Logo
Jeeify
MCQMediumJEE 2026Functions

JEE Mathematics 2026 Question with Solution

Let ff and gg be functions satisfying

f(x+y)=f(x)f(y),f(1)=7f(x+y) = f(x)f(y), \quad f(1) = 7 g(x+y)=g(xy),g(1)=1,g(x+y) = g(xy), \quad g(1) = 1,

for all x,y∈Nx,y \in \mathbb{N}. If

∑x=1n(f(x)g(x))=19607,\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607,

then nn is equal to

  • A

    55

  • B

    44

  • C

    66

  • D

    77

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

f(x+y)=f(x)f(y),f(1)=7f(x+y)=f(x)f(y), \quad f(1)=7 g(x+y)=g(xy),g(1)=1g(x+y)=g(xy), \quad g(1)=1

and

∑x=1nf(x)g(x)=19607\sum_{x=1}^{n} \frac{f(x)}{g(x)} = 19607

Find: nn

From

f(x+y)=f(x)f(y)f(x+y)=f(x)f(y)

and f(1)=7f(1)=7, we get the exponential form

f(x)=7xf(x)=7^x

for all x∈Nx \in \mathbb{N}.

Now use

g(x+y)=g(xy)g(x+y)=g(xy)

with x=y=1x=y=1:

g(2)=g(1)=1g(2)=g(1)=1

Similarly, this gives

g(x)=1for all x∈Ng(x)=1 \quad \text{for all } x\in\mathbb{N}

Therefore,

∑x=1nf(x)g(x)=∑x=1n7x\sum_{x=1}^{n} \frac{f(x)}{g(x)} = \sum_{x=1}^{n} 7^x

This is a geometric series, so

∑x=1n7x=7(7n−16)\sum_{x=1}^{n} 7^x = 7\left(\frac{7^n-1}{6}\right)

Given that this sum is 1960719607,

7(7n−16)=196077\left(\frac{7^n-1}{6}\right) = 19607

Hence,

7n−1=19607×67=168067^n - 1 = \frac{19607 \times 6}{7} = 16806

So,

7n=16807=757^n = 16807 = 7^5

Thus,

n=5n=5

Therefore, the correct option is A.

Using the functional equations step by step

Given: the two functional equations and the sum value.

Find: the value of nn.

For ff, write a few values using

f(x+y)=f(x)f(y)f(x+y)=f(x)f(y)

Then

f(2)=f(1+1)=f(1)f(1)=7⋅7=72f(2)=f(1+1)=f(1)f(1)=7\cdot 7=7^2 f(3)=f(2+1)=f(2)f(1)=72⋅7=73f(3)=f(2+1)=f(2)f(1)=7^2\cdot 7=7^3

Continuing similarly,

f(x)=7xf(x)=7^x

For gg, use

g(x+y)=g(xy)g(x+y)=g(xy)

and substitute x=y=1x=y=1:

g(2)=g(1)=1g(2)=g(1)=1

the solution concludes that repeating this idea gives

g(x)=1for all x∈Ng(x)=1 \quad \text{for all } x\in\mathbb{N}

So the sum becomes

∑x=1nf(x)g(x)=∑x=1n7x\sum_{x=1}^{n} \frac{f(x)}{g(x)} = \sum_{x=1}^{n} 7^x

Now use the geometric series formula:

∑x=1n7x=7(7n−1)6\sum_{x=1}^{n} 7^x = \frac{7(7^n-1)}{6}

Set this equal to 1960719607:

7(7n−1)6=19607\frac{7(7^n-1)}{6}=19607

Multiply by 66 and divide by 77:

7n−1=168067^n-1=16806

Therefore,

7n=168077^n=16807

Since

16807=7516807=7^5

we get

n=5n=5

Hence, the required value is 55 and the correct option is A.

Common mistakes

  • Assuming f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) means f(x)=x7f(x)=x^7 or some polynomial form. This is incorrect because the given relation is multiplicative over addition, which leads to an exponential pattern. Use repeated application with f(1)=7f(1)=7 to obtain f(x)=7xf(x)=7^x.

  • Using the wrong geometric series formula for ∑x=1n7x\sum_{x=1}^{n} 7^x. A common error is to write 7n−16\frac{7^n-1}{6}, which is the formula for a series starting at 707^0. Here the sum starts at x=1x=1, so use 7(7n−1)6\frac{7(7^n-1)}{6}.

  • Not using the functional equation for gg correctly. If you do not substitute simple values such as x=y=1x=y=1, you may miss that the solution simplifies to the constant function g(x)=1g(x)=1. First extract easy values, then simplify the given sum.

Practice more Functions questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Functions

Related questions