JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Functional Equations and Geometric Series

Q: 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.

Suppose f: ℝ → (0, ∞) be a differentiable function such that 5f(x + y) = f(x) · f(y), ∀ x, y ∈ ℝ. If f(3) = 320, then sum from n=0 to 5 of 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

$f : R \to (0, \infty)$ is differentiable, $5 f (x + y) = f (x) f (y)$ for all $x, y \in R$ , 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 $f (3) = 5 \cdot a^{3} = 320 ⟹ a^{3} = 64 ⟹ a = 4$ . 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 $f (x) = 5 a^{x}$ . 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can we write

g (x + y) = g (x) g (y) ⟹ 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: $f : R \to (0, \infty)$ is differentiable, $5 f (x + y) = f (x) f (y)$ for all $x, y \in R$ , 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 $1365$ , 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 $f (3) = 5 \cdot a^{3} = 320 ⟹ a^{3} = 64 ⟹ a = 4$ . 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.

Functional Equations and Geometric Series

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

We think you should solve this next.

Let $A = {1, 2, 3, 5, 8, 9}$ . Then the number of possible functions…

Let R be a relation on $N \times N$ defined by $(a, b) R (c, d)$ i…

The number of points on the curve $y = 54 x^{5} - 135 x^{4} - 70 x^{3} + 180 x^{2} + 210 x$ at whic…

Students also ask

Functional Equations and Geometric Series

By transforming the functional equation to g(x+y)=g(x)g(y)g(x+y)=g(x)g(y) with g(x)=f(x)/5g(x)=f(x)/5, we find f(n)=5⋅4nf(n)=5⋅4n, making the sum ∑n=055⋅4n=6825∑n=05​5⋅4n=6825.

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

Let A={1,2,3,5,8,9}A={1,2,3,5,8,9}. Then the number of possible functions…

Let R be a relation on N×NN×N defined by (a,b)R(c,d)(a,b)R(c,d) i…

The number of points on the curve y=54x5−135x4−70x3+180x2+210xy=54x5−135x4−70x3+180x2+210x at whic…

Students also ask

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 $\sum_{n = 0}^{5} 5 \cdot 4^{n} = 6825$ .

Let $A = {1, 2, 3, 5, 8, 9}$ . Then the number of possible functions…

Let R be a relation on $N \times N$ defined by $(a, b) R (c, d)$ i…

The number of points on the curve $y = 54 x^{5} - 135 x^{4} - 70 x^{3} + 180 x^{2} + 210 x$ at whic…