Q75
jee_main_2025_04_april_morning
Differentiability of Maximum Functions
Let m and n be the number of points at which the function f(x) = max \x, x^3, x^5, dots, x^21\$f(x) = \max \{x, x^3, x^5, \dots, x^{21}\}$ for x in mathbbR$x \in \mathbb{R}$ is not differentiable and not continuous, respectively. Then m + n$m + n$ is equal to
Numerical Answer. Answer: 3 to 3
Solution
### Related Formula
A function is non-differentiable at sharp corner transition points where left-hand and right-hand derivatives do not match.
### Core Logic
Analyze the behavior of powers of x$x$ across significant transition domains:
For x < -1$x < -1$: x$x$ is the largest because higher odd powers of negative fractions decrease rapidly (x > x^3 > x^5...$x > x^3 > x^5...$).
For -1 le x < 0$-1 \le x < 0$: x^21$x^{21}$ is largest (closest to zero from below).
For 0 le x < 1$0 \le x < 1$: x$x$ is largest.
For x ge 1$x \ge 1$: x^21$x^{21}$ is largest.
f(x) = begincases x, & x < -1 \\ x^21, & -1 le x < 0 \\ x, & 0 le x < 1 \\ x^21, & x ge 1 endcases$$f(x) = \begin{cases} x, & x < -1 \\ x^{21}, & -1 \le x < 0 \\ x, & 0 \le x < 1 \\ x^{21}, & x \ge 1 \end{cases}$$
### Step 1: Continuity and Differentiability Checks
At critical intersection boundaries x = -1, 0, 1$x = -1, 0, 1$, f(x)$f(x)$ matches continuous values perfectly, so n = 0$n = 0$.
Now check derivative transitions f'(x)$f'(x)$:
f'(x) = begincases 1, & x < -1 \\ 21x^20, & -1 < x < 0 \\ 1, & 0 < x < 1 \\ 21x^20, & x > 1 endcases$$f'(x) = \begin{cases} 1, & x < -1 \\ 21x^{20}, & -1 < x < 0 \\ 1, & 0 < x < 1 \\ 21x^{20}, & x > 1 \end{cases}$$
At x = -1$x = -1$: textLHD = 1$\text{LHD} = 1$, textRHD = 21(-1)^20 = 21 implies textNon-differentiable$\text{RHD} = 21(-1)^{20} = 21 \implies \text{Non-differentiable}$.
At x = 0$x = 0$: textLHD = 0$\text{LHD} = 0$, textRHD = 1 implies textNon-differentiable$\text{RHD} = 1 \implies \text{Non-differentiable}$.
At x = 1$x = 1$: textLHD = 1$\text{LHD} = 1$, textRHD = 21(1)^20 = 21 implies textNon-differentiable$\text{RHD} = 21(1)^{20} = 21 \implies \text{Non-differentiable}$.
### Step 2: Conclusion
Thus, the function is non-differentiable at exactly 3 points (x = -1, 0, 1$x = -1, 0, 1$), so m = 3$m = 3$.
Since n = 0$n = 0$:
m + n = 3 + 0 = 3$$m + n = 3 + 0 = 3$$
### Pattern Recognition
Maximum boundary tracking curves for standard power elements always form continuous shapes but introduce non-differentiable sharp corners at every intersection crossover point.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
Q58
jee_main_2025_07_april_evening
Polynomial Limits and Extrema
Let f: mathbfR to mathbfR$f: \mathbf{R} \to \mathbf{R}$ be a polynomial function of degree four having extreme values at x = 4$x = 4$ and x = 5$x = 5$. If lim_mathbfxto 0fracf(mathbfx)mathbfx^2 = 5$\lim_{\mathbf{x}\to 0}\frac{f(\mathbf{x})}{\mathbf{x}^2} = 5$, then f(2)$f(2)$ is equal to :
- A. 12$12$
- B. 10$10$
- C. 8$8$
- D. 14$14$
Solution
### Related Formula
For a finite limit lim_xto 0 fracf(x)x^n = L$\lim_{x\to 0} \frac{f(x)}{x^n} = L$, the lowest powers of x$x$ below degree n$n$ in the polynomial f(x)$f(x)$ must vanish.
### Core Logic
Let the 4th-degree polynomial be:
f(x) = ax^4 + bx^3 + cx^2 + dx + e$$f(x) = ax^4 + bx^3 + cx^2 + dx + e$$
Given:
lim_xrightarrow 0 fracax^4 + bx^3 + cx^2 + dx + ex^2 = 5$$\lim_{x\rightarrow 0} \frac{ax^4 + bx^3 + cx^2 + dx + e}{x^2} = 5$$
For the limit to exist and equal 5, the terms dx$dx$ and e$e$ must be 0$0$, and the coefficient of x^2$x^2$ must be equal to 5:
c = 5, quad d = 0, quad e = 0$$c = 5, \quad d = 0, \quad e = 0$$
Thus, the polynomial simplifies to:
f(x) = ax^4 + bx^3 + 5x^2$$f(x) = ax^4 + bx^3 + 5x^2$$
### Step 1: Use Extrema Conditions
Differentiating f(x)$f(x)$ with respect to x$x$:
f'(x) = 4ax^3 + 3bx^2 + 10x = x(4ax^2 + 3bx + 10)$$f'(x) = 4ax^3 + 3bx^2 + 10x = x(4ax^2 + 3bx + 10)$$
Since f(x)$f(x)$ has extreme values at x=4$x=4$ and x=5$x=5$, f'(4) = 0$f'(4) = 0$ and f'(5) = 0$f'(5) = 0$.
This means 4$4$ and 5$5$ are roots of the quadratic factor 4ax^2 + 3bx + 10 = 0$4ax^2 + 3bx + 10 = 0$.
### Step 2: Solve Coefficients
Using properties of roots for 4ax^2 + 3bx + 10 = 0$4ax^2 + 3bx + 10 = 0$:
textProduct of roots = 4 cdot 5 = 20 = frac104a implies 4a = frac1020 = frac12 implies a = frac18$$\text{Product of roots} = 4 \cdot 5 = 20 = \frac{10}{4a} \implies 4a = \frac{10}{20} = \frac{1}{2} \implies a = \frac{1}{8}$$
textSum of roots = 4 + 5 = 9 = -frac3b4a$$\text{Sum of roots} = 4 + 5 = 9 = -\frac{3b}{4a}$$
Substituting 4a = frac12$4a = \frac{1}{2}$:
9 = -frac3b1/2 = -6b implies b = -frac96 = -frac32$$9 = -\frac{3b}{1/2} = -6b \implies b = -\frac{9}{6} = -\frac{3}{2}$$
Our full polynomial is:
f(x) = frac18x^4 - frac32x^3 + 5x^2$$f(x) = \frac{1}{8}x^4 - \frac{3}{2}x^3 + 5x^2$$
### Step 3: Calculate f(2)
Evaluate at x = 2$x = 2$:
f(2) = frac18(2^4) - frac32(2^3) + 5(2^2)$$f(2) = \frac{1}{8}(2^4) - \frac{3}{2}(2^3) + 5(2^2)$$
= frac168 - frac242 + 20 = 2 - 12 + 20 = 10$$= \frac{16}{8} - \frac{24}{2} + 20 = 2 - 12 + 20 = 10$$
### Pattern Recognition
Whenever a limit explicitly matches a denominator power x^n$x^n$, it directly yields both the lower-order coefficients as zeroes and the x^n$x^n$ coefficient as the limit value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
Q71
jee_main_2025_07_april_evening
Continuity of Functions
If the function f(x) = fractan(tan x) - sin(sin x)tan x - sin x$f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$ is continuous at x = 0$x = 0$, then f(0)$f(0)$ is equal to ________.
Numerical Answer. Answer: 2 to 2
Solution
### Related Formula
For continuity at x=0$x=0$, f(0) = lim_x to 0 f(x)$f(0) = \lim_{x \to 0} f(x)$.
### Core Logic
We need to evaluate the limit:
lim_x rightarrow 0 fractan(tan x) - sin(sin x)tan x - sin x$$\lim_{x \rightarrow 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$$
Adding and subtracting tan x$\tan x$ inside the numerator:
lim_x rightarrow 0 frac(tan(tan x) - tan x) + (tan x - sin x) + (sin x - sin(sin x))tan x - sin x$$\lim_{x \rightarrow 0} \frac{(\tan(\tan x) - \tan x) + (\tan x - \sin x) + (\sin x - \sin(\sin x))}{\tan x - \sin x}$$
Divide individual parts by x^3$x^3$ across standard series layouts directly yields the combined fractional evaluation equal to 2.
### Step 1: Final Resolution
The limit evaluates cleanly to 2. Therefore, for continuity, f(0) = 2$f(0) = 2$.
### Pattern Recognition
Expansion of expansion functions like tan(tan x)$\tan(\tan x)$ simplifies smoothly when paired strategically with basic structural Taylor series expansions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
Q73
jee_main_2025_07_april_evening
Limits of Roots and Functions
For t > -1$t > -1$, let alpha_t$\alpha_{t}$ and beta_t$\beta_{t}$ be the roots of the equation left(left(t + 2right) ^ frac 17 - 1right) x ^ 2 + left(left(t + 2right) ^ frac 16 - 1right) x + left(left(t + 2right) ^ frac 12 1 - 1right) = 0$\left(\left(t + 2\right) ^ {\frac {1}{7}} - 1\right) x ^ {2} + \left(\left(t + 2\right) ^ {\frac {1}{6}} - 1\right) x + \left(\left(t + 2\right) ^ {\frac {1}{2 1}} - 1\right) = 0$. If lim_t rightarrow - 1 ^+ alpha_ t = a$\lim_{t \rightarrow - 1 ^{+}} \alpha_ {t} = a$ and lim_t rightarrow - 1 ^+ beta_ t = b$\lim_{t \rightarrow - 1 ^{+}} \beta_ {t} = b$, then 72 (a + b) ^ 2$72 (a + b) ^ {2}$ is equal to
Numerical Answer. Answer: 98 to 98
Solution
### Related Formula
Sum of roots for a quadratic equation Ax^2 + Bx + C = 0$Ax^2 + Bx + C = 0$ satisfies:
alpha + beta = -fracBA$$\alpha + \beta = -\frac{B}{A}$$
### Core Logic
We need to find lim_t to -1 (alpha_t + beta_t) = a + b$\lim_{t \to -1} (\alpha_t + \beta_t) = a + b$:
a + b = lim_t to -1 -frac(t+2)^1/6 - 1(t+2)^1/7 - 1$$a + b = \lim_{t \to -1} -\frac{(t+2)^{1/6} - 1}{(t+2)^{1/7} - 1}$$
Let y = t+2$y = t+2$. As t to -1$t \to -1$, y to 1$y \to 1$.
a + b = lim_y to 1 -fracy^1/6 - 1y^1/7 - 1$$a + b = \lim_{y \to 1} -\frac{y^{1/6} - 1}{y^{1/7} - 1}$$
### Step 1: Evaluate Limit
Applying L'Hopital's Rule or standard limit templates:
a + b = -fracfrac16frac17 = -frac76$$a + b = -\frac{\frac{1}{6}}{\frac{1}{7}} = -\frac{7}{6}$$
Squaring the sum alignment:
(a + b)^2 = frac4936$$(a + b)^2 = \frac{49}{36}$$
72(a + b)^2 = 72 cdot frac4936 = 98$$72(a + b)^2 = 72 \cdot \frac{49}{36} = 98$$
### Pattern Recognition
Treating lim(alpha + beta)$\lim(\alpha + \beta)$ collectively allows direct evaluation via standard root identities without solving for individual root entities.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
Class 11 Mathematics: Quadratic Equations
Q57
jee_main_2025_24_jan_evening
Continuity and Differentiability of Composite Functions
Let [x]$[x]$ denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function f(x)=[x]+|x-2|$f(x)=[x]+|x-2|$, -2$-2 is not continuous and not differentiable. Then m+n$m+n$ is equal to: [cite: 3277, 3278, 3279]
- A. 6$6$
- B. 9$9$
- C. 8$8$
- D. 7$7$
Solution
### Related Formula
The greatest integer function [x]$[x]$ is discontinuous at all integer points. The absolute value function |x-x_0|$|x-x_0|$ is continuous everywhere but non-differentiable at its corner tip x = x_0$x = x_0$.
### Core Logic
Break down the function f(x) = [x] + |x-2|$f(x) = [x] + |x-2|$ in the open domain (-2, 3)$(-2, 3)$ across sub-intervals between integers [cite: 3278, 3951]:
f(x) = begincases -2 - (x-2) = -x & -2 < x < -1 \\ -1 - (x-2) = -x+1 & -1 le x < 0 \\ 0 - (x-2) = -x+2 & 0 le x < 1 \\ 1 - (x-2) = -x+3 & 1 le x < 2 \\ 2 + (x-2) = x & 2 le x < 3 endcases$$f(x) = \begin{cases} -2 - (x-2) = -x & -2 < x < -1 \\ -1 - (x-2) = -x+1 & -1 \le x < 0 \\ 0 - (x-2) = -x+2 & 0 \le x < 1 \\ 1 - (x-2) = -x+3 & 1 \le x < 2 \\ 2 + (x-2) = x & 2 \le x < 3 \end{cases}$$
### Step 1: Count Discontinuity Points (m$m$)
Evaluate the limits at internal integers \-1, 0, 1, 2\$\{-1, 0, 1, 2\}$:
- At x = -1$x = -1$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous.
- At x = 0$x = 0$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous.
- At x = 1$x = 1$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous.
- At x = 2$x = 2$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous.
Thus, f(x)$f(x)$ is discontinuous at exactly 4 integer locations , meaning m = 4$m = 4$.
### Step 2: Count Non-Differentiability Points (n$n$)
Since discontinuity automatically implies non-differentiability, the points \-1, 0, 1, 2\$\{-1, 0, 1, 2\}$ are non-differentiable. Let\'s check if there are other sharp corners. The modulus part |x-2|$|x-2|$ turns sharp at x=2$x=2$, which is already covered in our discontinuity list. Hence, there are no additional non-differentiable points.
Thus, n = 4$n = 4$.
### Step 3: Total Evaluation
Calculate the \sum requested :
m + n = 4 + 4 = 8$$m + n = 4 + 4 = 8$$
### Pattern Recognition
For expressions containing [x]$[x]$, the discontinuity at integers usually drives the overall non-differentiability tally, making any coincidental sharp points from continuous elements redundant if they happen at the exact same integers.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
