Concept-wise Practice

range check MCQ Questions for Class 12

range check se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.

Practice Questions

5 questions tagged with range check.

\(\tan^{-1}x\) की सीमा से कौन सा मान संभव नहीं है?

Which value is not possible from the range of \(\tan^{-1}x\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{\pi}{2}\)

Step 1

Concept

The range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)), which does not include \(\frac{\pi}{2}\). Pay attention to open intervals.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{\pi}{2}\). The range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)), which does not include \(\frac{\pi}{2}\). Pay attention to open intervals.

Step 3

Exam Tip

\(\tan^{-1}x\) की सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है जिसमें \(\frac{\pi}{2}\) शामिल नहीं है। खुले अंतराल पर ध्यान दें।

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\(\cos^{-1}x\) की सीमा से कौन सा मान संभव नहीं है?

Which value is not possible from the range of \(\cos^{-1}x\)?

Explanation opens after your attempt
Correct Answer

A. \(-\frac{\pi}{3}\)

Step 1

Concept

The range of \(\cos^{-1}x\) is \([0,\pi]\), so a negative value is not possible. In such questions check only the output range.

Step 2

Why this answer is correct

The correct answer is A. \(-\frac{\pi}{3}\). The range of \(\cos^{-1}x\) is \([0,\pi]\), so a negative value is not possible. In such questions check only the output range.

Step 3

Exam Tip

\(\cos^{-1}x\) की सीमा \([0,\pi]\) है इसलिए ऋणात्मक मान संभव नहीं है। ऐसे प्रश्नों में केवल आउटपुट रेंज देखें।

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\(\sin^{-1}x\) की सीमा से कौन सा मान संभव नहीं है?

Which value is not possible from the range of \(\sin^{-1}x\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{2\pi}{3}\)

Step 1

Concept

The range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The value \(\frac{2\pi}{3}\) lies outside this range.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{2\pi}{3}\). The range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The value \(\frac{2\pi}{3}\) lies outside this range.

Step 3

Exam Tip

\(\sin^{-1}x\) की सीमा \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) है। \(\frac{2\pi}{3}\) इस सीमा के बाहर है।

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फलन \(f:[0,\infty\)\to \mathbb{R}), (f(x)=x-2) के लिए सही निष्कर्ष क्या है?

What is the correct conclusion for \(f:[0,\infty\)\to \mathbb{R}), (f(x)=x-2)?

Explanation opens after your attempt
Correct Answer

B. आच्छादक नहीं है क्योंकि ऋणात्मक मान नहीं मिलतेNot onto because negative values are not obtained

Step 1

Concept

Even when \(x\ge 0\), \(x^2\ge 0\).

Step 2

Why this answer is correct

The codomain \(\mathbb{R}\) contains negative values that are not obtained.

Step 3

Exam Tip

Getting some values is not enough; the whole codomain must be covered. चरण 1: \(x\ge 0\) होने पर भी \(x^2\ge 0\) ही रहता है। चरण 2: सहक्षेत्र \(\mathbb{R}\) में ऋणात्मक मान हैं जो नहीं मिलते। चरण 3: केवल कुछ मान मिलना काफी नहीं पूरे सहक्षेत्र का मिलना जरूरी है।

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फलन \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=x) यदि (x<0) और (f(x)=x+1) यदि \(x\ge0\) से परिभाषित किया गया है। सही कथन चुनिए।

The function \(f:\mathbb{R}\to\mathbb{R}\) is defined by (f(x)=x) if (x<0) and (f(x)=x+1) if \(x\ge0\). Choose the correct statement.

Explanation opens after your attempt
Correct Answer

A. (f) एक-एक है(f) is one-one

Step 1

Concept

For (x<0), the values are negative.

Step 2

Why this answer is correct

For \(x\ge0\), the values are (1) or more, so the ranges of the two parts do not overlap.

Step 3

Exam Tip

For piecewise functions check each part and the overlap of their ranges. चरण 1: (x<0) पर मान ऋणात्मक आते हैं। चरण 2: \(x\ge0\) पर मान (1) या उससे बड़े आते हैं, इसलिए दोनों भागों के परास आपस में नहीं मिलते। चरण 3: टुकड़ों में दिए फलन में हर भाग और उनके परास अलग-अलग जांचें।

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