If f(x)= cases2+2x,& -1≤ x lt 0 1-(x)/(3),& 0≤ x≤ 3 cases$f(x)=\begin{cases}2+2x,& -1\le x \lt 0\\ 1-\frac{x}{3},& 0\le x\le 3\end{cases}$; g(x)= cases-x,& -3≤ x≤ 0 x,& 0 lt x≤ 1 cases$g(x)=\begin{cases}-x,& -3\le x\le 0\\ x,& 0 \lt x\le 1\end{cases}$ , then range of (fog(x))$(fog(x))$ is

Related Formula

Core Logic

First, evaluate the constant m$m$:

The given functional equation is:

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):

Step 2: Solving for f(x)

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

Subtract the second from the first:

Step 3: Calculating Final Values

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

Finally:

Pattern Recognition

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

Chapter Mix

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

Related Formula

Core Logic

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

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

Given the composite function relation:

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

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

Step 1: Solve for f(x)

Using the quadratic formula for t$t$:

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

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).

Step 2: Final Calculation

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

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 ±$\pm$ ambiguity from the quadratic formula.

Chapter Mix

Class 11 Mathematics: Functions Class 11 Mathematics: Quadratic Equations

Core Logic

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

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)$f(x) = \frac{x^2 + 4x - 30}{x^2 - 8x + 18}$. Let's evaluate f(0)$f(0)$:

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

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

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

Related Formula

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:

This chain corresponds to exactly 5 ordered pairs:

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