/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 If \(f(x) \cdot f(y)=f(x)+f(y)+f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 20

If \(f(x) \cdot f(y)=f(x)+f(y)+f(x y)-2 \forall x, y \in R\) and if \(f(x)\) is not a constant function, then the value of \(f(1)\) is (A) 1 (B) 2 (C) 0 (D) \(-1\)

Short Answer

Expert verified
#Answer# The value of \(f(1)\) is \(2\).

Step by step solution

01

set x = 1 and y = 1

We will set \(x = 1\) and \(y = 1\) to get more information about the equation: $$ f(1) \cdot f(1)=f(1)+f(1)+f(1 \cdot 1)-2 $$
02

Simplify

Simplify the equation obtained in Step 1: $$ f(1)^2 = 3f(1)-2 $$
03

set x = y

Now we will set \(x = y\) to get more information about the equation: $$ f(x) \cdot f(x)=f(x)+f(x)+f(x^2)-2 $$
04

Simplify

Simplify the equation obtained in Step 3: $$ f(x)^2 = 2f(x) + f(x^2) - 2 $$
05

set x = 1

Now we will set \(x = 1\) in the simplified equation obtained in Step 4: $$ f(1)^2 = 2f(1) + f(1) - 2 $$
06

Compare

Compare the equations from Step 2 and Step 5: $$ f(1)^2 = 3f(1)-2 $$ $$ f(1)^2 = 2f(1) + f(1) - 2 $$
07

Solve

Both equations are equal, we will solve for \(f(1)\): $$ 3f(1) - 2 = 2f(1) + f(1) - 2 $$ $$ f(1) = 2 $$ Therefore, the value of \(f(1)\) is \(2\). This corresponds to the option (B) in the given choices.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The number of points where the function \(f(x)=\left(x^{2}-1\right)\left|x^{2}-x-2\right|+\sin (|x|)\) is not differentiable is (A) 0 (B) \(I\) (C) 2 (D) 3

Let \(f(x)\) be defined for all \(x \in R\) and the continuous. Let $\mathrm{f}(\mathrm{x}+\mathrm{y})-\mathrm{f}(\mathrm{x}-\mathrm{y})=4 \mathrm{xy} \forall \mathrm{x}, \mathrm{y}=\in \mathrm{R}$ and \(f(0)=0\) then (A) \(\mathrm{f}(\mathrm{x})\) is bounded (B) \(f(x)+f\left(\frac{1}{x}\right)=f\left(x+\frac{1}{x}\right)+2\) (C) \(\mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{f}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)+2\) (D) none of these

If \(\mathrm{y}=|1-| 2-|3-| 4-\mathrm{x}|||| ;\) then number of points where \(\mathrm{y}\) is not differentiable; is (A) 1 (B) 3 (C) 5 (D) \(>5\)

If \(f(x)=\operatorname{Max} \cdot\\{1,(\cos x+\sin x)\) \((\sin x-\cos x)\\} 0 \leq x \leq 5 \pi / 4\), then (A) \(\mathrm{f}(\mathrm{x})\) is not differentiable at \(\mathrm{x}=\pi / 6\) (B) \(f(x)\) is not differentiable at \(x=5 \pi / 6\) (C) \(f(x)\) is continuous for \(x \in[0,5 \pi / 4]\) (D) None of these

If $f(x)=\left\\{\begin{array}{ll}{[x]+\sqrt{\\{x\\}}} & x<1 \\\ \frac{1}{[x]+\\{x\\}^{2}} & x \geq 1\end{array}\right.\(, then [where \)[.]$ and \(\\{.\) represents greatest integer part and fractional part respectively.] (A) \(f(x)\) is continuous at \(x=1\) but not differentiable (B) \(f(x)\) is not continuous at \(x=1\) (C) \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) (D) \(\lim _{x \rightarrow 1} f(x)\) does not exist
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.