Let the domain of the function f(x) = log_2log_4log_6(3 + 4x - x^2)$f(x) = \log_{2}\log_{4}\log_{6}(3 + 4x - x^{2})$ be (a, b)$(a, b)$[cite: 663]. If int_0^b-a[x^2]dx = p - sqrtq - sqrtr$\int_{0}^{b-a}[x^{2}]dx = p - \sqrt{q} - \sqrt{r}$ [cite: 664], where p, q, r in mathbbN$p, q, r \in \mathbb{N}$ [cite: 664] and gcd(p, q, r) = 1$\gcd(p, q, r) = 1$ [cite: 668], and [cdot]$[\cdot]$ represents the greatest integer function [cite: 668], then p + q + r$p + q + r$ is equal to[cite: 668]:
A.10
B.8
C.11
D.9
Solution & Explanation
### Related Formula
Domain of log chain iterations: For log_2log_4(M) > 0$\log_2\log_4(M) > 0$, we require log_4(M) > 1 implies M > 4$\log_4(M) > 1 \implies M > 4$.
### Core Logic
Trace internal arguments outward sequentially [cite: 1393, 1394]:
log_4log_6(3 + 4x - x^2) > 0 implies log_6(3 + 4x - x^2) > 1$$\log_{4}\log_{6}(3 + 4x - x^2) > 0 \implies \log_{6}(3 + 4x - x^2) > 1$$ [cite: 1393, 1394]
3 + 4x - x^2 > 6^1 implies x^2 - 4x + 3 < 0$$3 + 4x - x^2 > 6^1 \implies x^2 - 4x + 3 < 0$$ [cite: 1395, 1396]
(x-1)(x-3) < 0 implies x in (1, 3)$$(x-1)(x-3) < 0 \implies x \in (1, 3)$$ [cite: 1397, 1398]
Thus, determine limits [cite: 1399]:
a = 1, quad b = 3 implies b - a = 2$$a = 1, \quad b = 3 \implies b - a = 2$$ [cite: 1399]
### Step 1: Setting up the greatest integer function integration
We need to evaluate int_0^2 [x^2] \, mathrmdx$\int_0^2 [x^2] \, \mathrm{d}x$[cite: 1400]. Identify step boundary switch locations inside range [0, 2]$[0, 2]$ [cite: 1400]:
- For x in [0, 1): [x^2] = 0$x \in [0, 1): [x^2] = 0$
- For x in [1, sqrt2): [x^2] = 1$x \in [1, \sqrt{2}): [x^2] = 1$
- For x in [sqrt2, sqrt3): [x^2] = 2$x \in [\sqrt{2}, \sqrt{3}): [x^2] = 2$
- For x in [sqrt3, 2): [x^2] = 3$x \in [\sqrt{3}, 2): [x^2] = 3$
Set up separate boundary component integrations [cite: 1400]:
int_0^2 [x^2] \, mathrmdx = int_0^1 0 \, mathrmdx + int_1^sqrt2 1 \, mathrmdx + int_sqrt2^sqrt3 2 \, mathrmdx + int_sqrt3^2 3 \, mathrmdx$$\int_0^2 [x^2] \, \mathrm{d}x = \int_0^1 0 \, \mathrm{d}x + \int_1^{\sqrt{2}} 1 \, \mathrm{d}x + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, \mathrm{d}x + \int_{\sqrt{3}}^2 3 \, \mathrm{d}x$$ [cite: 1400]
= 0 + (sqrt2 - 1) + 2(sqrt3 - sqrt2) + 3(2 - sqrt3)$$= 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$$ [cite: 1400]
= sqrt2 - 1 + 2sqrt3 - 2sqrt2 + 6 - 3sqrt3 = 5 - sqrt2 - sqrt3$$= \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 5 - \sqrt{2} - \sqrt{3}$$ [cite: 1400]
### Step 2: Matching coefficients
Compare values with requested answer template shape [cite: 1400]:
5 - sqrt2 - sqrt3 = p - sqrtq - sqrtr$$5 - \sqrt{2} - \sqrt{3} = p - \sqrt{q} - \sqrt{r}$$ [cite: 1400]
p = 5, quad q = 2, quad r = 3$$p = 5, \quad q = 2, \quad r = 3$$ [cite: 1400]
textFinal Sum = p + q + r = 5 + 2 + 3 = 10$$\text{Final Sum} = p + q + r = 5 + 2 + 3 = 10$$ [cite: 1400]
### Pattern Recognition
Integrals over greatest integer configurations change value exactly where the inner expression tracks through integer milestones. Mapping boundaries accurately resolves calculations smoothly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integrals
Keywords:#domain nested log step function definite integral#JEE Main 2025 Morning Q67#Integrals JEE Main 2025#Definite Integral of Greatest Integer Function
For xin(-fracpi2,fracpi2)$x\in(-\frac{\pi}{2},\frac{\pi}{2})$, if y(x)=intfracoperatornamecosec x+sin xoperatornamecosec x sec x+tan x sin^2 xdx$y(x)=\int\frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin^2 x}dx$ and lim_xrightarrow(fracpi2)^-y(x)=0$\lim_{x\rightarrow(\frac{\pi}{2})^-}y(x)=0$ then y(fracpi4)$y(\frac{\pi}{4})$ is equal to
### Related Formula
int frac1u^2 + a^2 du = frac1a tan^-1left(fracuaright) + C$$\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$$
### Core Logic
Simplify the given integrand by converting all trigonometric ratios into sin x$\sin x$ and cos x$\cos x$:
I = fracoperatornamecosec x + sin xoperatornamecosec x sec x + tan x sin^2 x$$I = \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x}$$
Numerator: frac1sin x + sin x = frac1 + sin^2 xsin x$\frac{1}{\sin x} + \sin x = \frac{1 + \sin^2 x}{\sin x}$
Denominator: frac1sin x cos x + fracsin xcos x cdot sin^2 x = frac1 + sin^4 xsin x cos x$\frac{1}{\sin x \cos x} + \frac{\sin x}{\cos x} \cdot \sin^2 x = \frac{1 + \sin^4 x}{\sin x \cos x}$
Divide Numerator by Denominator:
I = fracfrac1 + sin^2 xsin xfrac1 + sin^4 xsin x cos x = frac(1 + sin^2 x)cos x1 + sin^4 x$$I = \frac{\frac{1 + \sin^2 x}{\sin x}}{\frac{1 + \sin^4 x}{\sin x \cos x}} = \frac{(1 + \sin^2 x)\cos x}{1 + \sin^4 x}$$
### Step 1: Apply Substitution
Now rewrite the integral:
y(x) = int frac(1 + sin^2 x)cos x1 + sin^4 x dx$$y(x) = \int \frac{(1 + \sin^2 x)\cos x}{1 + \sin^4 x} dx$$
Let sin x = t$\sin x = t$, then cos x \, dx = dt$\cos x \, dx = dt$. The integral becomes:
y(x) = int frac1 + t^21 + t^4 dt$$y(x) = \int \frac{1 + t^2}{1 + t^4} dt$$
Divide numerator and denominator by t^2$t^2$:
y(x) = int frac1 + frac1t^2t^2 + frac1t^2 dt$$y(x) = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} dt$$
We know that t^2 + frac1t^2 = left(t - frac1tright)^2 + 2$t^2 + \frac{1}{t^2} = \left(t - \frac{1}{t}\right)^2 + 2$. Let u = t - frac1t$u = t - \frac{1}{t}$, then du = left(1 + frac1t^2right) dt$du = \left(1 + \frac{1}{t^2}\right) dt$.
y(x) = int fracduu^2 + (sqrt2)^2 = frac1sqrt2tan^-1left(fracusqrt2right) + C$$y(x) = \int \frac{du}{u^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C$$y(x) = frac1sqrt2tan^-1left(fract - 1/tsqrt2right) + C = frac1sqrt2tan^-1left(fracsin x - operatornamecosec xsqrt2right) + C$$y(x) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t - 1/t}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \operatorname{cosec} x}{\sqrt{2}}\right) + C$$
### Step 2: Solve for Constant C
We are given that lim_xrightarrow(pi/2)^- y(x) = 0$\lim_{x\rightarrow(\pi/2)^-} y(x) = 0$.
As x to fracpi2$x \to \frac{\pi}{2}$, sin x to 1$\sin x \to 1$ and operatornamecosec x to 1$\operatorname{cosec} x \to 1$.
Thus, sin x - operatornamecosec x to 0$\sin x - \operatorname{cosec} x \to 0$.
0 = frac1sqrt2tan^-1left(frac1 - 1sqrt2right) + C$$0 = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{1 - 1}{\sqrt{2}}\right) + C$$0 = frac1sqrt2tan^-1(0) + C Rightarrow C = 0$$0 = \frac{1}{\sqrt{2}}\tan^{-1}(0) + C \Rightarrow C = 0$$
### Step 3: Find y(pi/4)
Substitute x = fracpi4$x = \frac{\pi}{4}$ into the exact solution y(x) = frac1sqrt2tan^-1left(fracsin x - operatornamecosec xsqrt2right)$y(x) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \operatorname{cosec} x}{\sqrt{2}}\right)$:
At x = fracpi4$x = \frac{\pi}{4}$, sinleft(fracpi4right) = frac1sqrt2$\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and operatornamecosecleft(fracpi4right) = sqrt2$\operatorname{cosec}\left(\frac{\pi}{4}\right) = \sqrt{2}$.
yleft(fracpi4right) = frac1sqrt2tan^-1left(fracfrac1sqrt2 - sqrt2sqrt2right)$$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\frac{1}{\sqrt{2}} - \sqrt{2}}{\sqrt{2}}\right)$$= frac1sqrt2tan^-1left(fracfrac1 - 2sqrt2sqrt2right) = frac1sqrt2tan^-1left(frac-1/sqrt2sqrt2right)$$= \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\frac{1 - 2}{\sqrt{2}}}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{-1/\sqrt{2}}{\sqrt{2}}\right)$$= frac1sqrt2tan^-1left(-frac12right)$$= \frac{1}{\sqrt{2}}\tan^{-1}\left(-\frac{1}{2}\right)$$
### Pattern Recognition
A high-degree polynomial of sin x$\sin x$ or cos x$\cos x$ in the denominator matched with their derivative elements on top is the hallmark of the t+1/t$t+1/t$ or t-1/t$t-1/t$ structural substitution form. Dividing out the middle power (t^2$t^2$) aligns the differential instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integral Calculus
Q18jee_main_2024_29_jan_morningProperties of Definite Integrals
If the value of the integralint_-fracpi2^fracpi2left(fracx^2cos x1+pi^x+frac1+sin^2x1+e^sin x^2023right)dx=fracpi4(pi+a)-2$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2\cos x}{1+\pi^x}+\frac{1+\sin^2x}{1+e^{\sin x^{2023}}}\right)dx=\frac{\pi}{4}(\pi+a)-2$, then the value of a$a$ is
A.3$3$
B.-frac32$-\frac{3}{2}$
C.2$2$
D.frac32$\frac{3}{2}$
Solution
### Related Formula
textKing's Rule Variant: int_-a^a f(x) dx = int_0^a (f(x) + f(-x)) dx$$\text{King's Rule Variant: } \int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$$
### Core Logic
Let the integral be I$I$.
We apply the property int_-a^a f(x) dx = int_0^a (f(x)+f(-x)) dx$\int_{-a}^a f(x) dx = \int_0^a (f(x)+f(-x)) dx$ to each fractional term independently.
For the first term f_1(x) = fracx^2cos x1+pi^x$f_1(x) = \frac{x^2\cos x}{1+\pi^x}$:
f_1(x) + f_1(-x) = fracx^2cos x1+pi^x + frac(-x)^2cos(-x)1+pi^-x = x^2cos x left( frac11+pi^x + fracpi^xpi^x+1 right) = x^2cos x$$f_1(x) + f_1(-x) = \frac{x^2\cos x}{1+\pi^x} + \frac{(-x)^2\cos(-x)}{1+\pi^{-x}} = x^2\cos x \left( \frac{1}{1+\pi^x} + \frac{\pi^x}{\pi^x+1} \right) = x^2\cos x$$
For the second term f_2(x) = frac1+sin^2x1+e^sin x^2023$f_2(x) = \frac{1+\sin^2x}{1+e^{\sin x^{2023}}}$:
Observe the exponent: sin(-x)^2023 = sin(-x^2023) = -sin x^2023$\sin(-x)^{2023} = \sin(-x^{2023}) = -\sin x^{2023}$.
f_2(x) + f_2(-x) = frac1+sin^2x1+e^p + frac1+sin^2(-x)1+e^-p = (1+sin^2x) left( frac11+e^p + frace^pe^p+1 right) = 1+sin^2x$$f_2(x) + f_2(-x) = \frac{1+\sin^2x}{1+e^{p}} + \frac{1+\sin^2(-x)}{1+e^{-p}} = (1+\sin^2x) \left( \frac{1}{1+e^p} + \frac{e^p}{e^p+1} \right) = 1+\sin^2x$$
Thus, the integral radically simplifies to:
I = int_0^pi/2 (x^2cos x + 1 + sin^2 x) dx$$I = \int_0^{\pi/2} (x^2\cos x + 1 + \sin^2 x) dx$$
### Step 1: Execute Integration by Parts
Evaluate int_0^pi/2 x^2cos x \, dx$\int_0^{\pi/2} x^2\cos x \, dx$ using integration by parts:
u = x^2 Rightarrow du = 2x \, dx$$u = x^2 \Rightarrow du = 2x \, dx$$dv = cos x \, dx Rightarrow v = sin x$$dv = \cos x \, dx \Rightarrow v = \sin x$$int x^2cos x dx = x^2sin x - int 2xsin x dx$$\int x^2\cos x dx = x^2\sin x - \int 2x\sin x dx$$
Applying parts again to int 2xsin x dx$\int 2x\sin x dx$:
int 2xsin x dx = 2x(-cos x) - int 2(-cos x)dx = -2xcos x + 2sin x$$\int 2x\sin x dx = 2x(-\cos x) - \int 2(-\cos x)dx = -2x\cos x + 2\sin x$$
Substituting back:
left[ x^2sin x + 2xcos x - 2sin x right]_0^pi/2$$\left[ x^2\sin x + 2x\cos x - 2\sin x \right]_0^{\pi/2}$$
Evaluate at upper limit pi/2$\pi/2$:
(pi/2)^2(1) + 0 - 2(1) = fracpi^24 - 2$$(\pi/2)^2(1) + 0 - 2(1) = \frac{\pi^2}{4} - 2$$
Evaluate at lower limit 0:
0 + 0 - 0 = 0$0 + 0 - 0 = 0$
So, int_0^pi/2 x^2cos x \, dx = fracpi^24 - 2$\int_0^{\pi/2} x^2\cos x \, dx = \frac{\pi^2}{4} - 2$.
### Step 2: Execute Standard Integrals
Evaluate int_0^pi/2 (1 + sin^2 x) dx$\int_0^{\pi/2} (1 + \sin^2 x) dx$:
= int_0^pi/2 dx + int_0^pi/2 sin^2 x dx$$= \int_0^{\pi/2} dx + \int_0^{\pi/2} \sin^2 x dx$$= fracpi2 + left[ frac12 cdot fracpi2 right] = fracpi2 + fracpi4 = frac3pi4$$= \frac{\pi}{2} + \left[ \frac{1}{2} \cdot \frac{\pi}{2} \right] = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$$
### Step 3: Combine and Compare
Sum the components to get the total integral I$I$:
I = left(fracpi^24 - 2right) + frac3pi4 = fracpi^24 + frac3pi4 - 2$$I = \left(\frac{\pi^2}{4} - 2\right) + \frac{3\pi}{4} = \frac{\pi^2}{4} + \frac{3\pi}{4} - 2$$
Factor out fracpi4$\frac{\pi}{4}$:
I = fracpi4(pi + 3) - 2$$I = \frac{\pi}{4}(\pi + 3) - 2$$
The problem states I = fracpi4(pi + a) - 2$I = \frac{\pi}{4}(\pi + a) - 2$. Comparing the two expressions gives:
a = 3$a = 3$
### Pattern Recognition
Frightening denominators like 1+a^f(x)$1+a^{f(x)}$ in an integral from -L$-L$ to L$L$ where f(x)$f(x)$ is an odd function are almost exclusively designed to cancel out and leave 1 via the f(x)+f(-x)$f(x)+f(-x)$ expansion. Strip the ugly denominators immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integral Calculus
Let y = f(x)$y = f(x)$ be a thrice differentiable function in (-5, 5)$(-5, 5)$ . Let the tangents to the curve y = f(x)$y = f(x)$ at (1, f(1))$(1, f(1))$ and (3, f(3))$(3, f(3))$ make angles fracpi6$\frac{\pi}{6}$ and fracpi4$\frac{\pi}{4}$ , respectively with positive x-axis. If
27int_1^3left(left(f'(t)right)^2 + 1right)f''(t)dt = alpha + beta sqrt3 quad textwhere alpha, beta text are integers, then the value of alpha +beta text equals$$27\int_{1}^{3}\left(\left(f'(t)\right)^{2} + 1\right)f''(t)dt = \alpha + \beta \sqrt{3} \quad \text{where } \alpha, \beta \text{ are integers, then the value of } \alpha +\beta \text{ equals}$$
A.-14$-14$
B.26$26$
C.-16$-16$
D.36$36$
Solution
### Related Formula
textSlope of tangent at x=a text is f'(a) = tan theta$$\text{Slope of tangent at } x=a \text{ is } f'(a) = \tan \theta$$int u^n du = fracu^n+1n+1$$\int u^n du = \frac{u^{n+1}}{n+1}$$
### Core Logic
Given y=f(x)$y=f(x)$, the derivatives at the given points correspond to the slope of tangents:
left. fracdydx right|_x=1 = f'(1) = tanleft(fracpi6right) = frac1sqrt3$$\left. \frac{dy}{dx} \right|_{x=1} = f'(1) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$$left. fracdydx right|_x=3 = f'(3) = tanleft(fracpi4right) = 1$$\left. \frac{dy}{dx} \right|_{x=3} = f'(3) = \tan\left(\frac{\pi}{4}\right) = 1$$
We must evaluate the integral:
I = int_1^3 left( (f'(t))^2 + 1 right) f''(t) dt$$I = \int_{1}^{3} \left( (f'(t))^2 + 1 \right) f''(t) dt$$
### Step 1: Integration by Substitution
Let z = f'(t)$z = f'(t)$. Then dz = f''(t)dt$dz = f''(t)dt$.
The limits of integration change accordingly:
When t = 1$t = 1$, z = f'(1) = frac1sqrt3$z = f'(1) = \frac{1}{\sqrt{3}}$
When t = 3$t = 3$, z = f'(3) = 1$z = f'(3) = 1$
The integral becomes:
I = int_1/sqrt3^1 (z^2 + 1) dz$$I = \int_{1/\sqrt{3}}^{1} (z^2 + 1) dz$$I = left[ fracz^33 + z right]_1/sqrt3^1$$I = \left[ \frac{z^3}{3} + z \right]_{1/\sqrt{3}}^{1}$$
### Step 2: Evaluating the Definite Integral
Plug in the limits:
I = left( frac1^33 + 1 right) - left( frac13 cdot frac13sqrt3 + frac1sqrt3 right)$$I = \left( \frac{1^3}{3} + 1 \right) - \left( \frac{1}{3} \cdot \frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}} \right)$$I = frac43 - left( frac19sqrt3 + frac99sqrt3 right)$$I = \frac{4}{3} - \left( \frac{1}{9\sqrt{3}} + \frac{9}{9\sqrt{3}} \right)$$I = frac43 - frac109sqrt3 = frac43 - frac10sqrt327$$I = \frac{4}{3} - \frac{10}{9\sqrt{3}} = \frac{4}{3} - \frac{10\sqrt{3}}{27}$$
### Step 3: Finding alpha and beta
We are given that 27 cdot I = alpha + betasqrt3$27 \cdot I = \alpha + \beta\sqrt{3}$.
27 left( frac43 - frac10sqrt327 right) = 36 - 10sqrt3$$27 \left( \frac{4}{3} - \frac{10\sqrt{3}}{27} \right) = 36 - 10\sqrt{3}$$
Comparing with alpha + betasqrt3$\alpha + \beta\sqrt{3}$, we get:
alpha = 36$\alpha = 36$beta = -10$\beta = -10$
Therefore, alpha + beta = 36 - 10 = 26$\alpha + \beta = 36 - 10 = 26$.
### Pattern Recognition
Integrals containing a function derivative alongside its second derivative are classic substitution traps. Set u = f'(x)$u = f'(x)$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Integral Calculus
Class 12 Maths: Application of Derivatives
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