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Mathematics Real valued functions, domain and range of these functions MCQ Questions for Class 11 General

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Real valued functions, domain and range of these functions Practice Questions

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Question 401/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

फलन (f(x)=\sqrt{3x-12}) का प्रांत क्या है?

What is the domain of the function (f(x)=\sqrt{3x-12})?

Explanation opens after your attempt
Correct Answer

A. \(x\ge 4\)

Explanation

Simple Explanation

वर्गमूल के अंदर \(3x-12\ge 0\) होना चाहिए, इसलिए \(x\ge 4\)। परीक्षा में वर्गमूल के लिए अंदर का भाग हमेशा अशून्य या धनात्मक रखें। / The expression inside the square root must satisfy \(3x-12\ge 0\), so \(x\ge 4\). In exams, always keep the radicand non-negative.

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Question 402/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

फलन (f(x)=\sqrt{25-x-2}) का प्रांत चुनिए।

Choose the domain of (f(x)=\sqrt{25-x-2}).

Explanation opens after your attempt
Correct Answer

A. ([-5,5])

Explanation

Simple Explanation

यहां \(25-x^2\ge 0\), इसलिए \(x^2\le 25\) और \(x\in[-5,5]\)। ऐसे प्रश्नों में असमानता को सावधानी से हल करें। / Here \(25-x^2\ge 0\), so \(x^2\le 25\) and \(x\in[-5,5]\). In such questions, solve the inequality carefully.

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Question 403/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

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

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

Explanation opens after your attempt
Correct Answer

A. (x>2)

Explanation

Simple Explanation

हर में वर्गमूल है, इसलिए (x-2>0) चाहिए। इसलिए प्रांत (\(2,\infty\)) है। / The square root is in the denominator, so (x-2>0) is required. Hence the domain is (\(2,\infty\)).

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Question 404/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

फलन (f(x)=\frac{x+1}{x-2-9}) का प्रांत ज्ञात कीजिए।

Find the domain of (f(x)=\frac{x+1}{x-2-9}).

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}-{-3,3}\)

Explanation

Simple Explanation

हर \(x^2-9\ne 0\) होना चाहिए, इसलिए \(x\ne -3,3\)। परीक्षा में हर को शून्य बनाने वाले मान हटाएं। / The denominator must satisfy \(x^2-9\ne 0\), so \(x\ne -3,3\). In exams, remove values that make the denominator zero.

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Question 405/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

फलन (f(x)=\frac{\sqrt{x+1}}{x-4}) का प्रांत कौन सा है?

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

Explanation opens after your attempt
Correct Answer

A. \([-1,\infty\)-{4})

Explanation

Simple Explanation

वर्गमूल के लिए \(x+1\ge 0\) और हर के लिए \(x\ne 4\) चाहिए। इसलिए दोनों शर्तों का प्रतिच्छेद \([-1,\infty\)-{4}) है। / For the square root \(x+1\ge 0\) and for the denominator \(x\ne 4\) are required. Thus the intersection is \([-1,\infty\)-{4}).

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Question 406/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

फलन (f(x)=(x-2)2+5) का परिसर क्या है?

What is the range of (f(x)=(x-2)2+5)?

Explanation opens after your attempt
Correct Answer

A. \([5,\infty\))

Explanation

Simple Explanation

क्योंकि ((x-2)2\ge 0), न्यूनतम मान (5) है। अतः परिसर \([5,\infty\)) है। / Since ((x-2)2\ge 0), the minimum value is (5). Hence the range is \([5,\infty\)).

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Question 407/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

फलन (f(x)=7-(x+3)2) का परिसर चुनिए।

Choose the range of (f(x)=7-(x+3)2).

Explanation opens after your attempt
Correct Answer

A. (\(-\infty,7]\)

Explanation

Simple Explanation

क्योंकि ((x+3)2\ge 0), (7-(x+3)2\le 7)। अधिकतम मान (7) है। / Since ((x+3)2\ge 0), (7-(x+3)2\le 7). The maximum value is (7).

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Question 408/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

यदि (f(x)=\frac{2x+1}{x-3}), तो (f) का प्रांत क्या है?

If (f(x)=\frac{2x+1}{x-3}), what is the domain of (f)?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}-{3}\)

Explanation

Simple Explanation

हर \(x-3\ne 0\), इसलिए \(x\ne 3\)। रैशनल फलन में हर की शर्त सबसे पहले देखें। / The denominator \(x-3\ne 0\), so \(x\ne 3\). For rational functions, check the denominator condition first.

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Question 409/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

फलन (f(x)=\frac{2x+1}{x-3}) का परिसर क्या है?

What is the range of (f(x)=\frac{2x+1}{x-3})?

Explanation opens after your attempt
Correct Answer

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

Explanation

Simple Explanation

यदि \(y=\frac{2x+1}{x-3}\), तो \(x=\frac{3y+1}{y-2}\), अतः \(y\ne 2\)। इसलिए परिसर \(\mathbb{R}-{2}\) है। / If \(y=\frac{2x+1}{x-3}\), then \(x=\frac{3y+1}{y-2}\), so \(y\ne 2\). Hence the range is \(\mathbb{R}-{2}\).

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Question 410/600 Hard Mathematics Relations and Functions Real valued functions, domain and range of these functions Class 11 Level 33

फलन (f(x)=\frac{1}{x-2+4}) का परिसर ज्ञात कीजिए।

Find the range of (f(x)=\frac{1}{x-2+4}).

Explanation opens after your attempt
Correct Answer

A. (\(0,\frac{1}{4}]\)

Explanation

Simple Explanation

क्योंकि \(x^2+4\ge 4\), इसलिए (0<f(x)\le \frac{1}{4})। शून्य प्राप्त नहीं होता। / Since \(x^2+4\ge 4\), (0<f(x)\le \frac{1}{4}). The value (0) is never attained.

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