Let the circle be,
(x – a)2 + y2 = r2
Solving it with parabola
y2 = 4 – x we get
(x – a)2 + 4 – x = r2
x2 – x(2a + 1) + (a2 + 4 – r2) = 0 …(1)
D = 0
\(⇒\) 4r2 + 4a – 15 = 0
Clearly a ≥ r
So 4r2 + 4r – 15 ≤ 0
\(⇒\) rmax =\(\frac{3}{2}\) = a
Radius of circle C is \(\frac{3}{2}\)
Let the circle be,
(x – a)2 + y2 = r2
Solving it with parabola
y2 = 4 – x we get
(x – a)2 + 4 – x = r2
x2 – x(2a + 1) + (a2 + 4 – r2) = 0 …(1)
D = 0
\(⇒\) 4r2 + 4a – 15 = 0
Clearly a ≥ r
So 4r2 + 4r – 15 ≤ 0
\(⇒\) rmax = \(\frac{3}{2}\) = a
Radius of circle C is \(\frac{3}{2}\)
From (1) x2 – 4x + 4 = 0
\(⇒ x = 2 = α\)
Let the area of the region \( \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} \) be \( A \). Then \( 6A \) is equal to:
Find \( P(0<X<5) \).
A positive, singly ionized atom of mass number $ A_M $ is accelerated from rest by the voltage $ 192 \, \text{V} $. Thereafter, it enters a rectangular region of width $ w $ with magnetic field $ \vec{B}_0 = 0.1\hat{k} \, \text{T} $. The ion finally hits a detector at the distance $ x $ below its starting trajectory. Which of the following option(s) is(are) correct?
$ \text{(Given: Mass of neutron/proton = } \frac{5}{3} \times 10^{-27} \, \text{kg, charge of the electron = } 1.6 \times 10^{-19} \, \text{C).} $