JEE MainMathematics

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Sets, Relations and Functions: Mathematics | JEE Main

Q: Why can we write ?

Because and is given to be differentiable (and thus continuous), Cauchy's exponential functional equation has the unique family of solutions .

Suppose

f: \mathbb{R} \rightarrow (0, \infty)

be a differentiable function such that

5f(x + y) = f(x) \cdot f(y), \forall x, y \in \mathbb{R}

. If

f(3) = 320

, then

\sum_{n=0}^5 f(n)

is equal to :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A6875B6525C6825D6575

Step-by-step solutionView

Correct answer

By transforming the functional equation to $g(x+y)=g(x)g(y)$ with $g(x) = f(x)/5$ , we find $f(n) = 5 \cdot 4^n$ , making the sum $\sum_{n=0}^5 5 \cdot 4^n = 6825$ .

Option analysis

Why each option works or fails

A · 6875

The student computed the sum from $n=1$ to $n=5$ instead of $n=0$ to $n=5$ , but added an incorrect offset or miscalculated $5 \frac{4^6-1}{3} + 50$ . Evaluate the geometric series strictly from $n=0$ to $n=5$ using $S = \frac{a(r^k - 1)}{r - 1}$ with $a=f(0)=5$ and $k=6$ terms.

B · 6525

The student miscalculated the number of terms as 5 (computing $\sum_{n=1}^5 f(n)$ or $(4^5-1)/(4-1)$ instead of $n=0$ to $5$ ), yielding $5 \times 341 = 1705$ or made a related slip in evaluating powers of 4. Count the number of terms in $\sum_{n=0}^5$ : there are $5 - 0 + 1 = 6$ terms, so the power of the common ratio is $4^6$ , not $4^5$ .

C · 6825

Correct solution. Dividing the functional relation by 5 shows $g(x) = f(x)/5$ satisfies $g(x+y) = g(x)g(y)$ , so $f(x) = 5 \cdot a^x$ . Using $f(3)=320$ gives $a=4$ , leading to $\sum_{n=0}^5 5 \cdot 4^n = 5 \frac{4^6 - 1}{4 - 1} = 6825$ .

D · 6575

The student made an arithmetic error when evaluating $4^6 = 4096$ , using $4096 - 1 = 4095$ but incorrectly dividing $5 \times 4095 / 3$ as 6575 instead of 6825. Carefully divide $4095$ by $3$ to get $1365$ , and then multiply by $5$ to obtain $6825$ .

Reviewed route

Solution

StepWorking

01given

$f: \mathbb{R} \rightarrow (0, \infty)$ is differentiable, $5f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$ , and $f(3) = 320$ .

02goal

Evaluate $\sum_{n=0}^5 f(n) = f(0) + f(1) + f(2) + f(3) + f(4) + f(5)$ .

03approach

Define an auxiliary function $g(x) = \frac{f(x)}{5}$ . Then $g(x+y) = g(x)g(y)$ , which implies $g(x) = a^x$ . Hence $f(x) = 5 \cdot a^x$ . Use $f(3) = 320$ to solve for $a$ , then sum the resulting geometric progression from $n=0$ to $5$ .

04execute

From $f(x) = 5 \cdot a^x$ , we have $f(3) = 5 \cdot a^3 = 320 \implies a^3 = 64 \implies a = 4$ . Therefore, $f(n) = 5 \cdot 4^n$ .

05execute

Sum the geometric series: $\sum_{n=0}^5 f(n) = 5 \sum_{n=0}^5 4^n = 5 \cdot \frac{4^6 - 1}{4 - 1} = 5 \cdot \frac{4096 - 1}{3} = 5 \cdot \frac{4095}{3} = 5 \cdot 1365 = 6825$ .

✓verify

Check individual terms: $f(0)=5, f(1)=20, f(2)=80, f(3)=320, f(4)=1280, f(5)=5120$ . Sum = $5 + 20 + 80 + 320 + 1280 + 5120 = 6825$ . Correct.

Hints that build this answer step by step

If we define $g(x) = \frac{f(x)}{5}$ , what functional equation does $g(x)$ satisfy?
$g(x+y) = g(x)g(y)$
Given $g(x+y) = g(x)g(y)$ with differentiable and positive $f(x)$ , we have $g(x) = a^x$ and thus $f(x) = 5a^x$ . Given $f(3) = 320$ , what is the base $a$ ?
$a = 4$
What is the value of the finite geometric sum $\sum_{n=0}^5 f(n) = \sum_{n=0}^5 5 \cdot 4^n$ ?
$6825$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can we write

g(x+y) = g(x)g(y) \implies g(x) = a^x

Because $f(x) > 0$ and $f$ is given to be differentiable (and thus continuous), Cauchy's exponential functional equation has the unique family of solutions $g(x) = a^x$ .

Answer

By transforming the functional equation to $g(x+y)=g(x)g(y)$ with $g(x) = f(x)/5$ , we find $f(n) = 5 \cdot 4^n$ , making the sum $\sum_{n=0}^5 5 \cdot 4^n = 6825$ .

Why each option works or fails

A: 6875 - The student computed the sum from $n=1$ to $n=5$ instead of $n=0$ to $n=5$ , but added an incorrect offset or miscalculated $5 \frac{4^6-1}{3} + 50$ . Evaluate the geometric series strictly from $n=0$ to $n=5$ using $S = \frac{a(r^k - 1)}{r - 1}$ with $a=f(0)=5$ and $k=6$ terms.
B: 6525 - The student miscalculated the number of terms as 5 (computing $\sum_{n=1}^5 f(n)$ or $(4^5-1)/(4-1)$ instead of $n=0$ to $5$ ), yielding $5 \times 341 = 1705$ or made a related slip in evaluating powers of 4. Count the number of terms in $\sum_{n=0}^5$ : there are $5 - 0 + 1 = 6$ terms, so the power of the common ratio is $4^6$ , not $4^5$ .
C · correct: 6825 - Correct solution. Dividing the functional relation by 5 shows $g(x) = f(x)/5$ satisfies $g(x+y) = g(x)g(y)$ , so $f(x) = 5 \cdot a^x$ . Using $f(3)=320$ gives $a=4$ , leading to $\sum_{n=0}^5 5 \cdot 4^n = 5 \frac{4^6 - 1}{4 - 1} = 6825$ .
D: 6575 - The student made an arithmetic error when evaluating $4^6 = 4096$ , using $4096 - 1 = 4095$ but incorrectly dividing $5 \times 4095 / 3$ as 6575 instead of 6825. Carefully divide $4095$ by $3$ to get $1365$ , and then multiply by $5$ to obtain $6825$ .

Step-by-step solution

given: $f: \mathbb{R} \rightarrow (0, \infty)$ is differentiable, $5f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$ , and $f(3) = 320$ .
goal: Evaluate $\sum_{n=0}^5 f(n) = f(0) + f(1) + f(2) + f(3) + f(4) + f(5)$ .
approach: Define an auxiliary function $g(x) = \frac{f(x)}{5}$ . Then $g(x+y) = g(x)g(y)$ , which implies $g(x) = a^x$ . Hence $f(x) = 5 \cdot a^x$ . Use $f(3) = 320$ to solve for $a$ , then sum the resulting geometric progression from $n=0$ to $5$ .
execute: From $f(x) = 5 \cdot a^x$ , we have $f(3) = 5 \cdot a^3 = 320 \implies a^3 = 64 \implies a = 4$ . Therefore, $f(n) = 5 \cdot 4^n$ .
execute: Sum the geometric series: $\sum_{n=0}^5 f(n) = 5 \sum_{n=0}^5 4^n = 5 \cdot \frac{4^6 - 1}{4 - 1} = 5 \cdot \frac{4096 - 1}{3} = 5 \cdot \frac{4095}{3} = 5 \cdot 1365 = 6825$ .
verify: Check individual terms: $f(0)=5, f(1)=20, f(2)=80, f(3)=320, f(4)=1280, f(5)=5120$ . Sum = $5 + 20 + 80 + 320 + 1280 + 5120 = 6825$ . Correct.

Sets, Relations and Functions: Mathematics | JEE Main

What feels right?

By transforming the functional equation to $g(x+y)=g(x)g(y)$ with $g(x) = f(x)/5$ , we find $f(n) = 5 \cdot 4^n$ , making the sum $\sum_{n=0}^5 5 \cdot 4^n = 6825$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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