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The two square roots give \(4\le x\le10\) and the denominator gives \(x\ne6\). In exams take the intersection of all conditions.
Step 2
Why this answer is correct
The correct answer is A. \([4,10]\setminus{6}\). The two square roots give \(4\le x\le10\) and the denominator gives \(x\ne6\). In exams take the intersection of all conditions.
Step 3
Exam Tip
दोनों वर्गमूलों से \(4\le x\le10\) और हर से \(x\ne6\) चाहिए। परीक्षा में सभी शर्तों का प्रतिच्छेद लें।
A. प्रांत ( [-3,3] ), परिसर ( [0,3] )/Domain ( [-3,3] ), range ( [0,3] )
Step 1
Concept
For \(\sqrt{9-x^2}\) to be real, \(9-x^2\ge 0\), so \(x\in[-3,3]\) and \(f(x)\in[0,3]\). In exams, always keep the expression inside a square root \(\ge 0\).
Step 2
Why this answer is correct
The correct answer is A. प्रांत ( [-3,3] ), परिसर ( [0,3] ) / Domain ( [-3,3] ), range ( [0,3] ). For \(\sqrt{9-x^2}\) to be real, \(9-x^2\ge 0\), so \(x\in[-3,3]\) and \(f(x)\in[0,3]\). In exams, always keep the expression inside a square root \(\ge 0\).
Step 3
Exam Tip
\(\sqrt{9-x^2}\) वास्तविक होने के लिए \(9-x^2\ge 0\), इसलिए \(x\in[-3,3]\) और \(f(x)\in[0,3]\)। परीक्षा में वर्गमूल वाले फलन में अंदर की राशि को हमेशा \(\ge 0\) रखें।
The expression inside the square root must satisfy \(\frac{x-1}{x-7}\ge0\) and \(x\ne7\). A sign table gives ((-\infty,1]\cup\(7,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. ((-\infty,1]\cup\(7,\infty\)). The expression inside the square root must satisfy \(\frac{x-1}{x-7}\ge0\) and \(x\ne7\). A sign table gives ((-\infty,1]\cup\(7,\infty\)).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{x-1}{x-7}\ge0\) और \(x\ne7\) चाहिए। संकेत तालिका से ((-\infty,1]\cup\(7,\infty\)) मिलता है।
The square root is in the denominator, so ((x+2)(5-x)>0) is required. In exams do not include equality for a denominator radical.
Step 2
Why this answer is correct
The correct answer is A. ((-2,5)). The square root is in the denominator, so ((x+2)(5-x)>0) is required. In exams do not include equality for a denominator radical.
Step 3
Exam Tip
हर में वर्गमूल है इसलिए ((x+2)(5-x)>0) चाहिए। परीक्षा में denominator radical के लिए बराबरी शामिल न करें।
The inside expression is (x-2+4x-12=(x+6)(x-2)) and it must be \(\ge0\). In exams choose the outer intervals for an upward quadratic.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-6]\cup[2,\infty\)). The inside expression is (x-2+4x-12=(x+6)(x-2)) and it must be \(\ge0\). In exams choose the outer intervals for an upward quadratic.
Step 3
Exam Tip
अंदर की राशि (x-2+4x-12=(x+6)(x-2)) है और उसे \(\ge0\) होना चाहिए। परीक्षा में upward quadratic के लिए बाहरी अंतराल चुनें।
The square root needs \(18-2x-x^2\ge0\). This is a downward quadratic, so the closed interval between the roots is the domain.
Step 2
Why this answer is correct
The correct answer is A. \([-1-\sqrt{19},-1+\sqrt{19}]\). The square root needs \(18-2x-x^2\ge0\). This is a downward quadratic, so the closed interval between the roots is the domain.
Step 3
Exam Tip
वर्गमूल के लिए \(18-2x-x^2\ge0\) चाहिए। यह downward quadratic है इसलिए roots के बीच वाला बंद अंतराल डोमेन है।
(x-2-10x+21=(x-5)2-4), so the minimum is (-4). In exams complete the square to find the range of a quadratic.
Step 2
Why this answer is correct
The correct answer is A. \([-4,\infty\)). (x-2-10x+21=(x-5)2-4), so the minimum is (-4). In exams complete the square to find the range of a quadratic.
Step 3
Exam Tip
(x-2-10x+21=(x-5)2-4), इसलिए न्यूनतम (-4) है। परीक्षा में quadratic की रेंज के लिए पूर्ण वर्ग बनाएं।
The denominator (x-2-6x+13=(x-3)2+4), whose minimum value is (4). So the output is greater than (0) and up to \(\frac{1}{4}\).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\frac{1}{4}]\). The denominator (x-2-6x+13=(x-3)2+4), whose minimum value is (4). So the output is greater than (0) and up to \(\frac{1}{4}\).
Step 3
Exam Tip
हर (x-2-6x+13=(x-3)2+4) है, जिसकी न्यूनतम वैल्यू (4) है। इसलिए output (0) से बड़ा और \(\frac{1}{4}\) तक है।
\(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\). At (x=0), \(\frac{3}{7}\) is obtained and (1) is never obtained.
Step 2
Why this answer is correct
The correct answer is A. \([\frac{3}{7},1\)). \(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\). At (x=0), \(\frac{3}{7}\) is obtained and (1) is never obtained.
Step 3
Exam Tip
\(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\) है। (x=0) पर \(\frac{3}{7}\) मिलता है और (1) कभी नहीं मिलता।
The denominator (|x+4|+3) has minimum value (3) and can grow without bound. Hence the range is ( \(0,\frac{1}{3}]\).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\frac{1}{3}]\). The denominator (|x+4|+3) has minimum value (3) and can grow without bound. Hence the range is ( \(0,\frac{1}{3}]\).
Step 3
Exam Tip
हर (|x+4|+3) की न्यूनतम वैल्यू (3) है और यह अनंत तक बढ़ सकता है। इसलिए रेंज ( \(0,\frac{1}{3}]\) है।
Since \(|x+1|\ge0\), \(9-3|x+1|\le9\) and can go down without bound. In exams a negative coefficient changes the range direction.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,9]\). Since \(|x+1|\ge0\), \(9-3|x+1|\le9\) and can go down without bound. In exams a negative coefficient changes the range direction.
Step 3
Exam Tip
क्योंकि \(|x+1|\ge0\), इसलिए \(9-3|x+1|\le9\) और नीचे अनंत तक जा सकता है। परीक्षा में negative coefficient से range की दिशा बदलती है।
\(\sqrt{x-5}\ge0\), so \(4-\sqrt{x-5}\le4\). In exams if a minus sign is before the square root, outputs go downward.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,4]\). \(\sqrt{x-5}\ge0\), so \(4-\sqrt{x-5}\le4\). In exams if a minus sign is before the square root, outputs go downward.
Step 3
Exam Tip
\(\sqrt{x-5}\ge0\), इसलिए \(4-\sqrt{x-5}\le4\)। परीक्षा में वर्गमूल से पहले ऋण चिह्न हो तो output नीचे की ओर जाता है।
On the negative interval, the smallest value of \(x^2\) is (9) and the largest is (49). In exams check endpoints when (0) is not in the interval.
Step 2
Why this answer is correct
The correct answer is A. ([9,49]). On the negative interval, the smallest value of \(x^2\) is (9) and the largest is (49). In exams check endpoints when (0) is not in the interval.
Step 3
Exam Tip
ऋणात्मक अंतराल पर \(x^2\) की सबसे छोटी वैल्यू (9) और सबसे बड़ी (49) है। परीक्षा में interval में (0) न हो तो endpoints जांचें।
The domain contains (0), so the minimum is (0), and (6) is not included, so (36) is not included. In exams check how an open endpoint affects the range.
Step 2
Why this answer is correct
The correct answer is A. ([0,36)). The domain contains (0), so the minimum is (0), and (6) is not included, so (36) is not included. In exams check how an open endpoint affects the range.
Step 3
Exam Tip
डोमेन में (0) शामिल है, इसलिए न्यूनतम (0) है और (6) शामिल नहीं इसलिए (36) शामिल नहीं होगा। परीक्षा में खुले endpoint का असर रेंज पर देखें।
The function is decreasing; at (x=1) it gives (3), and as \(x\to5\) it approaches (-13), not included. In exams the range order reverses for a decreasing function.
Step 2
Why this answer is correct
The correct answer is A. ((-13,3]). The function is decreasing; at (x=1) it gives (3), and as \(x\to5\) it approaches (-13), not included. In exams the range order reverses for a decreasing function.
Step 3
Exam Tip
फलन घटता है, (x=1) पर (3) और \(x\to5\) पर (-13) मिलता है पर शामिल नहीं होता। परीक्षा में decreasing function में range का क्रम उलट जाता है।
The domain contains (x=5), so the minimum is (0), and the far endpoint (12) gives maximum (7). In exams check the zero point of modulus.
Step 2
Why this answer is correct
The correct answer is A. ([0,7]). The domain contains (x=5), so the minimum is (0), and the far endpoint (12) gives maximum (7). In exams check the zero point of modulus.
Step 3
Exam Tip
डोमेन में (x=5) शामिल है, इसलिए minimum (0) है और दूर endpoint (12) से maximum (7) है। परीक्षा में modulus का zero point देखें।
\(\sqrt{x+4}\) is an increasing function and gives (0) and (4) at the endpoints. In exams include endpoint values for a closed domain.
Step 2
Why this answer is correct
The correct answer is A. ([0,4]). \(\sqrt{x+4}\) is an increasing function and gives (0) and (4) at the endpoints. In exams include endpoint values for a closed domain.
Step 3
Exam Tip
\(\sqrt{x+4}\) बढ़ता हुआ फलन है और endpoints पर (0) तथा (4) देता है। परीक्षा में closed domain हो तो endpoint values शामिल करें।
On this domain the denominator is positive and the reciprocal decreases. (x=3) gives \(\frac{1}{2}\) and (x=9) gives \(\frac{1}{8}\).
Step 2
Why this answer is correct
The correct answer is A. \([\frac{1}{8},\frac{1}{2}]\). On this domain the denominator is positive and the reciprocal decreases. (x=3) gives \(\frac{1}{2}\) and (x=9) gives \(\frac{1}{8}\).
Step 3
Exam Tip
इस domain पर denominator धनात्मक है और reciprocal घटता है। (x=3) से \(\frac{1}{2}\) और (x=9) से \(\frac{1}{8}\) मिलता है।
The value of (x+2) lies in ([-4,-1]), so the reciprocal gives \([-1,-\frac{1}{4}]\). In exams handle the order carefully for negative denominators.
Step 2
Why this answer is correct
The correct answer is A. \([-1,-\frac{1}{4}]\). The value of (x+2) lies in ([-4,-1]), so the reciprocal gives \([-1,-\frac{1}{4}]\). In exams handle the order carefully for negative denominators.
Step 3
Exam Tip
(x+2) की वैल्यू ([-4,-1]) में है, इसलिए reciprocal \([-1,-\frac{1}{4}]\) देता है। परीक्षा में ऋणात्मक denominator पर क्रम सावधानी से रखें।
\(\frac{1}{x-9}\) is never (0), so output (4) cannot occur. In exams identify the excluded range value after a vertical shift.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{4}\). \(\frac{1}{x-9}\) is never (0), so output (4) cannot occur. In exams identify the excluded range value after a vertical shift.
Step 3
Exam Tip
\(\frac{1}{x-9}\) कभी (0) नहीं होता, इसलिए (4) output नहीं बन सकता। परीक्षा में vertical shift से excluded range value पहचानें।
\(\frac{6}{x+5}\) is never (0), so output (-2) is not obtained. In exams identify horizontal and vertical shifts of reciprocal functions separately.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-2}\). \(\frac{6}{x+5}\) is never (0), so output (-2) is not obtained. In exams identify horizontal and vertical shifts of reciprocal functions separately.
Step 3
Exam Tip
\(\frac{6}{x+5}\) कभी (0) नहीं होता, इसलिए (-2) output नहीं मिलता। परीक्षा में reciprocal function की horizontal और vertical shift अलग पहचानें।
In \(\frac{x}{x^2-9}\), the denominator is ((x-3)(x+3)), so \(x=\pm3\) are removed. In exams remove denominator zero values in rational functions.
Step 2
Why this answer is correct
The correct answer is A. (f(x)=\frac{x}{x-2-9}). In \(\frac{x}{x^2-9}\), the denominator is ((x-3)(x+3)), so \(x=\pm3\) are removed. In exams remove denominator zero values in rational functions.
Step 3
Exam Tip
\(\frac{x}{x^2-9}\) में हर ((x-3)(x+3)) है, इसलिए \(x=\pm3\) हटते हैं। परीक्षा में rational function में denominator zero values हटाएं।
\(\frac{2}{x-1}\) is never (0), so output (5) cannot occur. In exams identify the range of a vertically shifted reciprocal.
Step 2
Why this answer is correct
The correct answer is A. (f(x)=\frac{2}{x-1}+5). \(\frac{2}{x-1}\) is never (0), so output (5) cannot occur. In exams identify the range of a vertically shifted reciprocal.
Step 3
Exam Tip
\(\frac{2}{x-1}\) कभी (0) नहीं होता, इसलिए (5) output नहीं मिल सकता। परीक्षा में reciprocal की vertical shift वाली range पहचानें।
A. डोमेन \(\mathbb{R}\) और रेंज \([5,\infty\))/Domain \(\mathbb{R}\) and range \([5,\infty\))
Step 1
Concept
Since \(x^2+25\ge25\), (f(x)\ge5) and every real (x) is allowed. In exams first check the minimum value inside.
Step 2
Why this answer is correct
The correct answer is A. डोमेन \(\mathbb{R}\) और रेंज \([5,\infty\)) / Domain \(\mathbb{R}\) and range \([5,\infty\)). Since \(x^2+25\ge25\), (f(x)\ge5) and every real (x) is allowed. In exams first check the minimum value inside.
Step 3
Exam Tip
क्योंकि \(x^2+25\ge25\), इसलिए (f(x)\ge5) और हर वास्तविक (x) स्वीकार्य है। परीक्षा में पहले अंदर की न्यूनतम वैल्यू देखें।
A. डोमेन \(\mathbb{R}\setminus{-4,4}\) है/Domain is \(\mathbb{R}\setminus{-4,4}\)
Step 1
Concept
The denominator \(x^2-16\) becomes (0) when (x=-4) or (x=4). In exams remove values that make the denominator (0).
Step 2
Why this answer is correct
The correct answer is A. डोमेन \(\mathbb{R}\setminus{-4,4}\) है / Domain is \(\mathbb{R}\setminus{-4,4}\). The denominator \(x^2-16\) becomes (0) when (x=-4) or (x=4). In exams remove values that make the denominator (0).
Step 3
Exam Tip
हर \(x^2-16\) तब (0) होता है जब (x=-4) या (x=4)। परीक्षा में denominator को (0) बनाने वाली values हटाएं।
(|x-7|=11) gives (x-7=11) or (x-7=-11). In exams take both branches of a modulus equation.
Step 2
Why this answer is correct
The correct answer is A. (x=18) या (x=-4) / (x=18) or (x=-4). (|x-7|=11) gives (x-7=11) or (x-7=-11). In exams take both branches of a modulus equation.
Step 3
Exam Tip
(|x-7|=11) से (x-7=11) या (x-7=-11) मिलता है। परीक्षा में मापांक समीकरण की दोनों शाखाएं लें।
Since \(x^2\ge0\), \(x^2-15\ge-15\). In exams remember that the minimum value of the square function is (0).
Step 2
Why this answer is correct
The correct answer is A. \([-15,\infty\)). Since \(x^2\ge0\), \(x^2-15\ge-15\). In exams remember that the minimum value of the square function is (0).
Step 3
Exam Tip
क्योंकि \(x^2\ge0\), इसलिए \(x^2-15\ge-15\)। परीक्षा में square function की minimum value (0) याद रखें।
A. रेंज \([11,\infty\)) है और codomain \(\mathbb{R}\) है/Range is \([11,\infty\)) and codomain is \(\mathbb{R}\)
Step 1
Concept
In the notation, the second set is the codomain \(\mathbb{R}\), while actual outputs are \([11,\infty\)). In exams keep range and codomain separate.
Step 2
Why this answer is correct
The correct answer is A. रेंज \([11,\infty\)) है और codomain \(\mathbb{R}\) है / Range is \([11,\infty\)) and codomain is \(\mathbb{R}\). In the notation, the second set is the codomain \(\mathbb{R}\), while actual outputs are \([11,\infty\)). In exams keep range and codomain separate.
Step 3
Exam Tip
notation में दूसरा समुच्चय codomain \(\mathbb{R}\) है, जबकि वास्तविक outputs \([11,\infty\)) हैं। परीक्षा में range और codomain को अलग रखें।
The outputs are (5,2,0,3), written as ({0,2,3,5}) in set form. In exams write repeated or unordered values in set form.
Step 2
Why this answer is correct
The correct answer is A. ({0,2,3,5}). The outputs are (5,2,0,3), written as ({0,2,3,5}) in set form. In exams write repeated or unordered values in set form.
Step 3
Exam Tip
outputs (5,2,0,3) हैं और set में इन्हें ({0,2,3,5}) लिखा जाता है। परीक्षा में repeated या unordered values को set form में लिखें।
The expression inside the square root must satisfy \(\frac{6-x}{x+2}\ge0\) and \(x\ne-2\). A sign table gives ((-2,6]).
Step 2
Why this answer is correct
The correct answer is A. ((-2,6]). The expression inside the square root must satisfy \(\frac{6-x}{x+2}\ge0\) and \(x\ne-2\). A sign table gives ((-2,6]).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{6-x}{x+2}\ge0\) और \(x\ne-2\) चाहिए। संकेत तालिका से ((-2,6]) मिलता है।
The numerator square root gives \(x\ge2\) and the denominator square root gives (8-x>0). Hence the domain is ([2,8)).
Step 2
Why this answer is correct
The correct answer is A. ([2,8)). The numerator square root gives \(x\ge2\) and the denominator square root gives (8-x>0). Hence the domain is ([2,8)).
Step 3
Exam Tip
ऊपर के वर्गमूल से \(x\ge2\) और नीचे के वर्गमूल से (8-x>0) चाहिए। इसलिए डोमेन ([2,8)) है।
Since (x=2) is included, output (0) is included, but (x=18) is open, so (4) is not included. In exams track endpoint openness in the range.
Step 2
Why this answer is correct
The correct answer is A. ([0,4)). Since (x=2) is included, output (0) is included, but (x=18) is open, so (4) is not included. In exams track endpoint openness in the range.
Step 3
Exam Tip
(x=2) शामिल है इसलिए output (0) शामिल है, लेकिन (x=18) खुला है इसलिए (4) शामिल नहीं है। परीक्षा में endpoint की openness रेंज में भी देखें।
Since (x=20) is included, output (0) is included, and since (x=4) is open, output (4) is not included. In exams endpoints reverse for a decreasing square-root expression.
Step 2
Why this answer is correct
The correct answer is A. ([0,4)). Since (x=20) is included, output (0) is included, and since (x=4) is open, output (4) is not included. In exams endpoints reverse for a decreasing square-root expression.
Step 3
Exam Tip
(x=20) शामिल है इसलिए output (0) शामिल है, और (x=4) खुला है इसलिए output (4) शामिल नहीं है। परीक्षा में decreasing square-root expression में endpoints उलट जाते हैं।
