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Let f:Rrightarrow R and g:Rrightarrow R be defined as f(x)=begincaseslog_ex & , & x>0\\ e^-x & , & xle0endcases and g(x)=begincasesx & , & xge0\\ e^x & , & x<0endcases Then, g circ f: Rrightarrow R is:

Solution & Explanation

### Related Formula For composite functions, g(f(x)) is determined by substituting the range of f(x) into the appropriate domain intervals of g(y): g(f(x)) = begincases f(x) & , & f(x) ge 0 \\ e^f(x) & , & f(x) < 0 endcases ### Core Logic Let us analyze the definition of g(f(x)) branch-by-branch based on the domain of x: 1. **Case 1: x le 0** Here, f(x) = e^-x. Since x le 0, -x ge 0 implies e^-x ge 1 > 0. Since f(x) ge 0, we use the upper branch of g(y): g(f(x)) = f(x) = e^-x 2. **Case 2: x > 0** Here, f(x) = log_e x. - Subcase (a): If f(x) ge 0 implies log_e x ge 0 implies x ge 1. Then, g(f(x)) = f(x) = log_e x. - Subcase (b): If f(x) < 0 implies log_e x < 0 implies 0 < x < 1. Then, 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
Composition function graph for Q11 - JEE Main 2024 01 February Morning
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 and x_2 = e. g(f(0)) = e^-0 = 1 g(f(e)) = log_e e = 1 Since distinct inputs yield identical outputs (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, e^-x in [1, infty). - For 0 < x < 1, x in (0, 1). - For x ge 1, \log_e x in [0, infty). The union of these sets gives the total range as [0, infty). Since the codomain is given as mathbbR, textRange neq textCodomain, 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 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

Reference Study Guides

Relations And Functions

More Relations and Functions Previous-Year Questions — Page 6

Q53 jee_main_2025_24_jan_morning Functional Relations and Equations
Let f: mathbbR - \0\ to mathbbR be a function such that f(x) - 6fleft(frac1xright) = frac353x - frac52 If lim_x to 0 left( frac1alpha x + f(x) right) = beta for some alpha, beta in mathbbR, then alpha + 2beta is equal to :
  • A. 3
  • B. 5
  • C. 4
  • D. 6

Solution

### Related Formula For functional equations with inversion, substituting x to frac1x establishes a solvable system of algebraic equations to isolate f(x) directly. ### Core Logic The given equation is: f(x) - 6fleft(frac1xright) = frac353x - frac52 quad dots (1) Substitute x to frac1x in equation (1): fleft(frac1xright) - 6f(x) = frac35x3 - frac52 quad dots (2) ### Step 1: Eliminate f(1/x) Multiply equation (2) by 6 and add it to equation (1): left[ f(x) - 6fleft(frac1xright) right] + 6 left[ fleft(frac1xright) - 6f(x) right] = left( frac353x - frac52 right) + 6 left( frac35x3 - frac52 right) f(x) - 36f(x) = frac353x - frac52 + 70x - 15 -35f(x) = 70x + frac353x - frac352 Divide across by -35: f(x) = -2x - frac13x + frac12 ### Step 2: Evaluate the Limit We are given that the following limit evaluates to a finite constant beta: lim_x to 0 left( frac1alpha x + f(x) right) = beta lim_x to 0 left( frac1alpha x - 2x - frac13x + frac12 right) = beta lim_x to 0 left( left[ frac1alpha - frac13 right] frac1x - 2x + frac12 right) = beta For the limit to be a finite value, the coefficient of frac1x must vanish completely: frac1alpha - frac13 = 0 implies alpha = 3 When alpha = 3, the limit simplifies directly to the constant term: beta = lim_x to 0 left( -2x + frac12 right) = frac12 ### Step 3: Calculate Final Value Substitute the determined parameters alpha and beta: alpha + 2beta = 3 + 2left(frac12right) = 3 + 1 = 4 ### Pattern Recognition In limit problems involving fractional components where x to 0, any term like frac1x or higher negative powers must have a net coefficient of zero to guarantee existence of a finite limit value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Functions Class 11 Mathematics: Limits and Derivatives
Q55 jee_main_2025_24_jan_morning Symmetric Property of Functions
Let f(x) = frac2^x + 2 + 162^2x + 1 + 2^x + 4 + 32 . Then the value of 8left(fleft(frac115 ight) + fleft(frac215 ight) + ldots + fleft(frac5915 ight) ight) is equal to :
  • A. 118
  • B. 92
  • C. 102
  • D. 108

Solution

### Related Formula Many finite fractional sum questions involving functional terms rely on identifying an underlying symmetric summation invariant, typically of the form f(x) + f(k-x) = textconstant. ### Core Logic First simplify the expression for f(x) algebraically: f(x) = frac4 cdot 2^x + 162 cdot (2^x)^2 + 16 cdot 2^x + 32 Factor out 4 from the numerator and 2 from the denominator: f(x) = frac4(2^x + 4)2[(2^x)^2 + 8 cdot 2^x + 16] = frac2(2^x + 4)(2^x + 4)^2 = frac22^x + 4 ### Step 1: Establish Symmetry Pairings Let's check the value of f(x) + f(4-x): f(4-x) = frac22^4-x + 4 = frac2frac162^x + 4 = frac2 cdot 2^x16 + 4 cdot 2^x = frac2 cdot 2^x4(2^x + 4) = frac2^x2(2^x + 4) Now compute the sum directly: f(x) + f(4-x) = frac22^x + 4 + frac2^x2(2^x + 4) = frac4 + 2^x2(2^x + 4) = frac12 Hence, whenever two input arguments sum up to 4, the sum of their functional values is exactly frac12. ### Step 2: Group the Finite Series Terms Consider the terms inside the requested sequence: frac115 + frac5915 = frac6015 = 4 implies fleft(frac115 ight) + fleft(frac5915 ight) = frac12 frac215 + frac5815 = frac6015 = 4 implies fleft(frac215 ight) + fleft(frac5815 ight) = frac12 This complementary pairing continues up to: fleft(frac2915 ight) + fleft(frac3115 ight) = frac12 This yields exactly 29 distinct pairs. The single middle term left unpaired corresponds to: textMiddle Term = fleft(frac3015 ight) = f(2) = frac22^2 + 4 = frac28 = frac14 ### Step 3: Evaluate Final Expression Compute the total value by multiplying the grouped sum by 8: textTotal = 8 cdot left[ 29 cdot left(frac12right) + frac14 right] textTotal = 8 cdot frac292 + 8 cdot frac14 = 116 + 2 = 118 ### Pattern Recognition Whenever a symmetric set of arguments is presented inside a summation matching frackn + fracN-kn = textconstant, look for an algebraic reduction of f(x) that yields a uniform constant sum for symmetric pairs. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Functions
Q64 jee_main_2025_28_jan_evening Range and Onto Functions
Let f:[0,3]rightarrow A be defined by f(x)=2x^3-15x^2+36x+7 and g:[0,infty)rightarrow B be defined by g(x)=fracx^2025x^2025+1. If both the functions are onto and S=\xin Z:xin Atext or xin B\, then n(S) is equal to:
  • A. 30
  • B. 36
  • C. 29
  • D. 31

Solution

### Related Formula For a function to be onto, its Codomain must equal its Range. ### Core Logic First, find range A for f(x) = 2x^3 - 15x^2 + 36x + 7 over [0, 3]. Differentiating f(x): f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3) Critical points are x=2 and x=3. Evaluate f(x) at boundary and critical points: - f(0) = 7 - f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35 - f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34 Thus, Range A = [7, 35]. ### Step 1: Find Range B for g(x) Now look at g(x) = fracx^2025x^2025+1 = 1 - frac1x^2025+1 over [0, infty). - At x = 0, g(0) = 0. - As x to infty, g(x) to 1. Since g(x) is continuous and monotonically strictly increasing, Range B = [0, 1). ### Step 2: Find the Integer Count of Union Set S S = \x in mathbbZ : x in A text or x in B\ = mathbbZ cap (A cup B) A cup B = [0, 1) cup [7, 35] The integers in this set are: - From [0, 1): x = 0 - From [7, 35]: x = 7, 8, 9, dots, 35 Total number of integers n(S): n(S) = 1 + (35 - 7 + 1) = 1 + 29 = 30 ### Pattern Recognition The condition 'or' means union. Be careful not to include integers between 1 and 6 since they are not present in either continuous range segment. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Functions Class 12 Mathematics: Application of Derivatives
Q65 jee_main_2025_28_jan_evening Domain of Inverse Trigonometric Functions
Let [x] denote the greatest integer less than or equal to x. Then domain of f(x)=sec^-1(2[x]+1) is:
  • A. (-infty,-1]cup[0,infty)
  • B. (-infty,infty)
  • C. (-infty,-1]cup[1,infty)
  • D. (-infty,infty)-\0\

Solution

### Related Formula The domain of sec^-1(y) is given by |y| ge 1, which means: y le -1 quad textor quad y ge 1 ### Core Logic For f(x) = sec^-1(2[x]+1) to be defined: 2[x] + 1 le -1 quad textor quad 2[x] + 1 ge 1 ### Step 1: Solve individual inequalities Case 1: 2[x] + 1 le -1 implies 2[x] le -2 implies [x] le -1 This holds true for all x < 0, i.e., x in (-infty, 0). Case 2: 2[x] + 1 ge 1 implies 2[x] ge 0 implies [x] ge 0 This holds true for all x ge 0, i.e., x in [0, infty). ### Step 2: Take Union of the Solutions textDomain = (-infty, 0) cup [0, infty) = (-infty, infty) ### Pattern Recognition Since [x] covers all integer values and 2[x]+1 forms all odd integer values, the expression inside sec^-1 is always a non-zero integer. Non-zero integers always have absolute value ge 1. Hence, it is valid for all real numbers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Functions Class 12 Mathematics: Inverse Trigonometric Functions
Q70 jee_main_2025_28_jan_evening Functional Equations
Let f:R-\0\rightarrow(-infty,1] be a polynomial of degree 2, satisfying f(x)fleft(frac1xright)=f(x)+fleft(frac1xright). If f(K)=-2K then the sum of squares of all possible values of K is:
  • A. 1
  • B. 6
  • C. 7
  • D. 9

Solution

### Related Formula Standard result for functional equation of a polynomial satisfying f(x)f(1/x) = f(x) + f(1/x): f(x) = 1 pm x^n ### Core Logic Given that f(x) is a polynomial of degree 2, the identity implies: f(x) = 1 + x^2 quad textor quad f(x) = 1 - x^2 We are given the range is bounded above: (-infty, 1]. - For 1 + x^2, the range is [1, infty). - For 1 - x^2, the range is (-infty, 1]. Therefore, the correct functional form is f(x) = 1 - x^2. ### Step 1: Solve for K Given condition: f(K) = -2K 1 - K^2 = -2K implies K^2 - 2K - 1 = 0 Let the roots of this equation be K_1 and K_2. From quadratic properties (Vieta's formulas): K_1 + K_2 = 2 K_1 cdot K_2 = -1 ### Step 2: Calculate Sum of Squares We need the sum of squares of the values of K: K_1^2 + K_2^2 = (K_1 + K_2)^2 - 2K_1K_2 K_1^2 + K_2^2 = (2)^2 - 2(-1) = 4 + 2 = 6 ### Pattern Recognition The functional equation f(x)f(1/x)=f(x)+f(1/x) uniquely forces polynomials to be 1 pm x^n. Remembering this shortcut saves valuable time required to derive the template from general coefficients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 12 Mathematics: Functions

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