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Class 11 Mathematics Easy Quiz

Level 31 • 50/50 questions • 40 seconds per question.

Level readiness 50/50 Questions
Time Left 33:20 40 sec/question
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Question 1 / 50 0 score
Answered 0/50 Correct 0 Time 33:20

यदि (f(x)=x+3) एक वास्तविक मान वाला फलन है, तो इसका प्रांत क्या है?

If (f(x)=x+3) is a real valued function, what is its domain?

Explanation opens after your attempt
Correct Answer

A. सभी वास्तविक संख्याएँAll real numbers

Step 1

Concept

The linear function (f(x)=x+3) is defined for every real (x). In exams, the domain of a basic linear function is usually \(\mathbb{R}\).

Step 2

Why this answer is correct

The correct answer is A. सभी वास्तविक संख्याएँ / All real numbers. The linear function (f(x)=x+3) is defined for every real (x). In exams, the domain of a basic linear function is usually \(\mathbb{R}\).

Step 3

Exam Tip

रेखीय फलन (f(x)=x+3) हर वास्तविक (x) के लिए परिभाषित है। परीक्षा में हर रेखीय फलन का प्रांत सामान्यतः \(\mathbb{R}\) लें।

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फलन (f(x)=x-2) का परिसर क्या है जब प्रांत \(\mathbb{R}\) है?

What is the range of (f(x)=x-2) when the domain is \(\mathbb{R}\)?

Explanation opens after your attempt
Correct Answer

A. \([0,\infty\))

Step 1

Concept

For every real (x), \(x^2 \ge 0\). So the least value is (0), and the range is \([0,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. \([0,\infty\)). For every real (x), \(x^2 \ge 0\). So the least value is (0), and the range is \([0,\infty\)).

Step 3

Exam Tip

किसी भी वास्तविक (x) के लिए \(x^2 \ge 0\) होता है। इसलिए सबसे छोटा मान (0) है और परिसर \([0,\infty\)) है।

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फलन (f(x)=\sqrt{x}) का वास्तविक प्रांत क्या है?

What is the real domain of (f(x)=\sqrt{x})?

Explanation opens after your attempt
Correct Answer

A. \([0,\infty\))

Step 1

Concept

For the square root to be real, \(x \ge 0\) is required. Hence the domain is \([0,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. \([0,\infty\)). For the square root to be real, \(x \ge 0\) is required. Hence the domain is \([0,\infty\)).

Step 3

Exam Tip

वर्गमूल वास्तविक होने के लिए \(x \ge 0\) चाहिए। इसलिए प्रांत \([0,\infty\)) है।

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फलन (f(x)=\frac{1}{x}) का प्रांत क्या है?

What is the domain of (f(x)=\frac{1}{x})?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}-{0}\)

Step 1

Concept

The denominator is (x), so the function is not defined at (x=0). In exams, make sure the denominator is not (0).

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}-{0}\). The denominator is (x), so the function is not defined at (x=0). In exams, make sure the denominator is not (0).

Step 3

Exam Tip

हर में (x) है, इसलिए (x=0) पर फलन परिभाषित नहीं है। परीक्षा में हर को (0) न बनने दें।

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फलन (f(x)=5) का परिसर क्या है?

What is the range of (f(x)=5)?

Explanation opens after your attempt
Correct Answer

A. ({5})

Step 1

Concept

This is a constant function and gives only the value (5). The range of a constant function is a singleton set.

Step 2

Why this answer is correct

The correct answer is A. ({5}). This is a constant function and gives only the value (5). The range of a constant function is a singleton set.

Step 3

Exam Tip

यह स्थिर फलन है और हर (x) पर मान (5) ही देता है। स्थिर फलन का परिसर एकल समुच्चय होता है।

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यदि (f(x)=2x-1), तो (f(3)) का मान क्या है?

If (f(x)=2x-1), what is the value of (f(3))?

Explanation opens after your attempt
Correct Answer

A. (5)

Step 1

Concept

(f(3)=2(3)-1=5). To evaluate a function, substitute the given value in place of (x).

Step 2

Why this answer is correct

The correct answer is A. (5). (f(3)=2(3)-1=5). To evaluate a function, substitute the given value in place of (x).

Step 3

Exam Tip

(f(3)=2(3)-1=5) है। मान निकालते समय (x) की जगह दिए गए मान को रखें।

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फलन (f(x)=\sqrt{x-2}) का वास्तविक प्रांत क्या है?

What is the real domain of (f(x)=\sqrt{x-2})?

Explanation opens after your attempt
Correct Answer

A. \([2,\infty\))

Step 1

Concept

The expression inside the square root must satisfy \(x-2 \ge 0\). Thus \(x \ge 2\), so the domain is \([2,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. \([2,\infty\)). The expression inside the square root must satisfy \(x-2 \ge 0\). Thus \(x \ge 2\), so the domain is \([2,\infty\)).

Step 3

Exam Tip

वर्गमूल के अंदर \(x-2 \ge 0\) होना चाहिए। इसलिए \(x \ge 2\) और प्रांत \([2,\infty\)) है।

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फलन (f(x)=x-2+1) का परिसर क्या है जब \(x \in \mathbb{R}\)?

What is the range of (f(x)=x-2+1) when \(x \in \mathbb{R}\)?

Explanation opens after your attempt
Correct Answer

A. \([1,\infty\))

Step 1

Concept

Since \(x^2 \ge 0\), we have \(x^2+1 \ge 1\). The minimum value is (1).

Step 2

Why this answer is correct

The correct answer is A. \([1,\infty\)). Since \(x^2 \ge 0\), we have \(x^2+1 \ge 1\). The minimum value is (1).

Step 3

Exam Tip

क्योंकि \(x^2 \ge 0\), इसलिए \(x^2+1 \ge 1\) है। न्यूनतम मान (1) पर मिलता है।

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फलन (f(x)=|x|) का परिसर क्या है?

What is the range of (f(x)=|x|)?

Explanation opens after your attempt
Correct Answer

A. \([0,\infty\))

Step 1

Concept

(|x|) is never negative. Therefore its range is \([0,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. \([0,\infty\)). (|x|) is never negative. Therefore its range is \([0,\infty\)).

Step 3

Exam Tip

(|x|) कभी ऋणात्मक नहीं होता। इसलिए इसका परिसर \([0,\infty\)) है।

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फलन (f(x)=\frac{1}{x-4}) का प्रांत क्या है?

What is the domain of (f(x)=\frac{1}{x-4})?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}-{4}\)

Step 1

Concept

The denominator is (x-4), so at (x=4) it becomes (0). Hence (4) is excluded from the domain.

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}-{4}\). The denominator is (x-4), so at (x=4) it becomes (0). Hence (4) is excluded from the domain.

Step 3

Exam Tip

हर (x-4) है, इसलिए (x=4) पर हर (0) बनता है। इसीलिए (4) को प्रांत से हटाते हैं।

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यदि \(f:\mathbb{R}\to\mathbb{R}\) और (f(x)=3x), तो (f(-2)) क्या होगा?

If \(f:\mathbb{R}\to\mathbb{R}\) and (f(x)=3x), what is (f(-2))?

Explanation opens after your attempt
Correct Answer

A. (-6)

Step 1

Concept

(f(-2)=3(-2)=-6). Be careful with signs while substituting negative values.

Step 2

Why this answer is correct

The correct answer is A. (-6). (f(-2)=3(-2)=-6). Be careful with signs while substituting negative values.

Step 3

Exam Tip

(f(-2)=3(-2)=-6) है। ऋणात्मक मान रखते समय चिन्ह पर ध्यान दें।

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फलन (f(x)=\sqrt{5-x}) का वास्तविक प्रांत क्या है?

What is the real domain of (f(x)=\sqrt{5-x})?

Explanation opens after your attempt
Correct Answer

A. (\(-\infty,5]\)

Step 1

Concept

For the square root, \(5-x \ge 0\) is needed. This gives \(x \le 5\).

Step 2

Why this answer is correct

The correct answer is A. (\(-\infty,5]\). For the square root, \(5-x \ge 0\) is needed. This gives \(x \le 5\).

Step 3

Exam Tip

वर्गमूल के लिए \(5-x \ge 0\) चाहिए। इससे \(x \le 5\) मिलता है।

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फलन (f(x)=2x-2) का परिसर क्या है जब \(x \in \mathbb{R}\)?

What is the range of (f(x)=2x-2) when \(x \in \mathbb{R}\)?

Explanation opens after your attempt
Correct Answer

A. \([0,\infty\))

Step 1

Concept

Since \(x^2 \ge 0\), \(2x^2 \ge 0\). At (x=0), the value is (0).

Step 2

Why this answer is correct

The correct answer is A. \([0,\infty\)). Since \(x^2 \ge 0\), \(2x^2 \ge 0\). At (x=0), the value is (0).

Step 3

Exam Tip

\(x^2 \ge 0\) होने से \(2x^2 \ge 0\) है। (x=0) पर मान (0) मिलता है।

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फलन (f(x)=x-3) का प्रांत और परिसर क्या है?

What are the domain and range of (f(x)=x-3)?

Explanation opens after your attempt
Correct Answer

A. प्रांत \(\mathbb{R}\), परिसर \(\mathbb{R}\)Domain \(\mathbb{R}\), range \(\mathbb{R}\)

Step 1

Concept

The cubic function is defined for every real (x) and can take every real value. Hence both are \(\mathbb{R}\).

Step 2

Why this answer is correct

The correct answer is A. प्रांत \(\mathbb{R}\), परिसर \(\mathbb{R}\) / Domain \(\mathbb{R}\), range \(\mathbb{R}\). The cubic function is defined for every real (x) and can take every real value. Hence both are \(\mathbb{R}\).

Step 3

Exam Tip

घन फलन हर वास्तविक (x) के लिए परिभाषित है और हर वास्तविक मान ले सकता है। इसलिए दोनों \(\mathbb{R}\) हैं।

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यदि (f(x)=x+1) और प्रांत \(A=\{1,2,3\}\) है, तो परिसर क्या है?

If (f(x)=x+1) and the domain is \(A=\{1,2,3\}\), what is the range?

Explanation opens after your attempt
Correct Answer

A. ({2,3,4})

Step 1

Concept

Substituting (1,2,3) gives (2,3,4). For a finite domain, apply the function to each element.

Step 2

Why this answer is correct

The correct answer is A. ({2,3,4}). Substituting (1,2,3) gives (2,3,4). For a finite domain, apply the function to each element.

Step 3

Exam Tip

(1,2,3) रखने पर मान (2,3,4) मिलते हैं। सीमित प्रांत में हर तत्व पर फलन लगाएँ।

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फलन (f(x)=\frac{1}{x-2+1}) का वास्तविक प्रांत क्या है?

What is the real domain of (f(x)=\frac{1}{x-2+1})?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}\)

Step 1

Concept

\(x^2+1\) is never (0). Therefore the function is defined for every real (x).

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}\). \(x^2+1\) is never (0). Therefore the function is defined for every real (x).

Step 3

Exam Tip

\(x^2+1\) कभी (0) नहीं होता। इसलिए फलन हर वास्तविक (x) पर परिभाषित है।

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फलन (f(x)=\sqrt{x+3}) का प्रांत क्या है?

What is the domain of (f(x)=\sqrt{x+3})?

Explanation opens after your attempt
Correct Answer

A. \([-3,\infty\))

Step 1

Concept

The expression inside the square root must satisfy \(x+3 \ge 0\). Therefore \(x \ge -3\).

Step 2

Why this answer is correct

The correct answer is A. \([-3,\infty\)). The expression inside the square root must satisfy \(x+3 \ge 0\). Therefore \(x \ge -3\).

Step 3

Exam Tip

वर्गमूल के अंदर \(x+3 \ge 0\) होना चाहिए। इसलिए \(x \ge -3\) है।

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फलन (f(x)=|x-2|) का न्यूनतम मान क्या है?

What is the minimum value of (f(x)=|x-2|)?

Explanation opens after your attempt
Correct Answer

A. (0)

Step 1

Concept

(|x-2|) is never negative and becomes (0) at (x=2). Hence the minimum value is (0).

Step 2

Why this answer is correct

The correct answer is A. (0). (|x-2|) is never negative and becomes (0) at (x=2). Hence the minimum value is (0).

Step 3

Exam Tip

(|x-2|) का मान कभी ऋणात्मक नहीं होता और (x=2) पर (0) हो जाता है। इसलिए न्यूनतम मान (0) है।

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फलन (f(x)=|x-2|) का परिसर क्या है?

What is the range of (f(x)=|x-2|)?

Explanation opens after your attempt
Correct Answer

A. \([0,\infty\))

Step 1

Concept

A modulus value is (0) or positive. At (x=2), the minimum (0) is obtained.

Step 2

Why this answer is correct

The correct answer is A. \([0,\infty\)). A modulus value is (0) or positive. At (x=2), the minimum (0) is obtained.

Step 3

Exam Tip

मॉड्यूलस का मान (0) या धनात्मक होता है। (x=2) पर न्यूनतम (0) मिलता है।

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यदि (f(x)=4-x), तो (f(0)) क्या है?

If (f(x)=4-x), what is (f(0))?

Explanation opens after your attempt
Correct Answer

A. (4)

Step 1

Concept

(f(0)=4-0=4). Substitute the given (x) directly to find the function value.

Step 2

Why this answer is correct

The correct answer is A. (4). (f(0)=4-0=4). Substitute the given (x) directly to find the function value.

Step 3

Exam Tip

(f(0)=4-0=4) है। फलन का मान निकालने में दिए गए (x) को सीधे रखें।

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फलन (f(x)=\frac{2}{x+1}) का प्रांत क्या है?

What is the domain of (f(x)=\frac{2}{x+1})?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}-{-1}\)

Step 1

Concept

The denominator is (x+1), so it becomes (0) at (x=-1). This value is removed from the domain.

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}-{-1}\). The denominator is (x+1), so it becomes (0) at (x=-1). This value is removed from the domain.

Step 3

Exam Tip

हर (x+1) है, इसलिए (x=-1) पर हर (0) हो जाता है। इस मान को प्रांत से हटाते हैं।

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फलन (f(x)=\sqrt{9-x-2}) के वास्तविक प्रांत के लिए सही शर्त कौन सी है?

Which condition is correct for the real domain of (f(x)=\sqrt{9-x-2})?

Explanation opens after your attempt
Correct Answer

A. \(-3 \le x \le 3\)

Step 1

Concept

For the square root, \(9-x^2 \ge 0\) is needed. This gives \(x^2 \le 9\), that is \(-3 \le x \le 3\).

Step 2

Why this answer is correct

The correct answer is A. \(-3 \le x \le 3\). For the square root, \(9-x^2 \ge 0\) is needed. This gives \(x^2 \le 9\), that is \(-3 \le x \le 3\).

Step 3

Exam Tip

वर्गमूल के लिए \(9-x^2 \ge 0\) चाहिए। इससे \(x^2 \le 9\), यानी \(-3 \le x \le 3\) मिलता है।

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फलन (f(x)=\sqrt{9-x-2}) का वास्तविक प्रांत कौन सा है?

What is the real domain of (f(x)=\sqrt{9-x-2})?

Explanation opens after your attempt
Correct Answer

A. ([-3,3])

Step 1

Concept

The condition \(9-x^2 \ge 0\) gives ([-3,3]). Include the endpoints because the square root can be (0).

Step 2

Why this answer is correct

The correct answer is A. ([-3,3]). The condition \(9-x^2 \ge 0\) gives ([-3,3]). Include the endpoints because the square root can be (0).

Step 3

Exam Tip

शर्त \(9-x^2 \ge 0\) से ([-3,3]) मिलता है। सिरों को शामिल करें क्योंकि वर्गमूल (0) हो सकता है।

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फलन (f(x)=x-2-4) का न्यूनतम मान क्या है जब \(x \in \mathbb{R}\)?

What is the minimum value of (f(x)=x-2-4) when \(x \in \mathbb{R}\)?

Explanation opens after your attempt
Correct Answer

A. (-4)

Step 1

Concept

Since \(x^2 \ge 0\), \(x^2-4 \ge -4\). The minimum value is (-4) at (x=0).

Step 2

Why this answer is correct

The correct answer is A. (-4). Since \(x^2 \ge 0\), \(x^2-4 \ge -4\). The minimum value is (-4) at (x=0).

Step 3

Exam Tip

क्योंकि \(x^2 \ge 0\), इसलिए \(x^2-4 \ge -4\) है। न्यूनतम मान (x=0) पर (-4) है।

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फलन (f(x)=x-2-4) का परिसर क्या है?

What is the range of (f(x)=x-2-4)?

Explanation opens after your attempt
Correct Answer

A. \([-4,\infty\))

Step 1

Concept

The minimum value of \(x^2\) is (0), so the minimum value of \(x^2-4\) is (-4). Hence the range is \([-4,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. \([-4,\infty\)). The minimum value of \(x^2\) is (0), so the minimum value of \(x^2-4\) is (-4). Hence the range is \([-4,\infty\)).

Step 3

Exam Tip

\(x^2\) का न्यूनतम मान (0) है, इसलिए \(x^2-4\) का न्यूनतम मान (-4) है। अतः परिसर \([-4,\infty\)) है।

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यदि \(A=\{-1,0,1\}\) और (f(x)=x-2), तो परिसर क्या है?

If \(A=\{-1,0,1\}\) and (f(x)=x-2), what is the range?

Explanation opens after your attempt
Correct Answer

A. ({0,1})

Step 1

Concept

((-1)2=1), \(0^2=0\), and \(1^2=1\). The distinct output values are ({0,1}).

Step 2

Why this answer is correct

The correct answer is A. ({0,1}). ((-1)2=1), \(0^2=0\), and \(1^2=1\). The distinct output values are ({0,1}).

Step 3

Exam Tip

((-1)2=1), \(0^2=0\), और \(1^2=1\) हैं। अलग-अलग प्राप्त मान ({0,1}) हैं।

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फलन (f(x)=\frac{1}{x-2}) का प्रांत क्या है?

What is the domain of (f(x)=\frac{1}{x-2})?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}-{0}\)

Step 1

Concept

The denominator is \(x^2\), and it becomes (0) at (x=0). Therefore (0) is not in the domain.

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}-{0}\). The denominator is \(x^2\), and it becomes (0) at (x=0). Therefore (0) is not in the domain.

Step 3

Exam Tip

हर \(x^2\) है और (x=0) पर यह (0) बनता है। इसलिए (0) प्रांत में नहीं होगा।

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फलन (f(x)=\frac{1}{x-2}) का परिसर क्या है?

What is the range of (f(x)=\frac{1}{x-2})?

Explanation opens after your attempt
Correct Answer

A. (\(0,\infty\))

Step 1

Concept

When \(x \ne 0\), \(x^2>0\), so \(\frac{1}{x^2}>0\). The value (0) is never obtained.

Step 2

Why this answer is correct

The correct answer is A. (\(0,\infty\)). When \(x \ne 0\), \(x^2>0\), so \(\frac{1}{x^2}>0\). The value (0) is never obtained.

Step 3

Exam Tip

\(x \ne 0\) होने पर \(x^2>0\), इसलिए \(\frac{1}{x^2}>0\) है। मान (0) कभी नहीं मिलता।

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फलन (f(x)=x-2+4) का सबसे छोटा मान क्या है?

What is the least value of (f(x)=x-2+4)?

Explanation opens after your attempt
Correct Answer

A. (4)

Step 1

Concept

The least value of \(x^2\) is (0). Therefore the least value of \(x^2+4\) is (4).

Step 2

Why this answer is correct

The correct answer is A. (4). The least value of \(x^2\) is (0). Therefore the least value of \(x^2+4\) is (4).

Step 3

Exam Tip

\(x^2\) का सबसे छोटा मान (0) होता है। इसलिए \(x^2+4\) का सबसे छोटा मान (4) है।

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फलन (f(x)=x-2+4) का परिसर क्या है?

What is the range of (f(x)=x-2+4)?

Explanation opens after your attempt
Correct Answer

A. \([4,\infty\))

Step 1

Concept

Since \(x^2 \ge 0\), \(x^2+4 \ge 4\). Hence the range is \([4,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. \([4,\infty\)). Since \(x^2 \ge 0\), \(x^2+4 \ge 4\). Hence the range is \([4,\infty\)).

Step 3

Exam Tip

क्योंकि \(x^2 \ge 0\), इसलिए \(x^2+4 \ge 4\) है। अतः परिसर \([4,\infty\)) है।

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फलन (f(x)=\sqrt{x-2}) किसके बराबर है?

The function (f(x)=\sqrt{x-2}) is equal to what?

Explanation opens after your attempt
Correct Answer

A. (|x|)

Step 1

Concept

\(\sqrt{x^2}\) is always non-negative, so it equals (|x|). A common mistake is writing it directly as (x).

Step 2

Why this answer is correct

The correct answer is A. (|x|). \(\sqrt{x^2}\) is always non-negative, so it equals (|x|). A common mistake is writing it directly as (x).

Step 3

Exam Tip

\(\sqrt{x^2}\) का मान हमेशा गैर-ऋणात्मक होता है, इसलिए यह (|x|) के बराबर है। सामान्य गलती इसे सीधे (x) लिखना है।

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वास्तविक मान वाले फलन का अर्थ क्या है?

What does a real valued function mean?

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Correct Answer

A. जिसका हर मान \(\mathbb{R}\) में होIts every value lies in \(\mathbb{R}\)

Step 1

Concept

A real valued function has output values in the real numbers. The key focus is on the codomain or values.

Step 2

Why this answer is correct

The correct answer is A. जिसका हर मान \(\mathbb{R}\) में हो / Its every value lies in \(\mathbb{R}\). A real valued function has output values in the real numbers. The key focus is on the codomain or values.

Step 3

Exam Tip

वास्तविक मान वाला फलन वह है जिसके निर्गत मान वास्तविक संख्याएँ होते हैं। यहाँ मुख्य ध्यान सहप्रांत या मानों पर होता है।

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फलन \(f:A\to\mathbb{R}\) में (A) किसे दर्शाता है?

In the function \(f:A\to\mathbb{R}\), what does (A) represent?

Explanation opens after your attempt
Correct Answer

A. प्रांतDomain

Step 1

Concept

In \(f:A\to\mathbb{R}\), (A) is the input set. Hence (A) is called the domain.

Step 2

Why this answer is correct

The correct answer is A. प्रांत / Domain. In \(f:A\to\mathbb{R}\), (A) is the input set. Hence (A) is called the domain.

Step 3

Exam Tip

\(f:A\to\mathbb{R}\) में (A) इनपुट समुच्चय है। इसलिए (A) प्रांत कहलाता है।

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फलन \(f:A\to\mathbb{R}\) में \(\mathbb{R}\) सामान्यतः क्या दर्शाता है?

In the function \(f:A\to\mathbb{R}\), what does \(\mathbb{R}\) usually represent?

Explanation opens after your attempt
Correct Answer

A. सहप्रांतCodomain

Step 1

Concept

In \(f:A\to\mathbb{R}\), \(\mathbb{R}\) is the set where outputs are expected to lie. It is called the codomain.

Step 2

Why this answer is correct

The correct answer is A. सहप्रांत / Codomain. In \(f:A\to\mathbb{R}\), \(\mathbb{R}\) is the set where outputs are expected to lie. It is called the codomain.

Step 3

Exam Tip

\(f:A\to\mathbb{R}\) में \(\mathbb{R}\) वह समुच्चय है जिसमें मान जाने हैं। यह सहप्रांत कहलाता है।

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यदि (f(x)=x-2) और \(x \in {-2,2}\), तो कितने अलग-अलग मान मिलते हैं?

If (f(x)=x-2) and \(x \in {-2,2}\), how many distinct values are obtained?

Explanation opens after your attempt
Correct Answer

A. (1)

Step 1

Concept

((-2)2=4) and \(2^2=4\), so only one distinct value is obtained. Repeated values are written once in the range.

Step 2

Why this answer is correct

The correct answer is A. (1). ((-2)2=4) and \(2^2=4\), so only one distinct value is obtained. Repeated values are written once in the range.

Step 3

Exam Tip

((-2)2=4) और \(2^2=4\), इसलिए केवल एक अलग मान मिलता है। परिसर में दोहराए गए मान एक बार लिखते हैं।

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फलन (f(x)=\frac{x}{x-2}) का प्रांत क्या है?

What is the domain of (f(x)=\frac{x}{x-2})?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}-{2}\)

Step 1

Concept

The denominator is (x-2), and it becomes (0) at (x=2). Therefore (2) is excluded.

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}-{2}\). The denominator is (x-2), and it becomes (0) at (x=2). Therefore (2) is excluded.

Step 3

Exam Tip

हर (x-2) है और (x=2) पर हर (0) हो जाता है। इसलिए (2) को हटाते हैं।

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फलन (f(x)=\sqrt{2x-6}) का प्रांत क्या है?

What is the domain of (f(x)=\sqrt{2x-6})?

Explanation opens after your attempt
Correct Answer

A. \([3,\infty\))

Step 1

Concept

For the square root, \(2x-6 \ge 0\) is required. This gives \(x \ge 3\).

Step 2

Why this answer is correct

The correct answer is A. \([3,\infty\)). For the square root, \(2x-6 \ge 0\) is required. This gives \(x \ge 3\).

Step 3

Exam Tip

वर्गमूल के लिए \(2x-6 \ge 0\) होना चाहिए। इससे \(x \ge 3\) मिलता है।

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फलन (f(x)=\sqrt{4-2x}) का प्रांत क्या है?

What is the domain of (f(x)=\sqrt{4-2x})?

Explanation opens after your attempt
Correct Answer

A. (\(-\infty,2]\)

Step 1

Concept

The condition \(4-2x \ge 0\) gives \(x \le 2\). Thus the domain is (\(-\infty,2]\).

Step 2

Why this answer is correct

The correct answer is A. (\(-\infty,2]\). The condition \(4-2x \ge 0\) gives \(x \le 2\). Thus the domain is (\(-\infty,2]\).

Step 3

Exam Tip

शर्त \(4-2x \ge 0\) से \(x \le 2\) मिलता है। इसलिए प्रांत (\(-\infty,2]\) है।

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फलन (f(x)=|x|+2) का परिसर क्या है?

What is the range of (f(x)=|x|+2)?

Explanation opens after your attempt
Correct Answer

A. \([2,\infty\))

Step 1

Concept

The minimum value of (|x|) is (0). Therefore the minimum value of (|x|+2) is (2).

Step 2

Why this answer is correct

The correct answer is A. \([2,\infty\)). The minimum value of (|x|) is (0). Therefore the minimum value of (|x|+2) is (2).

Step 3

Exam Tip

(|x|) का न्यूनतम मान (0) है। इसलिए (|x|+2) का न्यूनतम मान (2) है।

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फलन (f(x)=3-|x|) का अधिकतम मान क्या है?

What is the maximum value of (f(x)=3-|x|)?

Explanation opens after your attempt
Correct Answer

A. (3)

Step 1

Concept

The minimum value of (|x|) is (0), so (3-|x|) is maximum at (3). This occurs at (x=0).

Step 2

Why this answer is correct

The correct answer is A. (3). The minimum value of (|x|) is (0), so (3-|x|) is maximum at (3). This occurs at (x=0).

Step 3

Exam Tip

(|x|) का न्यूनतम मान (0) है, इसलिए (3-|x|) अधिकतम (3) होगा। यह (x=0) पर मिलता है।

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फलन (f(x)=3-|x|) का परिसर क्या है?

What is the range of (f(x)=3-|x|)?

Explanation opens after your attempt
Correct Answer

A. (\(-\infty,3]\)

Step 1

Concept

As (|x|) becomes larger, (3-|x|) becomes smaller. The maximum is (3), and there is no lower bound.

Step 2

Why this answer is correct

The correct answer is A. (\(-\infty,3]\). As (|x|) becomes larger, (3-|x|) becomes smaller. The maximum is (3), and there is no lower bound.

Step 3

Exam Tip

(|x|) जितना बड़ा होगा, (3-|x|) उतना छोटा होगा। अधिकतम (3) है और नीचे कोई सीमा नहीं है।

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यदि (f(x)=\frac{x+1}{2}), तो (f(5)) क्या है?

If (f(x)=\frac{x+1}{2}), what is (f(5))?

Explanation opens after your attempt
Correct Answer

A. (3)

Step 1

Concept

(f(5)=\frac{5+1}{2}=3). In a fractional expression, evaluate the numerator carefully first.

Step 2

Why this answer is correct

The correct answer is A. (3). (f(5)=\frac{5+1}{2}=3). In a fractional expression, evaluate the numerator carefully first.

Step 3

Exam Tip

(f(5)=\frac{5+1}{2}=3) है। भिन्न वाले फलन में पहले कोष्ठक का मान निकालें।

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यदि (f(x)=x-2+2x+1), तो (f(1)) क्या है?

If (f(x)=x-2+2x+1), what is (f(1))?

Explanation opens after your attempt
Correct Answer

A. (4)

Step 1

Concept

(f(1)=12+2(1)+1=4). While substituting, evaluate each term separately.

Step 2

Why this answer is correct

The correct answer is A. (4). (f(1)=12+2(1)+1=4). While substituting, evaluate each term separately.

Step 3

Exam Tip

(f(1)=12+2(1)+1=4) है। प्रतिस्थापन में हर पद को अलग-अलग देखें।

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फलन (f(x)=(x-1)2) का परिसर क्या है?

What is the range of (f(x)=(x-1)2)?

Explanation opens after your attempt
Correct Answer

A. \([0,\infty\))

Step 1

Concept

The square ((x-1)2) is always (0) or positive. At (x=1), the minimum (0) is obtained.

Step 2

Why this answer is correct

The correct answer is A. \([0,\infty\)). The square ((x-1)2) is always (0) or positive. At (x=1), the minimum (0) is obtained.

Step 3

Exam Tip

वर्ग ((x-1)2) हमेशा (0) या धनात्मक है। (x=1) पर न्यूनतम (0) मिलता है।

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फलन (f(x)=(x+2)2+1) का परिसर क्या है?

What is the range of (f(x)=(x+2)2+1)?

Explanation opens after your attempt
Correct Answer

A. \([1,\infty\))

Step 1

Concept

The minimum value of ((x+2)2) is (0). Therefore the whole function has minimum value (1).

Step 2

Why this answer is correct

The correct answer is A. \([1,\infty\)). The minimum value of ((x+2)2) is (0). Therefore the whole function has minimum value (1).

Step 3

Exam Tip

((x+2)2) का न्यूनतम मान (0) है। इसलिए पूरा फलन न्यूनतम (1) देता है।

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फलन (f(x)=\frac{1}{x-2+4}) का प्रांत क्या है?

What is the domain of (f(x)=\frac{1}{x-2+4})?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}\)

Step 1

Concept

\(x^2+4\) is positive for every real (x). Therefore the denominator never becomes (0).

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}\). \(x^2+4\) is positive for every real (x). Therefore the denominator never becomes (0).

Step 3

Exam Tip

\(x^2+4\) हर वास्तविक (x) के लिए धनात्मक है। इसलिए हर कभी (0) नहीं बनता।

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यदि (f(x)=2\sqrt{x}), तो इसका वास्तविक प्रांत क्या है?

If (f(x)=2\sqrt{x}), what is its real domain?

Explanation opens after your attempt
Correct Answer

A. \([0,\infty\))

Step 1

Concept

For \(\sqrt{x}\) to be real, \(x \ge 0\) is needed. Multiplication by (2) does not change the domain.

Step 2

Why this answer is correct

The correct answer is A. \([0,\infty\)). For \(\sqrt{x}\) to be real, \(x \ge 0\) is needed. Multiplication by (2) does not change the domain.

Step 3

Exam Tip

\(\sqrt{x}\) वास्तविक होने के लिए \(x \ge 0\) चाहिए। (2) से गुणा करने पर प्रांत नहीं बदलता।

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यदि (f(x)=2\sqrt{x}), तो इसका परिसर क्या है?

If (f(x)=2\sqrt{x}), what is its range?

Explanation opens after your attempt
Correct Answer

A. \([0,\infty\))

Step 1

Concept

\(\sqrt{x}\ge 0\), so \(2\sqrt{x}\ge 0\). At (x=0), the value is (0).

Step 2

Why this answer is correct

The correct answer is A. \([0,\infty\)). \(\sqrt{x}\ge 0\), so \(2\sqrt{x}\ge 0\). At (x=0), the value is (0).

Step 3

Exam Tip

\(\sqrt{x}\ge 0\) होता है, इसलिए \(2\sqrt{x}\ge 0\) है। (x=0) पर मान (0) मिलता है।

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फलन (f(x)=\frac{1}{\sqrt{x-1}}) का प्रांत क्या है?

What is the domain of (f(x)=\frac{1}{\sqrt{x-1}})?

Explanation opens after your attempt
Correct Answer

A. (\(1,\infty\))

Step 1

Concept

The denominator is \(\sqrt{x-1}\), so (x-1>0) is required. Thus (x>1), and the domain is (\(1,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. (\(1,\infty\)). The denominator is \(\sqrt{x-1}\), so (x-1>0) is required. Thus (x>1), and the domain is (\(1,\infty\)).

Step 3

Exam Tip

हर में \(\sqrt{x-1}\) है, इसलिए (x-1>0) चाहिए। इस कारण (x>1) और प्रांत (\(1,\infty\)) है।

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यदि (f(x)=|x+1|-3), तो इसका न्यूनतम मान क्या है?

If (f(x)=|x+1|-3), what is its minimum value?

Explanation opens after your attempt
Correct Answer

A. (-3)

Step 1

Concept

The minimum value of (|x+1|) is (0). Therefore the minimum value of (|x+1|-3) is (-3).

Step 2

Why this answer is correct

The correct answer is A. (-3). The minimum value of (|x+1|) is (0). Therefore the minimum value of (|x+1|-3) is (-3).

Step 3

Exam Tip

(|x+1|) का न्यूनतम मान (0) है। इसलिए (|x+1|-3) का न्यूनतम मान (-3) है।

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FAQs

Class 11 Mathematics Quiz FAQs

How many questions are in this quiz?

This level is designed for 50 active questions. Currently 50 questions are available for the selected class and difficulty.

Is there a timer in this quiz?

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Can I open each question separately?

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