1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point
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= 2π ∫ π<br />
2 / 3<br />
2 t<br />
sec dt<br />
2<br />
t<br />
1+<br />
3tan<br />
2<br />
π/<br />
3<br />
2<br />
3 2du<br />
= 2π∫ =<br />
1/<br />
3<br />
2<br />
1+<br />
3u<br />
3<br />
4π . { } 3<br />
−1<br />
3<br />
3 tan u<br />
4π –1 –1 4π –1<br />
= (tan 3 – tan 1) = tan<br />
3<br />
3<br />
3<br />
⎛ 1 ⎞<br />
⎜ ⎟<br />
⎝ 2 ⎠<br />
π/<br />
3 π + 4x<br />
4π –1 ⎛ 1 ⎞<br />
∴ ∫ dx = tan ⎜ ⎟ .<br />
−π<br />
/ 3 ⎛ π ⎞<br />
2 − cos⎜|<br />
x | +<br />
3 ⎝ 2 ⎠<br />
⎟<br />
⎝ 3 ⎠<br />
14. An unbiased die, with faces numbered 1, 2, 3, 4, 5,<br />
6, is thrown n times and the list on n numbers<br />
showing up is noted. What is the probability that<br />
among the numbers 1, 2, 3, 4, 5, 6 only three<br />
numbers appear in this list ? [IIT-2001]<br />
Sol. Let us define at onto function F from A : [r1, r2 ...<br />
rn] to B : [1, 2, 3] where r1r2 .... rn are the readings<br />
of n throws and 1, 2, 3 are the numbers that appear<br />
in the n throws.<br />
Number of such functions,<br />
M = N – [n(1) – n(2) + n(3)]<br />
where N = total number of functions and<br />
n(t) = number of function having exactly t elements<br />
in the range.<br />
Now, N = 3 n , n(1) = 3.2 n , n(2) = 3, n(3) = 0<br />
⇒ M = 3 n – 3.2 n + 3<br />
Hence the total number of favourable cases<br />
= (3 n – 3.2 n + 3). 6 C3<br />
⇒ Required probability =<br />
Nuclear Physics<br />
1/<br />
n n 6<br />
( 3 − 3.<br />
2 + 3)<br />
× C3<br />
n<br />
6<br />
3<br />
Physics Facts<br />
15. Show that the value of tanx/tan3x, wherever defined<br />
never lies between 1/3 and 3. [IIT-1992]<br />
tan x<br />
Sol. y = =<br />
tan 3x<br />
x – tan<br />
<strong>Xtra</strong><strong>Edge</strong> for IIT-JEE 13 FEBRUARY <strong>2012</strong><br />
=<br />
=<br />
3tan<br />
tan x(<br />
1–<br />
3tan<br />
3tan<br />
1–<br />
3<br />
x – tan<br />
tan<br />
2<br />
2<br />
x<br />
3 – tan x<br />
⇒ x ≠ 0<br />
⇒ tan x ≠ 0<br />
0 ∞<br />
tan x<br />
1–<br />
3tan<br />
2<br />
3<br />
x)<br />
+ – +<br />
1/3 3<br />
Let tan x = t<br />
⇒ y =<br />
3 – t<br />
⇒ 3y – t 2 y = 1 – 3t 2<br />
⇒ 3y – 1 = t 2 y – 3t 2<br />
2<br />
1–<br />
3t<br />
⇒ 3y – 1 = t 2 (y – 3)<br />
⇒<br />
2<br />
3y<br />
– 1<br />
= t<br />
y – 3<br />
2<br />
x<br />
2<br />
3<br />
x<br />
x<br />
[Q tan 3x ≠ 0 ⇒ 3x ≠ 0]<br />
3y<br />
– 1<br />
⇒ ≥ 0, t<br />
y – 3<br />
2 ≥ 0 ∀ t ∈ R<br />
⇒ y ∈ (– ∞, 1/3) ∪ (3, ∞)<br />
Therefore, y is not defined in between (1/3, 3).<br />
<strong>1.</strong> Alpha particles are the same as helium nuclei and have the symbol .<br />
<strong>1.</strong> The atomic number is equal to the number of protons (2 for alpha)<br />
2. Deuterium ( ) is an isotope of hydrogen ( )<br />
3. The number of nucleons is equal to protons + neutrons (4 for alpha)<br />
4. Only charged particles can be accelerated in a particle accelerator such as a cyclotron or Van Der Graaf<br />
generator.<br />
5. Natural radiation is alpha ( ), beta ( ) and gamma (high energy x-rays)<br />
6. A loss of a beta particle results in an increase in atomic number.<br />
7. All nuclei weigh less than their parts. This mass defect is converted into binding energy. (E=mc 2 )<br />
8. Isotopes have different neutron numbers and atomic masses but the same number of protons (atomic<br />
numbers).<br />
9. Geiger counters, photographic plates, cloud and bubble chambers are all used to detect or observe radiation.