Let f:Rrightarrow R$f:R\rightarrow R$ and g:Rrightarrow R$g:R\rightarrow R$ be defined as
f(x)=begincaseslog_ex & , & x>0\\ e^-x & , & xle0endcases$f(x)=\begin{cases}\log_{e}x & , & x>0\\ e^{-x} & , & x\le0\end{cases}$
and
g(x)=begincasesx & , & xge0\\ e^x & , & x<0endcases$g(x)=\begin{cases}x & , & x\ge0\\ e^{x} & , & x<0\end{cases}$
Then, g circ f: Rrightarrow R$g \circ f: R\rightarrow R$ is:
A.textone-one but not onto$\text{one-one but not onto}$
B.textneither one-one nor onto$\text{neither one-one nor onto}$
C.textonto but not one-one$\text{onto but not one-one}$
D.textboth one-one and onto$\text{both one-one and onto}$
Solution & Explanation
### Related Formula
For composite functions, g(f(x))$g(f(x))$ is determined by substituting the range of f(x)$f(x)$ into the appropriate domain intervals of g(y)$g(y)$:
g(f(x)) = begincases f(x) & , & f(x) ge 0 \\ e^f(x) & , & f(x) < 0 endcases$g(f(x)) = \begin{cases} f(x) & , & f(x) \ge 0 \\ e^{f(x)} & , & f(x) < 0 \end{cases}$
### Core Logic
Let us analyze the definition of g(f(x))$g(f(x))$ branch-by-branch based on the domain of x$x$:
1. **Case 1: x le 0$x \le 0$**
Here, f(x) = e^-x$f(x) = e^{-x}$. Since x le 0$x \le 0$, -x ge 0 implies e^-x ge 1 > 0$-x \ge 0 \implies e^{-x} \ge 1 > 0$.
Since f(x) ge 0$f(x) \ge 0$, we use the upper branch of g(y)$g(y)$:
g(f(x)) = f(x) = e^-x$g(f(x)) = f(x) = e^{-x}$
2. **Case 2: x > 0$x > 0$**
Here, f(x) = log_e x$f(x) = \log_e x$.
- Subcase (a): If f(x) ge 0 implies log_e x ge 0 implies x ge 1$f(x) \ge 0 \implies \log_e x \ge 0 \implies x \ge 1$.
Then, g(f(x)) = f(x) = log_e x$g(f(x)) = f(x) = \log_e x$.
- Subcase (b): If f(x) < 0 implies log_e x < 0 implies 0 < x < 1$f(x) < 0 \implies \log_e x < 0 \implies 0 < x < 1$.
Then, g(f(x)) = e^f(x) = e^log_e x = x$g(f(x)) = e^{f(x)} = e^{\log_e x} = x$.
### Step 1: Constructing the Composition Function
Combining the branches obtained, the composite function is:
g(f(x)) = begincases e^-x & , & x le 0 \\ x & , & 0 < x < 1 \\ log_e x & , & x ge 1 endcases$g(f(x)) = \begin{cases} e^{-x} & , & x \le 0 \\ x & , & 0 < x < 1 \\ \log_e x & , & x \ge 1 \end{cases}$The graphic demonstrates the behavior of the piecewise composite function gof across its distinct linear and logarithmic domains.
### Step 2: Injectivity and Surjectivity Analysis
- **Injectivity (One-One Check):**
Let's test two different inputs: x_1 = 0$x_1 = 0$ and x_2 = e$x_2 = e$.
g(f(0)) = e^-0 = 1$g(f(0)) = e^{-0} = 1$g(f(e)) = log_e e = 1$g(f(e)) = \log_e e = 1$
Since distinct inputs yield identical outputs (g(f(0)) = g(f(e)) = 1$g(f(0)) = g(f(e)) = 1$), the function is **many-one** (not one-one).
- **Surjectivity (Onto Check):**
Evaluating the range across the branches:
- For x le 0$x \le 0$, e^-x in [1, infty)$e^{-x} \in [1, \infty)$.
- For 0 < x < 1$0 < x < 1$, x in (0, 1)$x \in (0, 1)$.
- For x ge 1$x \ge 1$, \log_e x in [0, infty)$\log_e x \in [0, \infty)$.
The union of these sets gives the total range as [0, infty)$[0, \infty)$. Since the codomain is given as mathbbR$\mathbb{R}$, textRange neq textCodomain$\text{Range} \neq \text{Codomain}$, so the function is **into** (not onto).
Therefore, the function is neither one-one nor onto.
### Pattern Recognition
Sees: Piecewise branch composition.
Shortcut: Sketching the graph quickly shows that a horizontal line at y=1$y=1$ intersects the function multiple times (not one-one) and no part of the graph goes below the x-axis (not onto).
Trap: Always determine the range of the inner function first to select the correct branch of the outer function.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Relations and Functions
Keywords:#composite functions mappings#JEE Main 2024 Morning Q11#one one onto function checks#piecewise functions calculus#composite function graph#piecewise curve representation#injective surjective check
More Relations and Functions Previous-Year Questions — Page 5
Q53jee_main_2025_07_april_eveningSet Inclusion and Regions
Let A = \(alpha ,beta)in mathbfRtimes mathbfR:|alpha -1|leq 4 text and |beta -5|leq 6\$A = \{(\alpha ,\beta)\in \mathbf{R}\times \mathbf{R}:|\alpha -1|\leq 4 \text{ and } |\beta -5|\leq 6\}$ and B = \(alpha , beta) in mathbfR times mathbfR: 16 (alpha - 2)^2 + 9 (beta - 6)^2 leq 144\$B = \{(\alpha , \beta) \in \mathbf{R} \times \mathbf{R}: 16 (\alpha - 2)^{2} + 9 (\beta - 6)^{2} \leq 144\}$. Then
A.B subset A$B \subset A$
B.A cup B = \(x, y) : -4 leq x leq 4, -1 leq y leq 11\$A \cup B = \{(x, y) : -4 \leq x \leq 4, -1 \leq y \leq 11\}$
C. neither A subset B$A \subset B$ nor B subset A$B \subset A$
D.A subset B$A \subset B$
Solution
### Related Formula
An ellipse equation is structured as:
frac(x-h)^2a^2 + frac(y-k)^2b^2 leq 1$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} \leq 1$
### Core Logic
Analyzing set A$A$:
|alpha - 1| le 4 implies -4 le alpha - 1 le 4 implies -3 le alpha le 5$|\alpha - 1| \le 4 \implies -4 \le \alpha - 1 \le 4 \implies -3 \le \alpha \le 5$|beta - 5| le 6 implies -6 le beta - 5 le 6 implies -1 le beta le 11$|\beta - 5| \le 6 \implies -6 \le \beta - 5 \le 6 \implies -1 \le \beta \le 11$
Thus, region A$A$ forms a rectangle bounded between x in [-3, 5]$x \in [-3, 5]$ and y in [-1, 11]$y \in [-1, 11]$.
Analyzing set B$B$:
16(alpha - 2)^2 + 9(beta - 6)^2 le 144$16(\alpha - 2)^2 + 9(\beta - 6)^2 \le 144$
Dividing by 144:
frac(alpha - 2)^29 + frac(beta - 6)^216 le 1$\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \le 1$
This represents the interior and boundary of an ellipse centered at (2, 6)$(2, 6)$ with semi-minor axis a = 3$a = 3$ and semi-major axis b = 4$b = 4$.
### Step 1: Spatial Inclusion Check
Let's check the extreme horizontal and vertical extents of the ellipse B$B$:
Horizontal extent: x in [2 - 3, 2 + 3] = [-1, 5]$x \in [2 - 3, 2 + 3] = [-1, 5]$
Vertical extent: y in [6 - 4, 6 + 4] = [2, 10]$y \in [6 - 4, 6 + 4] = [2, 10]$
Comparing with the boundaries of rectangle A$A$ (x in [-3, 5]$x \in [-3, 5]$ and y in [-1, 11]$y \in [-1, 11]$):
[-1, 5] subseteq [-3, 5]$[-1, 5] \subseteq [-3, 5]$[2, 10] subseteq [-1, 11]$[2, 10] \subseteq [-1, 11]$Set Inclusion and Regions diagram for Q53 - JEE Main 2025 Evening
Since all points of the ellipse lie perfectly inside the rectangular region, we conclusively find that B subset A$B \subset A$.
### Pattern Recognition
A bounding box check (finding h pm a$h \pm a$ and k pm b$k \pm b$) for conics is the fastest analytical shortcut to verify set inclusion without plotting extensive coordinates.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sets, Relations and Functions
Class 11 Mathematics: Conic Sections
Q54jee_main_2025_07_april_eveningRange of Rational Functions
If the range of the function f(x) = frac5 - xx^2 - 3x + 2$f(x) = \frac{5 - x}{x^2 - 3x + 2}$, x neq 1, 2$x \neq 1, 2$, is (-infty , alpha ] cup [ beta , infty)$(-\infty , \alpha ] \cup [ \beta , \infty)$, then alpha^2 +beta^2$\alpha^2 +\beta^2$ is equal to :
A.190$190$
B.192$192$
C.188$188$
D.194$194$
Solution
### Related Formula
For a quadratic equation Ax^2 + Bx + C = 0$Ax^2 + Bx + C = 0$ to yield real roots, its discriminant must satisfy:
D = B^2 - 4AC ge 0$D = B^2 - 4AC \ge 0$
### Core Logic
Set y = frac5 - xx^2 - 3x + 2$y = \frac{5 - x}{x^2 - 3x + 2}$:
y(x^2 - 3x + 2) = 5 - x$y(x^2 - 3x + 2) = 5 - x$yx^2 - 3xy + 2y + x - 5 = 0$yx^2 - 3xy + 2y + x - 5 = 0$
Rearranging into a standard quadratic equation in terms of x$x$:
yx^2 + (1 - 3y)x + (2y - 5) = 0$yx^2 + (1 - 3y)x + (2y - 5) = 0$
### Step 1: Discriminant Method
Case I: If y = 0$y = 0$, the equation simplifies to x - 5 = 0 implies x = 5$x - 5 = 0 \implies x = 5$, which is a valid part of the domain. Thus, 0$0$ belongs to the range.
Case II: If y neq 0$y \neq 0$, for x$x$ to be real, D ge 0$D \ge 0$:
(1 - 3y)^2 - 4(y)(2y - 5) ge 0$(1 - 3y)^2 - 4(y)(2y - 5) \ge 0$9y^2 + 1 - 6y - 8y^2 + 20y ge 0$9y^2 + 1 - 6y - 8y^2 + 20y \ge 0$y^2 + 14y + 1 ge 0$y^2 + 14y + 1 \ge 0$
### Step 2: Solving the Inequality
Completing the square for y^2 + 14y + 1 ge 0$y^2 + 14y + 1 \ge 0$:
(y + 7)^2 - 48 ge 0 implies (y + 7)^2 ge (4sqrt3)^2$(y + 7)^2 - 48 \ge 0 \implies (y + 7)^2 \ge (4\sqrt{3})^2$
This gives:
y le -7 - 4sqrt3 quad textor quad y ge -7 + 4sqrt3$y \le -7 - 4\sqrt{3} \quad \text{or} \quad y \ge -7 + 4\sqrt{3}$
Comparing with the interval (-infty , alpha ] cup [ beta , infty)$(-\infty , \alpha ] \cup [ \beta , \infty)$:
alpha = -7 - 4sqrt3$\alpha = -7 - 4\sqrt{3}$beta = -7 + 4sqrt3$\beta = -7 + 4\sqrt{3}$
### Step 3: Finding alpha^2 + beta^2
Using algebraic identities:
alpha^2 + beta^2 = (-7 - 4sqrt3)^2 + (-7 + 4sqrt3)^2$\alpha^2 + \beta^2 = (-7 - 4\sqrt{3})^2 + (-7 + 4\sqrt{3})^2$= 2(7^2 + (4sqrt3)^2) = 2(49 + 48) = 2(97) = 194$= 2(7^2 + (4\sqrt{3})^2) = 2(49 + 48) = 2(97) = 194$
### Pattern Recognition
For rational expressions of the form fractextLineartextQuadratic$\frac{\text{Linear}}{\text{Quadratic}}$, converting to a quadratic in x$x$ and forcing D ge 0$D \ge 0$ establishes the range boundaries elegantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sets, Relations and Functions
Q52jee_main_2025_24_jan_eveningOne-One and Onto Functions
The function f:(-infty,infty)rightarrow(-infty,1)$f:(-\infty,\infty)\rightarrow(-\infty,1)$, defined by f(x)=frac2^x-2^-x2^x+2^-x$f(x)=\frac{2^{x}-2^{-x}}{2^{x}+2^{-x}}$ is: [cite: 3251, 3252]
A.textOne-one but not onto$\text{One-one but not onto}$
B.textOnto but not one-one$\text{Onto but not one-one}$
C.textBoth one-one and onto$\text{Both one-one and onto}$
D.textNeither one-one nor onto$\text{Neither one-one nor onto}$
Solution
### Related Formula
A function is one-one if its derivative is strictly monotonic (always positive or always negative) across its domain. It is onto if its range equals its co-domain.
### Core Logic
Rewrite the function by multiplying the numerator and denominator by 2^x$2^x$:
f(x) = frac2^2x - 12^2x + 1 = 1 - frac22^2x + 1$f(x) = \frac{2^{2x} - 1}{2^{2x} + 1} = 1 - \frac{2}{2^{2x} + 1}$
### Step 1: Check One-One property
Differentiating f(x)$f(x)$ with respect to x$x$ :
f'(x) = frac2(2^2x + 1)^2 cdot 2 cdot 2^2x cdot ln 2 = frac4 cdot 2^2x cdot ln 2(2^2x + 1)^2$f'(x) = \frac{2}{(2^{2x} + 1)^2} \cdot 2 \cdot 2^{2x} \cdot \ln 2 = \frac{4 \cdot 2^{2x} \cdot \ln 2}{(2^{2x} + 1)^2}$
Since 2^2x > 0$2^{2x} > 0$ and ln 2 > 0$\ln 2 > 0$, f'(x) > 0$f'(x) > 0$ always. Thus, f(x)$f(x)$ is strictly increasing, confirming it is a one-one function.
### Step 2: Check Onto property
Analyze the limits at boundaries [cite: 3880, 3881]:
lim_x to -infty f(x) = 1 - frac20 + 1 = -1$\lim_{x \to -\infty} f(x) = 1 - \frac{2}{0 + 1} = -1$lim_x to infty f(x) = 1 - 0 = 1$\lim_{x \to \infty} f(x) = 1 - 0 = 1$
Thus, the range of the function is (-1, 1)$(-1, 1)$. Since the given co-domain is (-infty, 1)$(-\infty, 1)$ and textRange neq textCo-domain$\text{Range} \neq \text{Co-domain}$ , the function is not onto.
### Pattern Recognition
The expression given is a shifted form of the hyperbolic tangent function tanh(x ln 2)$\tanh(x \ln 2)$. Hyperbolic tangent always maps to (-1, 1)$(-1, 1)$, making its restriction against (-infty, 1)$(-\infty, 1)$ non-surjective (not onto).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Relations and Functions
Q58jee_main_2025_24_jan_eveningLinear Programming and Inequalities in Two Variables
Let the points (frac112,alpha)$(\frac{11}{2},\alpha)$ lie on or inside the \triangle with sides x+y=11$x+y=11$, x+2y=16$x+2y=16$ and 2x+3y=29$2x+3y=29$ Then the product of the smallest and the largest values of a is equal to: [cite: 3284, 3285, 3286, 3287, 3288, 3289]
A.22$22$
B.44$44$
C.33$33$
D.55$55$
Solution
### Related Formula
For a vertical line x = x_0$x = x_0$ crossing a bounded region, the valid coordinates of y$y$ sit between the boundary lines intersecting that specific line vertical plane.
### Core Logic
The point given is fixed at x = frac112 = 5.5$x = \frac{11}{2} = 5.5$[cite: 3285, 3925]. We evaluate the values of y$y$ along this vertical line segment across each boundary edge.
Linear Programming region graph for Q58 - JEE Main 2025 Evening
### Step 1: Evaluate Intersections
Substitute x = frac112$x = \frac{11}{2}$ into the three linear constraints:
1. From x + y = 11$x + y = 11$:
frac112 + y = 11 Rightarrow y = 11 - 5.5 = 5.5$\frac{11}{2} + y = 11 \Rightarrow y = 11 - 5.5 = 5.5$
2. From x + 2y = 16$x + 2y = 16$:
frac112 + 2y = 16 Rightarrow 2y = 16 - 5.5 = 10.5 Rightarrow y = 5.25$\frac{11}{2} + 2y = 16 \Rightarrow 2y = 16 - 5.5 = 10.5 \Rightarrow y = 5.25$
3. From 2x + 3y = 29$2x + 3y = 29$:
2left(frac112right) + 3y = 29 Rightarrow 11 + 3y = 29 Rightarrow 3y = 18 Rightarrow y = 6$2\left(\frac{11}{2}\right) + 3y = 29 \Rightarrow 11 + 3y = 29 \Rightarrow 3y = 18 \Rightarrow y = 6$
### Step 2: Define Extrema and Multiply
Checking the internal region of the \triangle bounded by these lines [cite: 3286, 3287], the valid range for alpha$\alpha$ along the section line is delimited by y = 5.5$y = 5.5$ and y = 6$y = 6$:
alpha_min = frac112 = 5.5$\alpha_{\min} = \frac{11}{2} = 5.5$alpha_max = 6$\alpha_{\max} = 6$
The product of the limits is :
alpha_min cdot alpha_max = frac112 times 6 = 33$\alpha_{\min} \cdot \alpha_{\max} = \frac{11}{2} \times 6 = 33$
### Pattern Recognition
Instead of drawing full coordinate diagrams or computing all three vertex points, evaluating values directly at the fixed coordinate constraint x = 5.5$x = 5.5$ saves time during multi-line area problems.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Linear Inequalities
Class 11 Mathematics: Straight Lines
Q71jee_main_2025_24_jan_eveningNumber of Functions under Constraints
Number of functions f:\1,2,dots,100\rightarrow\0,1\$f:\{1,2,\dots,100\}\rightarrow\{0,1\}$, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to \_\_\_\_.
Numerical Answer.Answer: 392
Solution
### Related Formula
Fundamental Counting Principle: If an operation can be performed in n_1$n_1$ ways, followed by a second operation in n_2$n_2$ ways, the total configurations equal n_1 times n_2$n_1 \times n_2$.
### Core Logic
The domain set contains integers from 1 to 100. We divide the mapping requirements across distinct subsets of this domain[cite: 4064, 4068, 4069].
Function mapping grid for Q71 - JEE Main 2025 Evening
### Step 1: Choose the single element from \1, 2, dots, 98\$\{1, 2, \dots, 98\}$
We must assign the image value 1 to exactly one positive integer from the \subset \1, 2, dots, 98\$\{1, 2, \dots, 98\}$. The number of ways to pick this single element is :
binom981 = 98 text ways$\binom{98}{1} = 98 \text{ ways}$
### Step 2: Mapping remaining elements
The remaining 97 elements in the \1, 2, dots, 98\$\{1, 2, \dots, 98\}$ \subset cannot map to 1, so they must map to 0. This leaves exactly 1 choice per remaining element.
For the final two elements in the domain, 99 and 100, there are no structural constraints [cite: 4068, 4069]:
- Element 99 can map to either 0 or 1 (2 options) .
- Element 100 can map to either 0 or 1 (2 options).
### Step 3: Total functions combination
Multiply the independent choices together :
textTotal functions = 98 times 2 times 2 = 392$\text{Total functions} = 98 \times 2 \times 2 = 392$ [cite: 4065, 4067]
### Pattern Recognition
Separate domains tightly into restricted blocks vs completely free components. Realizing that elements 99 and 100 behave independently with full co-domain targets leaves a clear product formulation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Relations and Functions
Class 11 Mathematics: Permutations and Combinations
More Relations and Functions Questions — jee_main_2024_01_february_morning
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