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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.

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