Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
86 views
in Calculus by (96.4k points)
closed by
Let f: [0, 1] → R be such that f(xy) = f(x)∙f(y) for all x, y ∈ [0, 1], and f(0) ≠ 0. If y = y(x) satisfies the differential equation\(\text{, }\!\!~\!\!\text{ }\frac{dy}{dx}=f\left( x \right)\text{ }\!\!~\!\!\)with y(0) = 1, then \(y\left( \frac{1}{4} \right)+y\left( \frac{3}{4} \right)\) is equal to:
1. 3
2. 4
3. 2
4. 5

1 Answer

0 votes
by (85.7k points)
selected by
 
Best answer
Correct Answer - Option 1 : 3

The given equation is:

f(xy) = f(x)·f(y) for all x,y ∈ [0, 1]     ….(i)

Put x = y = 0 in equation (i),

⇒ f (0) = f(0). f(0)

⇒ f(0).f(0) – f(0) = 0

⇒ f(0) [f(0) – 1] = 0

⇒ f(0) – 1 = 0

∴ f(0) = 1

Put y =0 in equation (i)

⇒ f(0) = f(x).f(0)

\(\Rightarrow f\left( x \right)=\frac{f\left( 0 \right)}{f\left( 0 \right)}\) 

∴ f(x) = 1

From question,

\(\Rightarrow \frac{dy}{dx}=f\left( x \right)\) 

On substituting value of ‘f(x)’,

\(\Rightarrow \frac{dy}{dx}=1\) 

⇒ dy = dx

On integrating both sides,

\(\Rightarrow \int dy=\int dx\) 

∴ y = x + C

From question,

⇒ y(0) = 1

⇒ 1 = 0 + C

∴ C = 1

Now,

∴ y = x + 1

From question,

\(\Rightarrow y\left( \frac{1}{4} \right)=\frac{1}{4}+1=\frac{1+4}{4}=\frac{5}{4}\) 

\(\Rightarrow y\left( \frac{3}{4} \right)=\frac{3}{4}+1=\frac{3+4}{4}=\frac{7}{4}\) 

\(\therefore ~y\left( \frac{1}{4} \right)+y\left( \frac{3}{4} \right)=\frac{5}{4}+\frac{7}{4}=\frac{12}{4}=3\) 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...