Top 50 Most Repeated APPLICATION OF DERIVATIVES PYQs | JEE MAINS
A curated collection of the most important questions from APPLICATION OF DERIVATIVES, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from APPLICATION OF DERIVATIVES, fully solved with step-by-step concepts to prepare for JEE MAINS.
At $x=1$, $y = 1^3 - 3(1) + 2 = 0$. Derivative $\frac{dy}{dx} = 3x^2 - 3$, so slope at $x=1$ is $3(1)^2 - 3 = 0$. Tangent is horizontal: $y = 0$....
Read Full Step-by-Step Solution →$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$; $f' < 0$ when
$dy/dx = 3x^2 - 3$. At $x=1$, slope = $3(1)^2 - 3 = 0$...
Read Full Step-by-Step Solution →Apply L'Hopital: derivatives yield $\frac{3/(1+3x)}{2\cos 2x} \to \frac{3}{2}$ as $x\to0$....
Read Full Step-by-Step Solution →Max area occurs when rectangle is square. Diagonal = 10 cm → side = 5√2 → area = 50....
Read Full Step-by-Step Solution →Max area when rectangle is square: side = 10 m → area = 100 m²....
Read Full Step-by-Step Solution →$dy = \frac{1}{2\sqrt{x}}dx$ gives the small change....
Read Full Step-by-Step Solution →f'(x) = 3x² - 12x + 9 = 0 → x = 1, 3; f''(1) < 0 → local max at x = 1....
Read Full Step-by-Step Solution →Step 1: The function $\boxed{f(x)}=\boxed{\frac{1}{x^2-4}}$ has a vertical asymptote at $x=2$.,Step 2: This is because the denominator of the function...
Read Full Step-by-Step Solution →Foundational check for Application of Derivatives in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →At $x=1$, $y = 1^3 - 3(1) + 2 = 0$. Derivative $\frac{dy}{dx} = 3x^2 - 3$, so slope at $x=1$ is $3(1)^2 - 3 = 0$. Tangent is horizontal: $y = 0$....
Read Full Step-by-Step Solution →$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$; $f' < 0$ when
$dy/dx = 3x^2 - 3$. At $x=1$, slope = $3(1)^2 - 3 = 0$...
Read Full Step-by-Step Solution →Apply L'Hopital: derivatives yield $\frac{3/(1+3x)}{2\cos 2x} \to \frac{3}{2}$ as $x\to0$....
Read Full Step-by-Step Solution →Max area occurs when rectangle is square. Diagonal = 10 cm → side = 5√2 → area = 50....
Read Full Step-by-Step Solution →Max area when rectangle is square: side = 10 m → area = 100 m²....
Read Full Step-by-Step Solution →$dy = \frac{1}{2\sqrt{x}}dx$ gives the small change....
Read Full Step-by-Step Solution →f'(x) = 3x² - 12x + 9 = 0 → x = 1, 3; f''(1) < 0 → local max at x = 1....
Read Full Step-by-Step Solution →Step 1: The function $\boxed{f(x)}=\boxed{\frac{1}{x^2-4}}$ has a vertical asymptote at $x=2$.,Step 2: This is because the denominator of the function...
Read Full Step-by-Step Solution →Foundational check for Application of Derivatives in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →At $x=1$, $y = 1^3 - 3(1) + 2 = 0$. Derivative $\frac{dy}{dx} = 3x^2 - 3$, so slope at $x=1$ is $3(1)^2 - 3 = 0$. Tangent is horizontal: $y = 0$....
Read Full Step-by-Step Solution →$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$; $f' < 0$ when
$dy/dx = 3x^2 - 3$. At $x=1$, slope = $3(1)^2 - 3 = 0$...
Read Full Step-by-Step Solution →Apply L'Hopital: derivatives yield $\frac{3/(1+3x)}{2\cos 2x} \to \frac{3}{2}$ as $x\to0$....
Read Full Step-by-Step Solution →Max area occurs when rectangle is square. Diagonal = 10 cm → side = 5√2 → area = 50....
Read Full Step-by-Step Solution →Max area when rectangle is square: side = 10 m → area = 100 m²....
Read Full Step-by-Step Solution →$dy = \frac{1}{2\sqrt{x}}dx$ gives the small change....
Read Full Step-by-Step Solution →f'(x) = 3x² - 12x + 9 = 0 → x = 1, 3; f''(1) < 0 → local max at x = 1....
Read Full Step-by-Step Solution →Step 1: The function $\boxed{f(x)}=\boxed{\frac{1}{x^2-4}}$ has a vertical asymptote at $x=2$.,Step 2: This is because the denominator of the function...
Read Full Step-by-Step Solution →Foundational check for Application of Derivatives in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →At $x=1$, $y = 1^3 - 3(1) + 2 = 0$. Derivative $\frac{dy}{dx} = 3x^2 - 3$, so slope at $x=1$ is $3(1)^2 - 3 = 0$. Tangent is horizontal: $y = 0$....
Read Full Step-by-Step Solution →$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$; $f' < 0$ when
$dy/dx = 3x^2 - 3$. At $x=1$, slope = $3(1)^2 - 3 = 0$...
Read Full Step-by-Step Solution →Apply L'Hopital: derivatives yield $\frac{3/(1+3x)}{2\cos 2x} \to \frac{3}{2}$ as $x\to0$....
Read Full Step-by-Step Solution →Max area occurs when rectangle is square. Diagonal = 10 cm → side = 5√2 → area = 50....
Read Full Step-by-Step Solution →Max area when rectangle is square: side = 10 m → area = 100 m²....
Read Full Step-by-Step Solution →$dy = \frac{1}{2\sqrt{x}}dx$ gives the small change....
Read Full Step-by-Step Solution →f'(x) = 3x² - 12x + 9 = 0 → x = 1, 3; f''(1) < 0 → local max at x = 1....
Read Full Step-by-Step Solution →Step 1: The function $\boxed{f(x)}=\boxed{\frac{1}{x^2-4}}$ has a vertical asymptote at $x=2$.,Step 2: This is because the denominator of the function...
Read Full Step-by-Step Solution →Foundational check for Application of Derivatives in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →At $x=1$, $y = 1^3 - 3(1) + 2 = 0$. Derivative $\frac{dy}{dx} = 3x^2 - 3$, so slope at $x=1$ is $3(1)^2 - 3 = 0$. Tangent is horizontal: $y = 0$....
Read Full Step-by-Step Solution →$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$; $f' < 0$ when
$dy/dx = 3x^2 - 3$. At $x=1$, slope = $3(1)^2 - 3 = 0$...
Read Full Step-by-Step Solution →Apply L'Hopital: derivatives yield $\frac{3/(1+3x)}{2\cos 2x} \to \frac{3}{2}$ as $x\to0$....
Read Full Step-by-Step Solution →Max area occurs when rectangle is square. Diagonal = 10 cm → side = 5√2 → area = 50....
Read Full Step-by-Step Solution →Max area when rectangle is square: side = 10 m → area = 100 m²....
Read Full Step-by-Step Solution →$dy = \frac{1}{2\sqrt{x}}dx$ gives the small change....
Read Full Step-by-Step Solution →f'(x) = 3x² - 12x + 9 = 0 → x = 1, 3; f''(1) < 0 → local max at x = 1....
Read Full Step-by-Step Solution →Step 1: The function $\boxed{f(x)}=\boxed{\frac{1}{x^2-4}}$ has a vertical asymptote at $x=2$.,Step 2: This is because the denominator of the function...
Read Full Step-by-Step Solution →Foundational check for Application of Derivatives in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →