Concept-wise Practice

inverse function MCQ Questions for Class 12

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

Practice Questions

40 questions tagged with inverse function.

Question 1/40 Expert Mathematics Relations and Functions Functions Class 12 Level 20

यदि (f(x)=\frac{x-1}{x+1}), तो \(x\ne -1\) के लिए (f^{-1}(x)) क्या है?

If (f(x)=\frac{x-1}{x+1}), then for \(x\ne -1\), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\frac{1+x}{1-x}\)

Step 1

Concept

Let \(y=\frac{x-1}{x+1}\).

Step 2

Why this answer is correct

From (y(x+1)=x-1), we get (x(y-1)=-(1+y)).

Step 3

Exam Tip

Hence \(x=\frac{1+y}{1-y}\), so (f^{-1}(x)=\frac{1+x}{1-x}). चरण 1: \(y=\frac{x-1}{x+1}\) मानें। चरण 2: (y(x+1)=x-1) से (x(y-1)=-(1+y)) मिलता है। चरण 3: इसलिए \(x=\frac{1+y}{1-y}\), अतः (f^{-1}(x)=\frac{1+x}{1-x})।

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Question 2/40 Expert Mathematics Relations and Functions Functions Class 12 Level 20

फलन \(f:\mathbb{R}\to\mathbb{R}\), (f(x)=3x-5) के लिए (f^{-1}(x)) क्या होगा?

For \(f:\mathbb{R}\to\mathbb{R}\), (f(x)=3x-5), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\frac{x+5}{3}\)

Step 1

Concept

Let (y=3x-5).

Step 2

Why this answer is correct

Solving for (x), we get \(x=\frac{y+5}{3}\).

Step 3

Exam Tip

Replace (y) by (x), so (f^{-1}(x)=\frac{x+5}{3}). चरण 1: मान लें (y=3x-5)। चरण 2: (x) को (y) के पदों में लिखने पर \(x=\frac{y+5}{3}\) मिलता है। चरण 3: अंत में (y) की जगह (x) रखने से (f^{-1}(x)=\frac{x+5}{3}) होगा।

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Question 3/40 Expert Mathematics Relations and Functions Functions Class 12 Level 19

यदि (f:\mathbb{R}\to\(0,\infty\)) को (f(x)=e^{2x+1}) से दिया गया है, तो (f^{-1}(x)) क्या है?

If (f:\mathbb{R}\to\(0,\infty\)) is given by (f(x)=e^{2x+1}), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\frac{\ln x-1}{2}\)

Step 1

Concept

Put \(y=e^{2x+1}\).

Step 2

Why this answer is correct

Then \(\ln y=2x+1\), so \(x=\frac{\ln y-1}{2}\).

Step 3

Exam Tip

Replacing (y) by (x) gives (f^{-1}(x)=\frac{\ln x-1}{2}). चरण 1: \(y=e^{2x+1}\) मानें। चरण 2: \(\ln y=2x+1\), इसलिए \(x=\frac{\ln y-1}{2}\)। चरण 3: चर बदलकर (f^{-1}(x)=\frac{\ln x-1}{2}) मिलता है।

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Question 4/40 Expert Mathematics Relations and Functions Functions Class 12 Level 19

यदि \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=x-2+1) और \(g:[1,\infty\)\to\mathbb{R}) को (g(x)=\sqrt{x-1}) से दिया गया है, तो कौन सा कथन सही है?

If \(f:\mathbb{R}\to\mathbb{R}\) is (f(x)=x-2+1) and \(g:[1,\infty\)\to\mathbb{R}) is (g(x)=\sqrt{x-1}), which statement is correct?

Explanation opens after your attempt
Correct Answer

B. (g) (f) के \(x\geq0\) तक सीमित रूप का प्रतिलोम है(g) is the inverse of (f) restricted to \(x\geq0\)

Step 1

Concept

(f(x)=x-2+1) is not one-one on all of \(\mathbb{R}\).

Step 2

Why this answer is correct

If its domain is restricted to \(x\geq0\), then \(y=x^2+1\) gives \(x=\sqrt{y-1}\).

Step 3

Exam Tip

To define an inverse, the domain often needs a suitable restriction. चरण 1: (f(x)=x-2+1) पूरे \(\mathbb{R}\) पर एकैकी नहीं है। चरण 2: यदि प्रांत \(x\geq0\) तक सीमित करें, तो \(y=x^2+1\) से \(x=\sqrt{y-1}\) मिलता है। चरण 3: प्रतिलोम बनाने के लिए कई बार प्रांत को उचित रूप से सीमित करना पड़ता है।

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Question 5/40 Expert Mathematics Relations and Functions Functions Class 12 Level 19

यदि \(f:\mathbb{R}\setminus{2}\to\mathbb{R}\setminus{3}\) को (f(x)=\frac{3x+1}{x-2}) से दिया गया है, तो (f^{-1}(x)) क्या है?

If \(f:\mathbb{R}\setminus{2}\to\mathbb{R}\setminus{3}\) is given by (f(x)=\frac{3x+1}{x-2}), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\frac{2x+1}{x-3}\)

Step 1

Concept

Put \(y=\frac{3x+1}{x-2}\) and isolate (x).

Step 2

Why this answer is correct

From (y(x-2)=3x+1), we get (x(y-3)=2y+1).

Step 3

Exam Tip

Hence \(x=\frac{2y+1}{y-3}\), and replacing (y) by (x) gives the inverse. चरण 1: \(y=\frac{3x+1}{x-2}\) मानकर (x) को अलग करें। चरण 2: (y(x-2)=3x+1) से (x(y-3)=2y+1) मिलता है। चरण 3: इसलिए \(x=\frac{2y+1}{y-3}\), और चर बदलने पर प्रतिलोम मिल जाता है।

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Question 6/40 Hard Mathematics Relations and Functions Functions Class 12 Level 21

यदि \(f:[3,\infty\)\to[1,\infty)) को (f(x)=x-2-6x+10) से परिभाषित किया गया है, तो (f^{-1}(x)) क्या होगा?

If \(f:[3,\infty\)\to[1,\infty)) is defined by (f(x)=x-2-6x+10), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(3+\sqrt{x-1}\)

Step 1

Concept

(f(x)=x-2-6x+10=(x-3)2+1).

Step 2

Why this answer is correct

From (y=(x-3)2+1), we get ((x-3)2=y-1).

Step 3

Exam Tip

Since the domain is \([3,\infty\)), \(x-3\geq0\), so the inverse is \(3+\sqrt{x-1}\). चरण 1: (f(x)=x-2-6x+10=(x-3)2+1)। चरण 2: (y=(x-3)2+1) से ((x-3)2=y-1) मिलता है। चरण 3: प्रांत \([3,\infty\)) है, इसलिए \(x-3\geq0\) और प्रतिलोम \(3+\sqrt{x-1}\) होगा।

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Question 7/40 Hard Mathematics Relations and Functions Functions Class 12 Level 21

यदि \(f:A\to B\) एकैकी और आच्छादी दोनों है, तो \(f^{-1}\) के बारे में कौन सा कथन सही है?

If \(f:A\to B\) is both one-one and onto, which statement about \(f^{-1}\) is correct?

Explanation opens after your attempt
Correct Answer

A. \(f^{-1}\) मौजूद है और \(B\to A\) फलन है\(f^{-1}\) exists and is a function from (B) to (A)

Step 1

Concept

One-one ensures each image has a unique original element.

Step 2

Why this answer is correct

Onto ensures every element of (B) is actually an image.

Step 3

Exam Tip

Therefore, the inverse function exists from (B) to (A). चरण 1: एकैकी होने से हर छवि का मूल तत्व अलग और निश्चित होता है। चरण 2: आच्छादी होने से (B) का हर तत्व किसी न किसी (A) तत्व की छवि है। चरण 3: इसलिए प्रतिलोम फलन \(B\to A\) के रूप में मौजूद होता है।

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Question 8/40 Hard Mathematics Relations and Functions Functions Class 12 Level 21

यदि (f(x)=\frac{x+2}{x-2}), तो (f^{-1}(x)) क्या होगा?

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

Explanation opens after your attempt
Correct Answer

A. \(\frac{2x+2}{x-1}\)

Step 1

Concept

Write \(y=\frac{x+2}{x-2}\).

Step 2

Why this answer is correct

(y(x-2)=x+2), so (x(y-1)=2y+2) and \(x=\frac{2y+2}{y-1}\).

Step 3

Exam Tip

Replace (y) by (x) at the end to get the inverse. चरण 1: \(y=\frac{x+2}{x-2}\) लिखें। चरण 2: (y(x-2)=x+2), इसलिए (x(y-1)=2y+2) और \(x=\frac{2y+2}{y-1}\)। चरण 3: अंत में (y) की जगह (x) लिखने से प्रतिलोम मिलता है।

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Question 9/40 Hard Mathematics Relations and Functions Functions Class 12 Level 21

यदि \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=2x-5) से परिभाषित किया गया है, तो (f^{-1}(x)) क्या होगा?

If \(f:\mathbb{R}\to\mathbb{R}\) is defined by (f(x)=2x-5), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

B. \(\frac{x+5}{2}\)

Step 1

Concept

Put (y=2x-5).

Step 2

Why this answer is correct

Solving for (x), we get \(x=\frac{y+5}{2}\).

Step 3

Exam Tip

While finding inverse, first write (y=f(x)), then isolate (x). चरण 1: (y=2x-5) मानिए। चरण 2: (x) के लिए हल करने पर \(x=\frac{y+5}{2}\) मिलता है। चरण 3: प्रतिलोम निकालते समय पहले (y) लिखें, फिर (x) को अलग करें।

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Question 10/40 Hard Mathematics Relations and Functions Functions Class 12 Level 19

यदि \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=2x-3-1) से दिया गया है, तो (f^{-1}(x)) क्या होगा?

If \(f:\mathbb{R}\to\mathbb{R}\) is given by (f(x)=2x-3-1), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt[3]{\frac{x+1}{2}}\)

Step 1

Concept

Put \(y=2x^3-1\).

Step 2

Why this answer is correct

Then \(y+1=2x^3\), so \(x=\sqrt[3]{\frac{y+1}{2}}\).

Step 3

Exam Tip

In inverse questions, isolate (x) first and then change the variable. चरण 1: \(y=2x^3-1\) मानें। चरण 2: \(y+1=2x^3\), इसलिए \(x=\sqrt[3]{\frac{y+1}{2}}\)। चरण 3: प्रतिलोम में पहले (x) को अलग करें, फिर चर बदलें।

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Question 11/40 Hard Mathematics Relations and Functions Functions Class 12 Level 19

यदि \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=3x-2) और \(g:\mathbb{R}\to\mathbb{R}\) को (g(x)=\frac{x+2}{3}) से दिया गया है, तो (g) और (f) के बीच क्या संबंध है?

If \(f:\mathbb{R}\to\mathbb{R}\) is given by (f(x)=3x-2) and \(g:\mathbb{R}\to\mathbb{R}\) by (g(x)=\frac{x+2}{3}), what is the relation between (g) and (f)?

Explanation opens after your attempt
Correct Answer

A. \(g=f^{-1}\)

Step 1

Concept

From (y=3x-2), we get \(x=\frac{y+2}{3}\).

Step 2

Why this answer is correct

This matches the form of (g(y)), so (g) is the inverse function.

Step 3

Exam Tip

You can also verify an inverse by checking (f(g(x))=x) or (g(f(x))=x). चरण 1: (y=3x-2) से \(x=\frac{y+2}{3}\) मिलता है। चरण 2: यही (g(y)) का रूप है, इसलिए (g) प्रतिलोम फलन है। चरण 3: प्रतिलोम जांचने के लिए (f(g(x))) या (g(f(x))) भी (x) होना चाहिए।

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Question 12/40 Hard Mathematics Relations and Functions Functions Class 12 Level 19

यदि \(f:\mathbb{R}\setminus{0}\to\mathbb{R}\setminus{0}\) को (f(x)=\frac{1}{x}) से दिया गया है, तो (f^{-1}(x)) क्या है?

If \(f:\mathbb{R}\setminus{0}\to\mathbb{R}\setminus{0}\) is given by (f(x)=\frac{1}{x}), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\frac{1}{x}\)

Step 1

Concept

Put \(y=\frac{1}{x}\).

Step 2

Why this answer is correct

This gives \(x=\frac{1}{y}\).

Step 3

Exam Tip

Hence the inverse function is the same as the original function. चरण 1: \(y=\frac{1}{x}\) मानें। चरण 2: इससे \(x=\frac{1}{y}\) मिलता है। चरण 3: इसलिए प्रतिलोम फलन मूल फलन जैसा ही है।

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Question 13/40 Hard Mathematics Relations and Functions Functions Class 12 Level 19

यदि \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=2x-5) से दिया गया है, तो (f^{-1}(x)) क्या होगा?

If \(f:\mathbb{R}\to\mathbb{R}\) is given by (f(x)=2x-5), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\frac{x+5}{2}\)

Step 1

Concept

Put (y=2x-5).

Step 2

Why this answer is correct

Solving for (x) gives \(x=\frac{y+5}{2}\).

Step 3

Exam Tip

While finding the inverse, replace (y) by (x) at the end. चरण 1: (y=2x-5) मानें। चरण 2: (x) के लिए हल करने पर \(x=\frac{y+5}{2}\) मिलता है। चरण 3: प्रतिलोम निकालते समय अंत में (y) की जगह (x) लिख दें।

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Question 14/40 Medium Mathematics Relations and Functions Functions Class 12 Level 21

यदि \(f:A\to B\) और \(g:B\to A\) ऐसे हैं कि \(g\circ f=I_A\) और \(f\circ g=I_B\), तो (g) के बारे में सही कथन क्या है?

If \(f:A\to B\) and \(g:B\to A\) satisfy \(g\circ f=I_A\) and \(f\circ g=I_B\), what is correct about (g)?

Explanation opens after your attempt
Correct Answer

A. \(g=f^{-1}\)

Step 1

Concept

Both compositions give identity functions.

Step 2

Why this answer is correct

This means (f) and (g) completely undo each other's action.

Step 3

Exam Tip

Therefore (g) is the inverse of (f), so \(g=f^{-1}\). चरण 1: दोनों संयुक्त फलन पहचान फलन दे रहे हैं। चरण 2: इसका अर्थ है कि (f) और (g) एक-दूसरे की क्रिया को पूरी तरह उलटते हैं। चरण 3: इसलिए (g), (f) का प्रतिलोम है और \(g=f^{-1}\) होगा।

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Question 15/40 Medium Mathematics Relations and Functions Functions Class 12 Level 21

यदि \(f:R\to R\), (f(x)=x-3-7), तो (f^{-1}(x)) क्या होगा?

If \(f:R\to R\), (f(x)=x-3-7), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt[3]{x+7}\)

Step 1

Concept

Write \(y=x^3-7\).

Step 2

Why this answer is correct

Then \(x^3=y+7\), so \(x=\sqrt[3]{y+7}\).

Step 3

Exam Tip

Replacing (y) by (x), (f^{-1}(x)=\sqrt[3]{x+7}). चरण 1: \(y=x^3-7\) लिखें। चरण 2: \(x^3=y+7\), इसलिए \(x=\sqrt[3]{y+7}\)। चरण 3: (y) को (x) से बदलने पर (f^{-1}(x)=\sqrt[3]{x+7})।

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Question 16/40 Medium Mathematics Relations and Functions Functions Class 12 Level 21

यदि \(f:R\to R\), (f(x)=x-2-6x), तो प्रतिलोम फलन क्यों नहीं बनता?

If \(f:R\to R\), (f(x)=x-2-6x), why does an inverse function not exist?

Explanation opens after your attempt
Correct Answer

A. क्योंकि यह एक-एकी नहीं हैBecause it is not one-one

Step 1

Concept

For an inverse function to exist, the original function must be one-one.

Step 2

Why this answer is correct

(f(0)=0) and (f(6)=0), while \(0\ne6\).

Step 3

Exam Tip

Since one image has two preimages, the inverse is not a well-defined function. चरण 1: प्रतिलोम फलन के लिए मूल फलन का एक-एकी होना जरूरी है। चरण 2: (f(0)=0) और (f(6)=0), जबकि \(0\ne6\)। चरण 3: एक ही छवि की दो पूर्वछवियाँ होने से प्रतिलोम फलन स्पष्ट नहीं बनता।

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Question 17/40 Medium Mathematics Relations and Functions Functions Class 12 Level 21

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

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

Explanation opens after your attempt
Correct Answer

A. \(\frac{2x-1}{5}\)

Step 1

Concept

Write \(y=\frac{5x+1}{2}\).

Step 2

Why this answer is correct

Then (2y=5x+1), so \(x=\frac{2y-1}{5}\).

Step 3

Exam Tip

Replacing (y) by (x), (f^{-1}(x)=\frac{2x-1}{5}). चरण 1: \(y=\frac{5x+1}{2}\) लिखें। चरण 2: (2y=5x+1), इसलिए \(x=\frac{2y-1}{5}\)। चरण 3: (y) को (x) से बदलने पर (f^{-1}(x)=\frac{2x-1}{5})।

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Question 18/40 Medium Mathematics Relations and Functions Functions Class 12 Level 21

यदि \(f:R\to R\), (f(x)=x-3-4), तो (f^{-1}(4)) क्या होगा?

If \(f:R\to R\), (f(x)=x-3-4), what is (f^{-1}(4))?

Explanation opens after your attempt
Correct Answer

A. (2)

Step 1

Concept

To find (f^{-1}(4)), solve \(x^3-4=4\).

Step 2

Why this answer is correct

\(x^3=8\), so (x=2).

Step 3

Exam Tip

For a cubic function, the real cube root can be taken directly. चरण 1: (f^{-1}(4)) के लिए \(x^3-4=4\) हल करें। चरण 2: \(x^3=8\), इसलिए (x=2)। चरण 3: घन फलन में वास्तविक घनमूल सीधे लिया जा सकता है।

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Question 19/40 Medium Mathematics Relations and Functions Functions Class 12 Level 21

यदि \(f:R\to R\), (f(x)=4x+7), तो (f^{-1}(23)) का मान क्या होगा?

If \(f:R\to R\), (f(x)=4x+7), what is the value of (f^{-1}(23))?

Explanation opens after your attempt
Correct Answer

A. (4)

Step 1

Concept

(f^{-1}(23)) means the value of (x) for which (f(x)=23).

Step 2

Why this answer is correct

From (4x+7=23), we get (4x=16), so (x=4).

Step 3

Exam Tip

To find an inverse value, equate the original function to the given value and solve. चरण 1: (f^{-1}(23)) का अर्थ है वह (x), जिसके लिए (f(x)=23)। चरण 2: (4x+7=23) से (4x=16), इसलिए (x=4)। चरण 3: प्रतिलोम मान निकालते समय मूल फलन को दिए गए मान के बराबर रखकर हल करें।

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Question 20/40 Medium Mathematics Relations and Functions Functions Class 12 Level 20

यदि \(f:A\to B\) और \(g:B\to A\) ऐसे हैं कि \(f\circ g=I_B\) और \(g\circ f=I_A\), तो (f) के बारे में सही कथन क्या है?

If \(f:A\to B\) and \(g:B\to A\) satisfy \(f\circ g=I_B\) and \(g\circ f=I_A\), what is correct about (f)?

Explanation opens after your attempt
Correct Answer

A. (f) उभयैक है और \(g=f^{-1}\)(f) is bijective and \(g=f^{-1}\)

Step 1

Concept

Both compositions give identity functions.

Step 2

Why this answer is correct

This means (f) and (g) completely undo each other's action.

Step 3

Exam Tip

Therefore (f) is bijective and (g) is the inverse of (f). चरण 1: दोनों संयुक्त फलन पहचान फलन दे रहे हैं। चरण 2: इसका अर्थ है कि (f) और (g) एक-दूसरे की क्रिया को पूरी तरह उलटते हैं। चरण 3: इसलिए (f) उभयैक है और (g), (f) का प्रतिलोम है।

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Question 21/40 Medium Mathematics Relations and Functions Functions Class 12 Level 20

यदि \(f:R\to R\), (f(x)=x-3+5), तो (f^{-1}(x)) क्या होगा?

If \(f:R\to R\), (f(x)=x-3+5), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt[3]{x-5}\)

Step 1

Concept

Write \(y=x^3+5\).

Step 2

Why this answer is correct

Then \(x^3=y-5\), so \(x=\sqrt[3]{y-5}\).

Step 3

Exam Tip

Replacing (y) by (x), (f^{-1}(x)=\sqrt[3]{x-5}). चरण 1: \(y=x^3+5\) लिखें। चरण 2: \(x^3=y-5\), इसलिए \(x=\sqrt[3]{y-5}\)। चरण 3: (y) को (x) से बदलने पर (f^{-1}(x)=\sqrt[3]{x-5})।

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Question 22/40 Medium Mathematics Relations and Functions Functions Class 12 Level 20

यदि \(f:R\to R\), (f(x)=x-2+2x), तो प्रतिलोम फलन क्यों नहीं बनता?

If \(f:R\to R\), (f(x)=x-2+2x), why does an inverse function not exist?

Explanation opens after your attempt
Correct Answer

A. क्योंकि यह एक-एकी नहीं हैBecause it is not one-one

Step 1

Concept

For an inverse function to exist, the original function must be one-one.

Step 2

Why this answer is correct

(f(0)=0) and (f(-2)=0), while \(0\ne-2\).

Step 3

Exam Tip

Since one image has two preimages, the inverse is not a well-defined function. चरण 1: प्रतिलोम फलन के लिए मूल फलन का एक-एकी होना जरूरी है। चरण 2: (f(0)=0) और (f(-2)=0), जबकि \(0\ne-2\)। चरण 3: एक छवि की दो पूर्वछवियाँ होने से प्रतिलोम फलन स्पष्ट नहीं बनता।

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Question 23/40 Medium Mathematics Relations and Functions Functions Class 12 Level 20

यदि (f(x)=\frac{3x-2}{4}), तो (f^{-1}(x)) क्या होगा?

If (f(x)=\frac{3x-2}{4}), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\frac{4x+2}{3}\)

Step 1

Concept

Write \(y=\frac{3x-2}{4}\).

Step 2

Why this answer is correct

Then (4y=3x-2), so \(x=\frac{4y+2}{3}\).

Step 3

Exam Tip

Replacing (y) by (x), (f^{-1}(x)=\frac{4x+2}{3}). चरण 1: \(y=\frac{3x-2}{4}\) लिखें। चरण 2: (4y=3x-2), इसलिए \(x=\frac{4y+2}{3}\)। चरण 3: (y) को (x) से बदलने पर (f^{-1}(x)=\frac{4x+2}{3})।

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Question 24/40 Medium Mathematics Relations and Functions Functions Class 12 Level 20

यदि \(f:R\to R\), (f(x)=x-3+2), तो (f^{-1}(10)) क्या होगा?

If \(f:R\to R\), (f(x)=x-3+2), what is (f^{-1}(10))?

Explanation opens after your attempt
Correct Answer

A. (2)

Step 1

Concept

To find (f^{-1}(10)), solve \(x^3+2=10\).

Step 2

Why this answer is correct

\(x^3=8\), so (x=2).

Step 3

Exam Tip

For a cubic function, the real cube root can be taken directly. चरण 1: (f^{-1}(10)) के लिए \(x^3+2=10\) हल करें। चरण 2: \(x^3=8\), इसलिए (x=2)। चरण 3: घन फलन में वास्तविक घनमूल सीधे लिया जा सकता है।

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Question 25/40 Medium Mathematics Relations and Functions Functions Class 12 Level 20

यदि \(f:R\to R\), (f(x)=2x+5), तो (f^{-1}(13)) का मान क्या होगा?

If \(f:R\to R\), (f(x)=2x+5), what is the value of (f^{-1}(13))?

Explanation opens after your attempt
Correct Answer

A. (4)

Step 1

Concept

(f^{-1}(13)) means the value of (x) for which (f(x)=13).

Step 2

Why this answer is correct

From (2x+5=13), we get (2x=8), so (x=4).

Step 3

Exam Tip

While finding an inverse value, equate the original function to the given value. चरण 1: (f^{-1}(13)) का अर्थ है वह (x), जिसके लिए (f(x)=13)। चरण 2: (2x+5=13) से (2x=8), इसलिए (x=4)। चरण 3: प्रतिलोम मान निकालते समय मूल फलन को दिए गए मान के बराबर रखें।

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Question 26/40 Medium Mathematics Relations and Functions Functions Class 12 Level 19

यदि \(f:A\to B\) और \(g:B\to A\) ऐसे हैं कि \(g\circ f=I_A\) और \(f\circ g=I_B\), तो (g) क्या है?

If \(f:A\to B\) and \(g:B\to A\) satisfy \(g\circ f=I_A\) and \(f\circ g=I_B\), what is (g)?

Explanation opens after your attempt
Correct Answer

A. (f) का प्रतिलोमInverse of (f)

Step 1

Concept

Both compositions give identity functions.

Step 2

Why this answer is correct

This means (f) and (g) undo each other's action.

Step 3

Exam Tip

Therefore \(g=f^{-1}\). चरण 1: दोनों संयुक्त फलन पहचान फलन दे रहे हैं। चरण 2: इसका अर्थ है कि (f) और (g) एक-दूसरे की क्रिया को उलटते हैं। चरण 3: इसलिए \(g=f^{-1}\) होगा।

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Question 27/40 Medium Mathematics Relations and Functions Functions Class 12 Level 19

यदि \(f:R\to R\), (f(x)=x-3), तो (f^{-1}(x)) क्या होगा?

If \(f:R\to R\), (f(x)=x-3), what is (f^{-1}(x))?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt[3]{x}\)

Step 1

Concept

Write \(y=x^3\).

Step 2

Why this answer is correct

Then \(x=\sqrt[3]{y}\).

Step 3

Exam Tip

Replacing (y) by (x), (f^{-1}(x)=\sqrt[3]{x}). चरण 1: \(y=x^3\) लिखें। चरण 2: \(x=\sqrt[3]{y}\) मिलेगा। चरण 3: (y) को (x) से बदलने पर (f^{-1}(x)=\sqrt[3]{x})।

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Question 28/40 Medium Mathematics Relations and Functions Functions Class 12 Level 19

यदि (f(x)=x-2) को (R) से (R) में माना जाए, तो प्रतिलोम फलन क्यों नहीं बनता?

If (f(x)=x-2) is considered from (R) to (R), why does an inverse function not exist?

Explanation opens after your attempt
Correct Answer

A. क्योंकि यह एक-एकी नहीं हैBecause it is not one-one

Step 1

Concept

For an inverse function to exist, the original function must be one-one.

Step 2

Why this answer is correct

Here (f(1)=f(-1)=1), while \(1\ne-1\).

Step 3

Exam Tip

Since one image has two preimages, the inverse is not a well-defined function. चरण 1: प्रतिलोम फलन के लिए मूल फलन का एक-एकी होना जरूरी है। चरण 2: यहाँ (f(1)=f(-1)=1), जबकि \(1\ne-1\)। चरण 3: एक ही छवि की दो पूर्वछवियाँ होने से प्रतिलोम फलन स्पष्ट नहीं बनता।

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Question 29/40 Medium Mathematics Relations and Functions Functions Class 12 Level 19

यदि (f(x)=\frac{2x+1}{3}), तो (f^{-1}(x)) क्या होगा?

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

Explanation opens after your attempt
Correct Answer

A. \(\frac{3x-1}{2}\)

Step 1

Concept

Write \(y=\frac{2x+1}{3}\).

Step 2

Why this answer is correct

Then (3y=2x+1), so \(x=\frac{3y-1}{2}\).

Step 3

Exam Tip

Replacing (y) by (x), (f^{-1}(x)=\frac{3x-1}{2}). चरण 1: \(y=\frac{2x+1}{3}\) लिखें। चरण 2: (3y=2x+1), इसलिए \(x=\frac{3y-1}{2}\)। चरण 3: (y) को (x) से बदलने पर (f^{-1}(x)=\frac{3x-1}{2})।

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Question 30/40 Medium Mathematics Relations and Functions Functions Class 12 Level 19

यदि \(f:R\to R\), (f(x)=x-3-1), तो (f^{-1}(7)) क्या होगा?

If \(f:R\to R\), (f(x)=x-3-1), what is (f^{-1}(7))?

Explanation opens after your attempt
Correct Answer

A. (2)

Step 1

Concept

To find (f^{-1}(7)), solve \(x^3-1=7\).

Step 2

Why this answer is correct

\(x^3=8\), so (x=2).

Step 3

Exam Tip

For a cubic function, the real cube root can be taken directly. चरण 1: (f^{-1}(7)) के लिए \(x^3-1=7\) हल करें। चरण 2: \(x^3=8\), इसलिए (x=2)। चरण 3: घन फलन में वास्तविक घनमूल सीधे लिया जा सकता है।

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