \(x\le 4\). In exams dividing by a negative coefficient reverses the inequality.
Step 3
Exam Tip
वर्गमूल के लिए \(12-3x\ge 0\) से \(x\le 4\) मिलता है। परीक्षा में ऋणात्मक गुणांक से भाग देने पर असमानता उलटती है।
The expression inside the square root must satisfy ((x-2)(x-6)\ge 0). In exams use a sign table to choose the outer intervals.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,2]\cup[6,\infty\)). The expression inside the square root must satisfy ((x-2)(x-6)\ge 0). In exams use a sign table to choose the outer intervals.
Step 3
Exam Tip
वर्गमूल के अंदर ((x-2)(x-6)\ge 0) होना चाहिए। परीक्षा में संकेत तालिका से बाहरी अंतराल चुनें।
The minimum value of (|x+2|) is (0), so the output is at least (1). In exams the multiplier (3) does not change the minimum when modulus is (0).
Step 2
Why this answer is correct
The correct answer is A. \([1,\infty\)). The minimum value of (|x+2|) is (0), so the output is at least (1). In exams the multiplier (3) does not change the minimum when modulus is (0).
Step 3
Exam Tip
(|x+2|) की न्यूनतम वैल्यू (0) है, इसलिए आउटपुट कम से कम (1) होगा। परीक्षा में गुणक (3) न्यूनतम को नहीं बदलता जब मापांक (0) हो।
\(\sqrt{x-1}\ge 0\), so \(2-\sqrt{x-1}\le 2\) and it can go down without bound. In exams a negative sign reverses the direction of the range.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,2]\). \(\sqrt{x-1}\ge 0\), so \(2-\sqrt{x-1}\le 2\) and it can go down without bound. In exams a negative sign reverses the direction of the range.
Step 3
Exam Tip
\(\sqrt{x-1}\ge 0\), इसलिए \(2-\sqrt{x-1}\le 2\) और नीचे अनंत तक जा सकता है। परीक्षा में ऋण चिह्न रेंज की दिशा बदलता है।
The denominator \(x^2+1\) has minimum value (1), so the maximum output is (1) and (0) is not reached. In exams (0) may be a limit, not a value, for reciprocals.
Step 2
Why this answer is correct
The correct answer is A. ( (0,1]). The denominator \(x^2+1\) has minimum value (1), so the maximum output is (1) and (0) is not reached. In exams (0) may be a limit, not a value, for reciprocals.
Step 3
Exam Tip
हर \(x^2+1\) की न्यूनतम वैल्यू (1) है, इसलिए अधिकतम आउटपुट (1) है और (0) नहीं मिलता। परीक्षा में reciprocal में (0) सीमा हो सकता है, वैल्यू नहीं।
The range of \(\frac{2}{x^2+4}\) is ( \(0,\frac{1}{2}]\), so adding (3) gives ( \(3,\frac{7}{2}]\). In exams apply the vertical shift to the range.
Step 2
Why this answer is correct
The correct answer is A. ( \(3,\frac{7}{2}]\). The range of \(\frac{2}{x^2+4}\) is ( \(0,\frac{1}{2}]\), so adding (3) gives ( \(3,\frac{7}{2}]\). In exams apply the vertical shift to the range.
Step 3
Exam Tip
\(\frac{2}{x^2+4}\) की रेंज ( \(0,\frac{1}{2}]\) है, इसलिए (3) जोड़ने पर ( \(3,\frac{7}{2}]\) मिलेगा। परीक्षा में रेंज पर vertical shift लगाएं।
The interval ([-3,2]) contains (0), so the minimum is (0) and the maximum is (9). In exams check both endpoints and the critical point.
Step 2
Why this answer is correct
The correct answer is A. ([0,9]). The interval ([-3,2]) contains (0), so the minimum is (0) and the maximum is (9). In exams check both endpoints and the critical point.
Step 3
Exam Tip
अंतराल ([-3,2]) में (0) है, इसलिए न्यूनतम (0) और अधिकतम (9) है। परीक्षा में endpoints और critical point दोनों देखें।
On ([2,5]), \(x^2\) is increasing, so endpoint values are (4) and (25). In exams include endpoints for a closed interval.
Step 2
Why this answer is correct
The correct answer is A. ([4,25]). On ([2,5]), \(x^2\) is increasing, so endpoint values are (4) and (25). In exams include endpoints for a closed interval.
Step 3
Exam Tip
([2,5]) पर \(x^2\) बढ़ता है, इसलिए सिरों पर वैल्यू (4) और (25) मिलती हैं। परीक्षा में बंद अंतराल के endpoint शामिल करें।
The function is decreasing, so at (x=0) it gives (2) and at (x=4) it gives (-10). In exams the order may reverse for a decreasing function.
Step 2
Why this answer is correct
The correct answer is A. ([-10,2]). The function is decreasing, so at (x=0) it gives (2) and at (x=4) it gives (-10). In exams the order may reverse for a decreasing function.
Step 3
Exam Tip
फलन घटता है, इसलिए (x=0) पर (2) और (x=4) पर (-10) मिलता है। परीक्षा में घटते फलन में क्रम बदल सकता है।
The domain contains (x=1), so the minimum is (0) and the maximum is (3) at an endpoint. In exams always check the zero point of modulus.
Step 2
Why this answer is correct
The correct answer is A. ([0,3]). The domain contains (x=1), so the minimum is (0) and the maximum is (3) at an endpoint. In exams always check the zero point of modulus.
Step 3
Exam Tip
डोमेन में (x=1) शामिल है, इसलिए न्यूनतम (0) है और अधिकतम endpoint पर (3) है। परीक्षा में मापांक का zero point जरूर देखें।
On positive (x), \(\frac{1}{x}\) decreases, so the range is \([\frac{1}{4},1]\). In exams reverse the order of endpoint values for a decreasing reciprocal.
Step 2
Why this answer is correct
The correct answer is A. \([\frac{1}{4},1]\). On positive (x), \(\frac{1}{x}\) decreases, so the range is \([\frac{1}{4},1]\). In exams reverse the order of endpoint values for a decreasing reciprocal.
Step 3
Exam Tip
धनात्मक (x) पर \(\frac{1}{x}\) घटता है, इसलिए रेंज \([\frac{1}{4},1]\) है। परीक्षा में घटते reciprocal में endpoint values का क्रम बदलें।
\(\frac{1}{x-2}\) can never be (0). In exams a horizontal shift keeps (0) excluded from the reciprocal range.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{0}\). \(\frac{1}{x-2}\) can never be (0). In exams a horizontal shift keeps (0) excluded from the reciprocal range.
Step 3
Exam Tip
\(\frac{1}{x-2}\) कभी (0) नहीं हो सकता। परीक्षा में horizontal shift से reciprocal की रेंज से हटने वाली वैल्यू (0) ही रहती है।
\(\frac{3}{x+1}\) is never (0), so the output (-4) is not obtained. In exams a vertical shift changes the excluded range value.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-4}\). \(\frac{3}{x+1}\) is never (0), so the output (-4) is not obtained. In exams a vertical shift changes the excluded range value.
Step 3
Exam Tip
\(\frac{3}{x+1}\) कभी (0) नहीं होता, इसलिए (-4) आउटपुट नहीं मिलेगा। परीक्षा में vertical shift से excluded range value बदलती है।
Inside, (x-2+6x+13=(x+3)2+4), so the minimum is (4) and the range is \([2,\infty\)). In exams complete the square inside first.
Step 2
Why this answer is correct
The correct answer is A. \([2,\infty\)). Inside, (x-2+6x+13=(x+3)2+4), so the minimum is (4) and the range is \([2,\infty\)). In exams complete the square inside first.
Step 3
Exam Tip
अंदर (x-2+6x+13=(x+3)2+4) है, इसलिए न्यूनतम (4) और रेंज \([2,\infty\)) है। परीक्षा में पहले अंदर का पूर्ण वर्ग बनाएं।
Since \(x^2\ge 0\), the output can be (0), but it never reaches (1). In exams \(\frac{x^2}{x^2+1}=1-\frac{1}{x^2+1}\) is useful.
Step 2
Why this answer is correct
The correct answer is A. ([0,1)). Since \(x^2\ge 0\), the output can be (0), but it never reaches (1). In exams \(\frac{x^2}{x^2+1}=1-\frac{1}{x^2+1}\) is useful.
Step 3
Exam Tip
\(x^2\ge 0\) से आउटपुट (0) मिल सकता है, पर (1) कभी नहीं मिलता। परीक्षा में \(\frac{x^2}{x^2+1}=1-\frac{1}{x^2+1}\) उपयोगी है।
\(\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}\). The minimum \(\frac{1}{2}\) is obtained, but (1) is not obtained.
Step 2
Why this answer is correct
The correct answer is A. \([\frac{1}{2},1\)). \(\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}\). The minimum \(\frac{1}{2}\) is obtained, but (1) is not obtained.
Step 3
Exam Tip
\(\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}\) है। न्यूनतम \(\frac{1}{2}\) मिलता है, लेकिन (1) नहीं मिलता।
The denominator (|x|+2) has minimum value (2) and can grow without bound. So the output is greater than (0) and up to \(\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\frac{1}{2}]\). The denominator (|x|+2) has minimum value (2) and can grow without bound. So the output is greater than (0) and up to \(\frac{1}{2}\).
Step 3
Exam Tip
हर (|x|+2) की न्यूनतम वैल्यू (2) है और यह अनंत तक बढ़ सकता है। इसलिए आउटपुट (0) से बड़ा और \(\frac{1}{2}\) तक है।
\(\sqrt{x-2}\) needs \(x-2\ge 0\), so all real (x) are not allowed. In exams check the domain restriction in square root options.
Step 2
Why this answer is correct
The correct answer is C. (f(x)=\sqrt{x-2}). \(\sqrt{x-2}\) needs \(x-2\ge 0\), so all real (x) are not allowed. In exams check the domain restriction in square root options.
Step 3
Exam Tip
\(\sqrt{x-2}\) के लिए \(x-2\ge 0\) चाहिए, इसलिए सभी वास्तविक (x) नहीं चलेंगे। परीक्षा में वर्गमूल वाले विकल्प पर domain restriction जांचें।
\(\frac{1}{x}\) is never (0), so the output cannot be (2). In exams identify the vertical shift of a reciprocal.
Step 2
Why this answer is correct
The correct answer is A. (f(x)=\frac{1}{x}+2). \(\frac{1}{x}\) is never (0), so the output cannot be (2). In exams identify the vertical shift of a reciprocal.
Step 3
Exam Tip
\(\frac{1}{x}\) कभी (0) नहीं होता, इसलिए (2) आउटपुट नहीं बनता। परीक्षा में reciprocal के vertical shift को पहचानें।
A. डोमेन \(\mathbb{R}\) और रेंज \([3,\infty\))/Domain \(\mathbb{R}\) and range \([3,\infty\))
Step 1
Concept
Since \(x^2+9\ge 9\), the square root is at least (3) and the domain is all real numbers. In exams check the minimum value inside.
Step 2
Why this answer is correct
The correct answer is A. डोमेन \(\mathbb{R}\) और रेंज \([3,\infty\)) / Domain \(\mathbb{R}\) and range \([3,\infty\)). Since \(x^2+9\ge 9\), the square root is at least (3) and the domain is all real numbers. In exams check the minimum value inside.
Step 3
Exam Tip
\(x^2+9\ge 9\), इसलिए वर्गमूल कम से कम (3) है और डोमेन सभी वास्तविक है। परीक्षा में अंदर की न्यूनतम वैल्यू देखें।
A. डोमेन \(\mathbb{R}\setminus{-1,1}\) है/Domain is \(\mathbb{R}\setminus{-1,1}\)
Step 1
Concept
The denominator \(x^2-1\) becomes (0) when \(x=\pm 1\). In exams remove denominator zero values in rational functions.
Step 2
Why this answer is correct
The correct answer is A. डोमेन \(\mathbb{R}\setminus{-1,1}\) है / Domain is \(\mathbb{R}\setminus{-1,1}\). The denominator \(x^2-1\) becomes (0) when \(x=\pm 1\). In exams remove denominator zero values in rational functions.
Step 3
Exam Tip
हर \(x^2-1\) तब (0) होता है जब \(x=\pm 1\)। परीक्षा में rational function में denominator zero values हटाएं।
A. रेंज \([4,\infty\)) है और codomain \(\mathbb{R}\) है/Range is \([4,\infty\)) and codomain is \(\mathbb{R}\)
Step 1
Concept
The second set in the function notation is the codomain, while actual outputs are \([4,\infty\)). In exams identify codomain from arrow notation.
Step 2
Why this answer is correct
The correct answer is A. रेंज \([4,\infty\)) है और codomain \(\mathbb{R}\) है / Range is \([4,\infty\)) and codomain is \(\mathbb{R}\). The second set in the function notation is the codomain, while actual outputs are \([4,\infty\)). In exams identify codomain from arrow notation.
Step 3
Exam Tip
फलन में दिया गया दूसरा समुच्चय codomain है, जबकि वास्तविक आउटपुट \([4,\infty\)) हैं। परीक्षा में arrow notation से codomain पहचानें।
For (x=1,2,4,5), the outputs are \(0,1,\sqrt{3},2\). In exams use only the given inputs in a set domain.
Step 2
Why this answer is correct
The correct answer is A. \({0,1,\sqrt{3},2}\). For (x=1,2,4,5), the outputs are \(0,1,\sqrt{3},2\). In exams use only the given inputs in a set domain.
Step 3
Exam Tip
(x=1,2,4,5) पर outputs \(0,1,\sqrt{3},2\) मिलते हैं। परीक्षा में set domain में केवल दिए गए inputs लें।
The expression inside the square root must satisfy \(\frac{x-2}{x+1}\ge 0\) and \(x\ne -1\). A sign table gives (\(-\infty,-1\)\cup[2,\infty)).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-1\)\cup[2,\infty)). The expression inside the square root must satisfy \(\frac{x-2}{x+1}\ge 0\) and \(x\ne -1\). A sign table gives (\(-\infty,-1\)\cup[2,\infty)).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{x-2}{x+1}\ge 0\) और \(x\ne -1\) चाहिए। संकेत तालिका से (\(-\infty,-1\)\cup[2,\infty)) मिलता है।
