JEE Main 2026 (22 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $f$ and $g$ be functions satisfying $f (x + y) = f (x) f (y), f (1) = 7$ and $g (x + y) = g (x y), g (1) = 1$ , for all $x, y \in N$ . If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)} ight) = 19607 $, t h e n$ n$ is equal to:

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Look at the structure of the functional equations. The first equation

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

is a classic Cauchy functional equation that maps addition to multiplication, which is the hallmark of exponential functions. For the second equation

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

, try testing small integer values like

y = 1

to see if the function behaves in a specific way.

Hint 2

Hint 2: The Arsenal. Once you identify

f (x) = a^{x}

and determine

g (x)

is a constant function, the summation

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

transforms into a standard geometric series. Recall the formula for the sum of

n

terms of a geometric progression:

S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}

Hint 3

Hint 3: The Execution. Substitute your derived functions into the summation to get

\sum_{x = 1}^{n} 7^{x} = 19607

. This expands to

7^{1} + 7^{2} + \dots + 7^{n} = 19607

. Use the geometric series sum formula with

a = 7

and

r = 7

to set up the equation

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

and solve for

n

Visualized Solution

Analyze the Functional Equation for $f (x)$

Given functional equation: $f (x + y) = f (x) f (y)$
This is a standard Cauchy functional equation for exponential functions.
The general solution for such a function is $f (x) = a^{x}$ where $a$ is a constant.

Determine the Specific Form of $f (x)$

Given: $f (1) = 7$
Substitute $x = 1$ into $f (x) = a^{x}$ :
$a^{1} = 7 ⟹ a = 7$
Therefore, $f (x) = 7^{x}$ for all $x \in N$ .

Analyze the Functional Equation for $g (x)$

Given functional equation: $g (x + y) = g (x y)$
Substitute $y = 1$ into the equation:
$g (x + 1) = g (x \cdot 1) = g (x)$
This implies that the function $g (x)$ is constant for all $x \in N$ .

Determine the Specific Form of $g (x)$

Given: $g (1) = 1$
Since $g (x + 1) = g (x)$ , then $g (2) = g (1) = 1$ , $g (3) = g (2) = 1$ , and so on.
Therefore, $g (x) = 1$ for all $x \in N$ .

Formulate the Summation Expression

The given sum is: $\sum_{x = 1}^{n} (\frac{f ( x )}{g ( x )}) = 19607$
Substitute $f (x) = 7^{x}$ and $g (x) = 1$ :
$\sum_{x = 1}^{n} \frac{7 ^{x}}{1} = \sum_{x = 1}^{n} 7^{x} = 19607$

Identify the Geometric Progression

The series is: $7^{1} + 7^{2} + 7^{3} + \dots + 7^{n}$
This is a G.P. with first term $a = 7$ and common ratio $r = 7$ .
Sum of $n$ terms of a G.P.: $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$

Apply the G.P. Sum Formula

Substitute $a = 7$ and $r = 7$ into the sum formula:
$\frac{7 ( 7 ^{n} - 1 )}{7 - 1} = 19607$
$\frac{7 ( 7 ^{n} - 1 )}{6} = 19607$

Simplify the Equation

$7^{n} - 1 = \frac{19607 \times 6}{7}$
$7^{n} - 1 = 2801 \times 6$
$7^{n} - 1 = 16806$
$7^{n} = 16807$

Solve for $n$

We need to find $n$ such that $7^{n} = 16807$ .
Check powers of $7$ :
$7^{1} = 7, 7^{2} = 49, 7^{3} = 343, 7^{4} = 2401, 7^{5} = 16807$
So, $7^{n} = 7^{5} ⟹ n = 5$ .

Final Conclusion

Key Takeaway:
1. $f (x + y) = f (x) f (y)$ with $f (1) = a$ implies $f (x) = a^{x}$ .
2. $g (x + y) = g (x y)$ with $g (1) = c$ implies $g (x) = c$ for all $x \in N$ .
3. The final answer is $n = 5$ .

00:00 / 00:00

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Functional Equation $f (x)$

The functional equation

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

is a classic cornerstone of functional analysis. Whenever a function converts addition inside the argument into multiplication outside, it points directly to an exponential form.

Because exponents satisfy the property

a^{x + y} = a^{x} \cdot a^{y}

, the general solution is

f (x) = a^{x}

. Given the condition

f (1) = 7

, we substitute

x = 1

to find

a^{1} = 7

, which yields

a = 7

Thus, our function is defined as:

f (x) = 7^{x}

The Mystery of the Constant $g (x)$

Now, we address the second riddle:

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

. This equation appears unusual, but we can resolve it through strategic substitution.

If we set

y = 1

, the equation becomes

g (x + 1) = g (x \cdot 1)

, which simplifies to

g (x + 1) = g (x)

. This implies that the function does not change its value as we increment

x

, identifying it as a constant function.

Since we are given

g (1) = 1

, it follows that for all natural numbers

x

g (x) = 1

The Summation Journey

We now combine these pieces into the given summation:

x = 1 \sum n \frac{f ( x )}{g ( x )} = 19607

Substituting our findings

f (x) = 7^{x}

and

g (x) = 1

, the expression simplifies to:

x = 1 \sum n 7^{x} = 19607

Expanding this, we obtain

7^{1} + 7^{2} + 7^{3} + \dots + 7^{n} = 19607

. This is a Geometric Progression (G.P.) where the first term

a = 7

and the common ratio

r = 7

The sum of

n

terms of a G.P. is given by the formula:

S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}

Plugging in our values, we get:

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

This simplifies to:

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

The Final Calculation

To isolate

7^{n}

, we multiply both sides by

6

and divide by

7

7^{n} - 1 = \frac{19607 \times 6}{7}

Dividing

19607

7

gives

2801

. Then, calculating

2801 \times 6

results in

16806

Adding

1

to both sides, we arrive at:

7^{n} = 16807

We evaluate the powers of

7

7^{1} = 7

7^{2} = 49

7^{3} = 343

7^{4} = 2401

, and

7^{5} = 16807

. Therefore, the final answer is: $n = 5$

Common Traps & FAQs

Wait, why do we assume

f (x) = a^{x}

for

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

That is a great observation! This is a famous functional equation known as the Cauchy functional equation. When a function satisfies

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

, it essentially transforms the additive property of the input into the multiplicative property of the output. The only continuous functions that satisfy this property are exponential functions of the form

f (x) = a^{x}

. You can verify this by checking

a^{x + y} = a^{x} \cdot a^{y}

, which holds true for all

x, y

How did you know

g (x)

was a constant function?

Excellent question! When you are faced with a functional equation like

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

, the best strategy is to test simple values. If we set

y = 1

, we get

g (x + 1) = g (x \cdot 1) = g (x)

. This tells us that the value of the function at

x + 1

is exactly the same as the value at

x

. Since this holds for all natural numbers, the function cannot change its value as

x

increases. Therefore, it must be a constant function.

Is there a trick to solving

7^{n} = 16807

quickly?

Instead of using logarithms, which might be overkill here, just use the property of powers. We know

7^{1} = 7

7^{2} = 49

7^{3} = 343

, and

7^{4} = 2401

. Since

16807

ends in a

7

, we can guess that

7^{n}

must end in a

7

(since

7 \times 7 = 49

9 \times 7 = 63

3 \times 7 = 21

1 \times 7 = 7

). Multiplying

2401

7

gives exactly

16807

. So,

n = 5

is the logical conclusion!

Animated Solution for Mathematics - Sequence and Series: Let $f$ and $g$ be functions satisfying $f (x + y) = f (x) f (y), f (1) = 7$ and $g (x + y) = g (x y), g (1) = 1$ , for all $x, y \in N$ . If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)} ight) = 19607 $, t h e n$ n$ is equal to:

Select Answer:

Visualized Solution

Analyze the Functional Equation for $f (x)$

Determine the Specific Form of $f (x)$

Analyze the Functional Equation for $g (x)$

Determine the Specific Form of $g (x)$

Formulate the Summation Expression

Identify the Geometric Progression

Apply the G.P. Sum Formula

Simplify the Equation

Solve for $n$

Final Conclusion

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Functional Equation $f (x)$

The Mystery of the Constant $g (x)$

The Summation Journey

The Final Calculation

Similar Questions

Let $f (x)$ be a function such that $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in N$ . If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) = 3279$ , then the value of $n$ is

Find the natural number 'a' for which $\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1)$ , where the function 'f' satisfies the relation $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and further $f (1) = 2$ .

If $f (x + y) = f (x) f (y)$ and $\sum_{x = 1}^{\infty} f (x) = 2$ , $x, y \in N$ where $N$ is the set of all natural numbers, then the value of $\frac{f ( 4 )}{f ( 2 )}$ is:

Let $\sum_{k = 1}^{10} f (a + k) = 16 (2^{10} - 1)$ , where the function $f$ satisfies $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and $f (1) = 2$ . then the natural number 'a' is

If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

Let $729, 81, 9, 1, \dots$ be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence. If $2 \sum_{n = 1}^{40} (P_{n})^{1/ n} = \frac{3 ^{α} - 1}{3 ^{β}}$ and $g cd (α, β) = 1$ , then $α + β$ is equal to

If for $x, y \in R, x > 0$ \\ $y = lo g_{10} x + lo g_{10} x^{1/3} + lo g_{10} x^{1/9} + \dots$ upto $\infty$ terms\\ and $\frac{2 + 4 + 6 + \dots + 2 y}{3 + 6 + 9 + \dots + 3 y} = \frac{4}{l o g _{10} x}$ , then the ordered pair $(x, y)$ is equal to :

Let $S = N \cup {0}$ . Define a relation R from S to R by: $R = {(x, y) : lo g_{e} y = x lo g_{e} (\frac{2}{5}), x \in S, y \in R}$ . Then, the sum of all the elements in the range of R is equal to

If $0 < θ, ϕ < \frac{π}{2}$ , $x = \sum_{n = 0}^{\infty} cos^{2 n} θ$ , $y = \sum_{n = 0}^{\infty} sin^{2 n} ϕ$ and $z = \sum_{n = 0}^{\infty} cos^{2 n} θ \cdot sin^{2 n} ϕ$ then

The $4^{t h}$ term of GP is 500 and its common ratio is $\frac{1}{m}, m \in N$ . Let $S_{n}$ denote the sum of the first $n$ terms of this GP. If $S_{6} > S_{5} + 1$ and $S_{7} < S_{6} + \frac{1}{2}$ , then the number of possible values of $m$ is

Select Answer:

Visualized Solution

Analyze the Functional Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Determine the Specific Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Analyze the Functional Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Determine the Specific Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Formulate the Summation Expression

Identify the Geometric Progression

Apply the G.P. Sum Formula

Simplify the Equation

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Final Conclusion

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Functional Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Mystery of the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

The Summation Journey

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a function such that f(x+y)=f(x)⋅f(y) for all x,y∈N. If f(1)=3 and ∑k=1n​f(k)=3279, then the value of n is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x+y)=f(x)f(y) and ∑x=1∞​f(x)=2, x,y∈N where N is the set of all natural numbers, then the value of f(2)f(4)​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∑k=110​f(a+k)=16(210−1), where the function f satisfies f(x+y)=f(x)f(y) for all natural numbers x,y and f(1)=2. then the natural number 'a' is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an​=43​−(43​)2+(43​)3+⋯+(−1)n−1(43​)n and bn​=1−an​, then find the least natural number n0​ such that bn​>an​ for all n≥n0​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x,y∈R,x>0\\ y=log10​x+log10​x1/3+log10​x1/9+… upto ∞ terms\\ and 3+6+9+⋯+3y2+4+6+⋯+2y​=log10​x4​, then the ordered pair (x,y) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S=N∪{0}. Define a relation R from S to R by: R={(x,y):loge​y=xloge​(52​),x∈S,y∈R}. Then, the sum of all the elements in the range of R is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 0<θ,ϕ<2π​, x=∑n=0∞​cos2nθ, y=∑n=0∞​sin2nϕ and z=∑n=0∞​cos2nθ⋅sin2nϕ then

Analyze the Functional Equation for $f (x)$

Determine the Specific Form of $f (x)$

Analyze the Functional Equation for $g (x)$

Determine the Specific Form of $g (x)$

Solve for $n$

Analyzing the Functional Equation $f (x)$

The Mystery of the Constant $g (x)$

Let $f (x)$ be a function such that $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in N$ . If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) = 3279$ , then the value of $n$ is

If $f (x + y) = f (x) f (y)$ and $\sum_{x = 1}^{\infty} f (x) = 2$ , $x, y \in N$ where $N$ is the set of all natural numbers, then the value of $\frac{f ( 4 )}{f ( 2 )}$ is:

Let $\sum_{k = 1}^{10} f (a + k) = 16 (2^{10} - 1)$ , where the function $f$ satisfies $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and $f (1) = 2$ . then the natural number 'a' is

If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

If for $x, y \in R, x > 0$ \\ $y = lo g_{10} x + lo g_{10} x^{1/3} + lo g_{10} x^{1/9} + \dots$ upto $\infty$ terms\\ and $\frac{2 + 4 + 6 + \dots + 2 y}{3 + 6 + 9 + \dots + 3 y} = \frac{4}{l o g _{10} x}$ , then the ordered pair $(x, y)$ is equal to :

Let $S = N \cup {0}$ . Define a relation R from S to R by: $R = {(x, y) : lo g_{e} y = x lo g_{e} (\frac{2}{5}), x \in S, y \in R}$ . Then, the sum of all the elements in the range of R is equal to

If $0 < θ, ϕ < \frac{π}{2}$ , $x = \sum_{n = 0}^{\infty} cos^{2 n} θ$ , $y = \sum_{n = 0}^{\infty} sin^{2 n} ϕ$ and $z = \sum_{n = 0}^{\infty} cos^{2 n} θ \cdot sin^{2 n} ϕ$ then