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max-min MCQ Questions for Class 11

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2 questions tagged with max-min.

Question 1/2 Medium Mathematics Trigonometric Functions Trigonometric functions and their properties Class 11 Level 72

\(\sin \theta-\cos \theta\) का न्यूनतम मान क्या है?

What is the minimum value of \(\sin \theta-\cos \theta\)?

Explanation opens after your attempt
Correct Answer

A. \(-\sqrt{2}\)

Step 1

Concept

The amplitude of \( \sin \theta-\cos \theta\) is ( \sqrt{12+(-1)2}=\sqrt{2}). In exams the minimum is the negative amplitude.

Step 2

Why this answer is correct

The correct answer is A. \(-\sqrt{2}\). The amplitude of \( \sin \theta-\cos \theta\) is ( \sqrt{12+(-1)2}=\sqrt{2}). In exams the minimum is the negative amplitude.

Step 3

Exam Tip

\(\sin \theta-\cos \theta\) का amplitude (\sqrt{12+(-1)2}=\sqrt{2}) है। परीक्षा में minimum हमेशा negative amplitude होगा।

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Question 2/2 Medium Mathematics Trigonometric Functions Trigonometric functions and their properties Class 11 Level 72

\(\sin \theta+\cos \theta\) का अधिकतम मान क्या है?

What is the maximum value of \(\sin \theta+\cos \theta\)?

Explanation opens after your attempt
Correct Answer

B. \(\sqrt{2}\)

Step 1

Concept

( \sin \theta+\cos \theta=\sqrt{2}\sin\left\(\theta+\frac{\pi}{4}\right\)), so the maximum is \( \sqrt{2}\). In exams use maximum of \(a\sin x+b\cos x\) as \( \sqrt{a^2+b^2}\).

Step 2

Why this answer is correct

The correct answer is B. \(\sqrt{2}\). ( \sin \theta+\cos \theta=\sqrt{2}\sin\left\(\theta+\frac{\pi}{4}\right\)), so the maximum is \( \sqrt{2}\). In exams use maximum of \(a\sin x+b\cos x\) as \( \sqrt{a^2+b^2}\).

Step 3

Exam Tip

(\sin \theta+\cos \theta=\sqrt{2}\sin\left\(\theta+\frac{\pi}{4}\right\)), इसलिए अधिकतम \(\sqrt{2}\) है। परीक्षा में \(a\sin x+b\cos x\) का maximum \(\sqrt{a^2+b^2}\) लें।

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