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Q17 jee_main_2026_24_january_evening Functional Equations

Let

f

be a function such that

3f(x) + 2f((m)/(19x)) = 5x

,

x \neq 0

, where

m = Σi=1⁹ (i)²

. Then

f(5) - f(2)

is equal to

A. $-9$ $-9$
B. $36$ $36$
C. $18$ $18$
D. $9$ $9$

Solution

Related Formula

Sum of squares: Σi=1ⁿ i² = (n(n+1)(2n+1))/(6)

Core Logic

First, evaluate the constant $m$ $m$:

m = Σi=1⁹ i² = (9 \times 10 \times 19)/(6) = 15 \times 19

The given functional equation is:

3f(x) + 2f((15 \times 19)/(19x)) = 5x

3f(x) + 2f((15)/(x)) = 5x (1)

Step 1: Forming System of Equations

To eliminate $f((15)/(x))$ $f\left(\frac{15}{x}\right)$, replace $x$ $x$ with $(15)/(x)$ $\frac{15}{x}$ in equation (1):

3f((15)/(x)) + 2f(x) = 5((15)/(x)) = (75)/(x) (2)

Step 2: Solving for f(x)

Multiply equation (1) by 3 and equation (2) by 2:

9f(x) + 6f((15)/(x)) = 15x

4f(x) + 6f((15)/(x)) = (150)/(x)

Subtract the second from the first:

9f(x) - 4f(x) = 15x - (150)/(x)

5f(x) = 15x - (150)/(x) f(x) = 3x - (30)/(x)

Step 3: Calculating Final Values

Evaluate $f(5)$ $f(5)$ and $f(2)$ $f(2)$:

f(5) = 3(5) - (30)/(5) = 15 - 6 = 9

f(2) = 3(2) - (30)/(2) = 6 - 15 = -9

Finally:

f(5) - f(2) = 9 - (-9) = 18

Pattern Recognition

Functional equations of the form $af(x) + bf((k)/(x)) = g(x)$ strictly require the classic substitution $x \to (k)/(x)$ $x \to \frac{k}{x}$ to create a straightforward $2 \times 2$ $2 \times 2$ algebraic system of equations.

Chapter Mix

Class 11 Maths: Functions Class 11 Maths: Sequences and Series

Q1 jee_main_2026_28_january_morning Composite Functions

If

g(x)=3x²+2x-3

,

f(0)=-3

and

4g(f(x))=3x²-32x+72

, then

f(g(2))

is equal to:

A. $(25)/(6)$ $\frac{25}{6}$
B. $-(25)/(6)$ $-\frac{25}{6}$
C. $(7)/(2)$ $\frac{7}{2}$
D. $-(7)/(2)$ $-\frac{7}{2}$

Solution

Related Formula

g(f(x)) = 3(f(x))² + 2f(x) - 3

Core Logic

First, evaluate $g(2)$ $g(2)$:

g(2) = 3(2)² + 2(2) - 3 = 12 + 4 - 3 = 13

We need to find $f(g(2)) = f(13)$ $f(g(2)) = f(13)$.

Given the composite function relation:

4g(f(x)) = 3x² - 32x + 72

Substitute $g(t)$ $g(t)$ expansion:

4[3(f(x))² + 2f(x) - 3] = 3x² - 32x + 72

Let $f(x) = t$ $f(x) = t$:

12t² + 8t - 12 = 3x² - 32x + 72

12t² + 8t - (3x² - 32x + 84) = 0

Step 1: Solve for f(x)

Using the quadratic formula for $t$ $t$:

t = f(x) = -8 \pm \sqrt(64 - 4(12)(-(3x² - 32x + 84)))24

f(x) = -8 \pm \sqrt(64 + 48(3x² - 32x + 84))24

f(x) = (-8 \pm 4(3x - 16))/(24)

Since $f(0) = -3$ $f(0) = -3$:

f(0) = (-8 \pm 4(-16))/(24) = (-8 \pm (-64))/(24)

Choosing the positive sign gives $(-8 - 64)/(24) = -3$ $\frac{-8 - 64}{24} = -3$, so we take the positive sign branch (where the inner term was $3x-16$ $3x-16$, wait, $+4(3x-16)$ $+4(3x-16)$ with $x=0$ $x=0$ is $-64$ $-64$. So $+$ $+$ sign works).

f(x) = (-8 + 4(3x - 16))/(24)

Step 2: Final Calculation

Evaluate $f(13)$ $f(13)$:

f(13) = (-8 + 4(3(13) - 16))/(24) = (-8 + 4(23))/(24)

f(13) = (-8 + 92)/(24) = (84)/(24) = (7)/(2)

Pattern Recognition

Composite equations resolving to quadratics in $f(x)$ $f(x)$ typically require boundary conditions (like $f(0)=-3$ $f(0)=-3$) to eliminate the $\pm$ $\pm$ ambiguity from the quadratic formula.

Chapter Mix

Class 11 Mathematics: Functions Class 11 Mathematics: Quadratic Equations

Q8 jee_main_2026_28_january_evening One-One and Many-One Functions

Given below are two statements: Statement I: The function

f: R arrow R

defined by

f(x) = (x)/(1 + |x|)

is one-one. Statement II: The function

f: R arrow R

defined by

f(x) = x² + 4x - 30x² - 8x + 18

is many-one. In the light of the above statements, choose the correct answer from the options given below :

A. Both Statement I and Statement II are false.
B. Both Statement I and Statement II are true.
C. Statement I is false but Statement II is true.
D. Statement I is true but Statement II is false.

Solution

Core Logic

Statement I: $f(x) = (x)/(1+|x|)$ $f(x) = \frac{x}{1+|x|}$.

f(x) = cases (x)/(1+x) & x \geq 0 (x)/(1-x) & x < 0 cases

The derivative $f'(x) = (1)/((1+|x|)²) > 0$ $f'(x) = \frac{1}{(1+|x|)^2} > 0$ for all $x$ $x$. Since it is strictly increasing, $f(x)$ $f(x)$ is one-one. Statement I is true.

Execution

Statement II: $f(x) = (x² + 4x - 30)/(x² - 8x + 18)$ . Let's evaluate $f(0)$ $f(0)$:

f(0) = (-30)/(18) = -(5)/(3)

Set $f(x) = -(5)/(3)$ $f(x) = -\frac{5}{3}$ to find if there are other roots:

(x² + 4x - 30)/(x² - 8x + 18) = -(5)/(3)

3x² + 12x - 90 = -5x² + 40x - 90

8x² - 28x = 0 \Rightarrow 4x(2x - 7) = 0

$x = 0$ $x = 0$ or $x = (7)/(2)$ $x = \frac{7}{2}$ (Note: the PDF says on solving $x=0, -1$ $x=0, -1$, but algebraic check shows $x=0, 7/2$ $x=0, 7/2$. Regardless, it maps to multiple points). Since $f(0) = f(7/2) = -(5)/(3)$ $f(0) = f(7/2) = -\frac{5}{3}$, the function maps distinct inputs to the same output. It is many-one. Statement II is true.

Step 1: Final Conclusion

Both Statement I and Statement II are true.

Pattern Recognition

Checking $x=0$ $x=0$ in a rational function $y = P(x)/Q(x)$ $y = P(x)/Q(x)$ provides a quick horizontal line test benchmark. Equating the function to $f(0)$ $f(0)$ immediately reveals if it's many-one without computing full derivatives.

Chapter Mix

Class 12 Maths: Relations and Functions

Q15 jee_main_2026_28_january_evening Signum Function Properties

The sum of all the elements in the range of

f(x) = Sgn( x) + Sgn( x) + Sgn( x) + Sgn( x), x \neq (nπ)/(2), n in Z

, where

Sgn(t) = cases 1, & if t > 0 -1 & if t < 0 cases

, is

A. $4$ $4$
B. $2$ $2$
C. $-2$ $-2$
D. $0$ $0$

Solution

Core Logic

Analyze the signs of trigonometric functions in each quadrant: Quadrant I: $x in (0, π/2)$ $x \in (0, \pi/2)$. $, , ,$ $\sin, \cos, \tan, \cot$ are all positive. $y = 1 + 1 + 1 + 1 = 4$ $y = 1 + 1 + 1 + 1 = 4$

Quadrant II: $x in (π/2, π)$ $x \in (\pi/2, \pi)$. $\sin$ positive; $, ,$ $\cos, \tan, \cot$ negative. $y = 1 - 1 - 1 - 1 = -2$ $y = 1 - 1 - 1 - 1 = -2$

Quadrant III: $x in (π, 3π/2)$ $x \in (\pi, 3\pi/2)$. $,$ $\tan, \cot$ positive; $,$ $\sin, \cos$ negative. $y = -1 - 1 + 1 + 1 = 0$ $y = -1 - 1 + 1 + 1 = 0$

Quadrant IV: $x in (3π/2, 2π)$ $x \in (3\pi/2, 2\pi)$. $\cos$ positive; $, ,$ $\sin, \tan, \cot$ negative. $y = -1 + 1 - 1 - 1 = -2$ $y = -1 + 1 - 1 - 1 = -2$

Execution

The set of unique values produced by $f(x)$ $f(x)$ represents its range. Range = $-2, 0, 4$ $\{-2, 0, 4\}$

The sum of all elements in the range is: $-2 + 0 + 4 = 2$ $-2 + 0 + 4 = 2$

Pattern Recognition

When applying signum to all four base trigonometric identities, simply count the number of positive mappings per quadrant (all positive = 4, sine only = -2, tan/cot only = 0, cos only = -2).

Chapter Mix

Class 11 Maths: Trigonometric Functions Class 12 Maths: Relations and Functions

Q54 jee_main_2025_02_april_evening Relations

Let

A = 1, 2, 3, , 100

and

R

be a relation on

A

such that

R = (a, b) : a = 2b + 1

. Let

(a₁, a₂), (a₂, a₃), (a₃, a₄), , (ak, ak+1)

be a sequence of

k

elements of

R

such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer

k

, for which such a sequence exists, is equal to:

A. 6
B. 7
C. 5
D. 8

Solution

Related Formula

Chain definition: aᵢ = 2 aᵢ₊₁ + 1 for i = 1, 2, , k

Core Logic

To find the longest sequence of connected pairs, we trace the relation backward starting from the smallest elements in $A$ $A$.

Step 1: Trace the relations backward

To maximize $k$ $k$, we want the chain of elements to go down as low as possible. Let the final element in the chain be $ak+1 in A$ $a_{k+1} \in \mathrm{A}$. Since $ak = 2 ak+1 + 1$ $a_k = 2 a_{k+1} + 1$:

If $ak+1 = 1 ak = 3$ $a_{k+1} = 1 \implies a_k = 3$
If $ak+1 = 2 ak = 5$ $a_{k+1} = 2 \implies a_k = 5$

Let's test the chain starting with $ak+1 = 1$ $a_{k+1} = 1$:

$ak = 2(1) + 1 = 3$ $a_k = 2(1) + 1 = 3$
$ak-1 = 2(3) + 1 = 7$ $a_{k-1} = 2(3) + 1 = 7$
$ak-2 = 2(7) + 1 = 15$ $a_{k-2} = 2(7) + 1 = 15$
$ak-3 = 2(15) + 1 = 31$ $a_{k-3} = 2(15) + 1 = 31$
$ak-4 = 2(31) + 1 = 63$ $a_{k-4} = 2(31) + 1 = 63$
$ak-5 = 2(63) + 1 = 127$ $a_{k-5} = 2(63) + 1 = 127$ (but $127 A$ $127 \notin \mathrm{A}$!)

Thus, the longest chain within the set $A$ $A$ has 6 elements:

63, 31, 15, 7, 3, 1

This chain corresponds to exactly 5 ordered pairs:

(63, 31), (31, 15), (15, 7), (7, 3), (3, 1)

So the maximum number of pairs in the sequence is $k = 5$ $k = 5$.

Step 2: Check alternative chains

If we start with $ak+1 = 2$ $a_{k+1} = 2$:

$ak+1 = 2$ $a_{k+1} = 2$
$ak = 5$ $a_k = 5$
$ak-1 = 11$ $a_{k-1} = 11$
$ak-2 = 23$ $a_{k-2} = 23$
$ak-3 = 47$ $a_{k-3} = 47$
$ak-4 = 95$ $a_{k-4} = 95$
$ak-5 = 191 > 100$ $a_{k-5} = 191 > 100$

Again, the maximum number of pairs is $k = 5$ $k = 5$. Thus, the largest integer $k$ $k$ is 5.

Pattern Recognition

Recursive scaling: Tracing exponential chains of the form $xₙ₊₁ = c xₙ + d$ $x_{n+1} = c x_n + d$ shows that the elements grow very quickly. Calculating the limits of growth determines the maximum possible depth of the sequence.

Chapter Mix

Class 11 Mathematics: Relations and Functions

Solution & Explanation

Related Formula

Core Logic

Step 1: Testing x = 11

Step 2: Testing x = 10

Pattern Recognition

Chapter Mix

Reference Study Guides

More Sets Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Forming System of Equations

Step 2: Solving for f(x)

Step 3: Calculating Final Values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for f(x)

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Core Logic

Execution

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Trace the relations backward

Step 2: Check alternative chains

Pattern Recognition

Chapter Mix

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