Calculus

Application of Derivatives

Hard

applied

The points on the ellipse $4x^2 + y^2 = 4$ that are farthest from the point (1,0) are;

$\left(\frac{1}{3}, \pm\frac{4\sqrt{2}}{3}\right)$

$\left(\frac{-1}{3}, \pm\frac{4}{3}\right)$

$\left(\frac{-1}{3}, \pm\frac{4\sqrt{2}}{3}\right)$

$\left(\frac{1}{3}, \pm\frac{\sqrt{2}}{3}\right)$

✅ Correct Option: 3

On the ellipse $y^2=4(1-x^2)$ , so the squared distance is $D^2=(x-1)^2+4-4x^2=-3x^2-2x+5$ . Maximising this quadratic gives $x=-\frac13$ , and then $y^2=4\left(1-\frac19\right)=\frac{32}{9}$ , so $y=\pm\frac{4\sqrt2}{3}$ .

2026: 15 May Shift 1

Maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 30$ is

2025: 16 May Shift 1

If the interval in which f(x) = $\frac{x}{4}$ + $\frac{4}{x}$ , x ≠ 0 is strictly increasing is (-∞, a) ∪ (b, ∞), then

2025: 14 May Shift 2

The function $f(x) = kx^3 + 6kx^2 + 18x + 17$ is increasing on $\mathbb{R}$ (set of real numbers) if:

2025: 27 May Shift 1

A balloon which always remains spherical, has a variable diameter $\frac{3}{2}(5x+7)$ . Then the rate of change of its volume with respect to x is

2025: 19 May Shift 1

For the function $f(x) = x^{1/x}$ , $x > 0$ , which of the following are correct?

(A) $x = 0$ is the only point where extremum may occur.

(B) The given function is maximum at $x = e$ .

(D) The maximum value of the function $f(x)$ is $e^{1/e}$ .

Choose the correct answer from the options given below:

2025: 29 May Shift 2

Let $f(x) = 4x^3 - 18x^2 + 27x - 5$ , $x \in R$ . Then which of the following statements are TRUE?

(A) $f''(x) = 24x - 36$

(B) f has local maxima at $x = \frac{3}{2}$ but no minima

(D) f has both maxima and minima

Choose the correct answer from the options given below:

The points on the ellipse 4x2+y2=44x^2 + y^2 = 44x2+y2=4 that are farthest from the point (1,0) are;

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The points on the ellipse $4x^2 + y^2 = 4$ that are farthest from the point (1,0) are;