1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point 1. Xtra Edge February 2012 - Career Point
Sol. F is mid-point of BC i.e., F ≡ and AE ⊥ DE (given) A(i + j + k) λ B(i) E 1 F(2i) D C(3i) ^ 3 ^ i+ i 2 = 2 ^ i Let E divides AF in λ : 1. The position vector of E is given by ^ ^ ^ ^ 2λ i+ 1( i+ j+ k) ⎛ 2λ + 1⎞ ^ 1 ^ = ⎜ ⎟ i + j + λ + 1 ⎝ λ + 1 ⎠ λ + 1 Now, volume of the tetrahedron 1 = (area of the base) (height) 3 ⇒ 2 2 3 1 = (area of the ∆ABC) (DE) 3 1 → → But area of the ∆ABC = | ( BC × BA) | 2 1 ^ ^ ^ 1 ^ k λ + = | 2 i × ( j+ k) | = | i × j+ i× k) | 2 ^ ^ = | k – j) | = 2 2 2 1 Therefore, = ( 2) (DE) 3 3 ⇒ DE = 2 Since ∆ADE is a right angle triangle, AD 2 = AE 2 + DE 2 ⇒ (4) 2 = AE 2 + (2) 2 ⇒ AE 2 = 12 But → 2λ + 1 ^ 1 ^ 1 ^ ^ ^ ^ AE = i + j + k – ( i + j+ k) λ + 1 λ + 1 λ + 1 ^ = 1 i λ ^ – λ + 1 j λ ^ – λ + 1 k λ λ + ⇒ 2 | | → 1 AE = ( λ + 1 2 2 ) 3λ Therefore, 12 = 2 ( λ + 1) ⇒ 4(λ + 1) 2 = λ 2 ⇒ 4λ 2 + 4 + 8λ = λ 2 ⇒ 3λ 2 + 8λ + 4 = 0 ⇒ 3λ 2 + 6λ + 2λ + 4 = 0 ^ ^ [λ 2 + λ 2 + λ 2 ] = ^ ^ 1 2 3λ ( λ + 1) 2 ⇒ 3λ(λ + 2) + 2(λ + 2) = 0 ⇒ (3λ + 2) (λ + 2) = 0 ⇒ λ = – 2/3, λ = – 2 Therefore, when λ = – 2/3, position vector of E is given by ⎛ 2λ + 1⎞ ^ 1 ^ 1 ^ ⎜ ⎟ i + j + k ⎝ λ + 1 ⎠ λ + 1 λ + 1 2.(– 2 / 3) + 1 ^ 1 ^ 1 ^ = i + j + k – 2 / 3 + 1 – 2 / 3 + 1 – 2 / 3 + 1 – 4 / 3 + 1 ^ 1 ^ 1 ^ = i + j + k – 2 + 3 – 2 + 3 – 2 + 3 3 – 4 + 3 3 3 = 3 1/ 3 ^ 1 ^ 1 ^ i + j + k 1/ 3 1/ 3 = – ^ i + XtraEdge for IIT-JEE 12 FEBRUARY 2012 ^ 3 j + ^ 3k and when λ = – 2 Position vector of E is given by, 2× (– 2) + 1 ^ 1 ^ 1 ^ – 4 + 1 ^ i + j + k = i – – 2 + 1 – 2 + 1 – 2 + 1 – 1 ^ j – ^ k ^ = 3i – ^ j – ^ k Therefore, – ^ ^ ^ i + 3 j + 3k and + ^ 3i – ^ j – ^ k are the position vector of E. 13. Evaluate ∫ π Sol. Let, −π / 3 / 3 3 π + 4x dx [IIT-2004] ⎛ π ⎞ 2 − cos⎜| x | + ⎟ ⎝ 3 ⎠ π / 3 πdx π / 3 x dx I = ∫ + 4 −π / 3 ⎛ π ⎞ ∫− π/ 3 ⎛ π ⎞ 2 − cos⎜| x | + ⎟ 2 − cos⎜| x | + ⎟ ⎝ 3 ⎠ ⎝ 3 ⎠ a ⎡ 0, f ( −x) = − f ( x) ⎤ Using f ( x) dx= ⎢ a ⎥ ∫− a ⎢2 − = ⎥ ⎣ ∫ f ( x) dx, f ( x) f ( x) 0 ⎦ ∴ I = 2 ∫ π 0 / 3 πdx + 0 ⎛ π ⎞ 2 − cos⎜| x | + ⎟ ⎝ 3 ⎠ ⎧ ⎪ ⎨as ⎪ ⎪⎩ ∫ π/ 3 −π / 3 π/ 3 dx I = 2π∫ 0 2 − cos( x + π / 3) 2π = 2π∫ dt 2 t / 3 π / 3 − cos ⎫ 3 x dx ⎪ is odd ⎬ ⎛ π ⎞ 2 − cos⎜| x | + ⎟ ⎪ ⎝ 3 ⎠ ⎪⎭ π , where x + = t 3 3
= 2π ∫ π 2 / 3 2 t sec dt 2 t 1+ 3tan 2 π/ 3 2 3 2du = 2π∫ = 1/ 3 2 1+ 3u 3 4π . { } 3 −1 3 3 tan u 4π –1 –1 4π –1 = (tan 3 – tan 1) = tan 3 3 3 ⎛ 1 ⎞ ⎜ ⎟ ⎝ 2 ⎠ π/ 3 π + 4x 4π –1 ⎛ 1 ⎞ ∴ ∫ dx = tan ⎜ ⎟ . −π / 3 ⎛ π ⎞ 2 − cos⎜| x | + 3 ⎝ 2 ⎠ ⎟ ⎝ 3 ⎠ 14. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list on n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in this list ? [IIT-2001] Sol. Let us define at onto function F from A : [r1, r2 ... rn] to B : [1, 2, 3] where r1r2 .... rn are the readings of n throws and 1, 2, 3 are the numbers that appear in the n throws. Number of such functions, M = N – [n(1) – n(2) + n(3)] where N = total number of functions and n(t) = number of function having exactly t elements in the range. Now, N = 3 n , n(1) = 3.2 n , n(2) = 3, n(3) = 0 ⇒ M = 3 n – 3.2 n + 3 Hence the total number of favourable cases = (3 n – 3.2 n + 3). 6 C3 ⇒ Required probability = Nuclear Physics 1/ n n 6 ( 3 − 3. 2 + 3) × C3 n 6 3 Physics Facts 15. Show that the value of tanx/tan3x, wherever defined never lies between 1/3 and 3. [IIT-1992] tan x Sol. y = = tan 3x x – tan XtraEdge for IIT-JEE 13 FEBRUARY 2012 = = 3tan tan x( 1– 3tan 3tan 1– 3 x – tan tan 2 2 x 3 – tan x ⇒ x ≠ 0 ⇒ tan x ≠ 0 0 ∞ tan x 1– 3tan 2 3 x) + – + 1/3 3 Let tan x = t ⇒ y = 3 – t ⇒ 3y – t 2 y = 1 – 3t 2 ⇒ 3y – 1 = t 2 y – 3t 2 2 1– 3t ⇒ 3y – 1 = t 2 (y – 3) ⇒ 2 3y – 1 = t y – 3 2 x 2 3 x x [Q tan 3x ≠ 0 ⇒ 3x ≠ 0] 3y – 1 ⇒ ≥ 0, t y – 3 2 ≥ 0 ∀ t ∈ R ⇒ y ∈ (– ∞, 1/3) ∪ (3, ∞) Therefore, y is not defined in between (1/3, 3). 1. Alpha particles are the same as helium nuclei and have the symbol . 1. The atomic number is equal to the number of protons (2 for alpha) 2. Deuterium ( ) is an isotope of hydrogen ( ) 3. The number of nucleons is equal to protons + neutrons (4 for alpha) 4. Only charged particles can be accelerated in a particle accelerator such as a cyclotron or Van Der Graaf generator. 5. Natural radiation is alpha ( ), beta ( ) and gamma (high energy x-rays) 6. A loss of a beta particle results in an increase in atomic number. 7. All nuclei weigh less than their parts. This mass defect is converted into binding energy. (E=mc 2 ) 8. Isotopes have different neutron numbers and atomic masses but the same number of protons (atomic numbers). 9. Geiger counters, photographic plates, cloud and bubble chambers are all used to detect or observe radiation.
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Sol. F is mid-point of BC i.e., F ≡<br />
and AE ⊥ DE (given)<br />
A(i + j + k)<br />
λ<br />
B(i)<br />
E<br />
1<br />
F(2i)<br />
D<br />
C(3i)<br />
^<br />
3 ^<br />
i+ i<br />
2<br />
= 2 ^ i<br />
Let E divides AF in λ : <strong>1.</strong> The position vector of E<br />
is given by<br />
^<br />
^<br />
^<br />
^<br />
2λ<br />
i+<br />
1(<br />
i+<br />
j+<br />
k)<br />
⎛ 2λ<br />
+ 1⎞<br />
^ 1 ^<br />
= ⎜ ⎟ i + j +<br />
λ + 1 ⎝ λ + 1 ⎠ λ + 1<br />
Now, volume of the tetrahedron<br />
1<br />
= (area of the base) (height)<br />
3<br />
⇒<br />
2 2<br />
3<br />
1<br />
= (area of the ∆ABC) (DE)<br />
3<br />
1 → →<br />
But area of the ∆ABC = | ( BC × BA)<br />
|<br />
2<br />
1 ^ ^ ^<br />
1 ^<br />
k<br />
λ +<br />
= | 2 i × ( j+<br />
k)<br />
| = | i × j+<br />
i×<br />
k)<br />
|<br />
2<br />
^<br />
^<br />
= | k – j)<br />
| = 2<br />
2 2 1<br />
Therefore, = ( 2)<br />
(DE)<br />
3 3<br />
⇒ DE = 2<br />
Since ∆ADE is a right angle triangle,<br />
AD 2 = AE 2 + DE 2<br />
⇒ (4) 2 = AE 2 + (2) 2<br />
⇒ AE 2 = 12<br />
But → 2λ<br />
+ 1 ^ 1 ^ 1 ^ ^ ^ ^<br />
AE = i + j + k – ( i + j+<br />
k)<br />
λ + 1 λ + 1 λ + 1<br />
^<br />
=<br />
1 i<br />
λ<br />
^<br />
–<br />
λ + 1 j<br />
λ<br />
^<br />
–<br />
λ + 1 k<br />
λ<br />
λ +<br />
⇒<br />
2<br />
| |<br />
→ 1<br />
AE =<br />
( λ + 1<br />
2<br />
2<br />
)<br />
3λ<br />
Therefore, 12 =<br />
2<br />
( λ + 1)<br />
⇒ 4(λ + 1) 2 = λ 2<br />
⇒ 4λ 2 + 4 + 8λ = λ 2<br />
⇒ 3λ 2 + 8λ + 4 = 0<br />
⇒ 3λ 2 + 6λ + 2λ + 4 = 0<br />
^<br />
^<br />
[λ 2 + λ 2 + λ 2 ] =<br />
^<br />
^<br />
1<br />
2<br />
3λ<br />
( λ + 1)<br />
2<br />
⇒ 3λ(λ + 2) + 2(λ + 2) = 0<br />
⇒ (3λ + 2) (λ + 2) = 0<br />
⇒ λ = – 2/3, λ = – 2<br />
Therefore, when λ = – 2/3, position vector of E is<br />
given by<br />
⎛ 2λ<br />
+ 1⎞<br />
^ 1 ^ 1 ^<br />
⎜ ⎟ i + j + k<br />
⎝ λ + 1 ⎠ λ + 1 λ + 1<br />
2.(–<br />
2 / 3)<br />
+ 1 ^ 1 ^ 1 ^<br />
=<br />
i + j + k<br />
– 2 / 3 + 1 – 2 / 3 + 1 – 2 / 3 + 1<br />
– 4 / 3 + 1 ^ 1 ^ 1 ^<br />
= i + j + k<br />
– 2 + 3 – 2 + 3 – 2 + 3<br />
3<br />
– 4 + 3<br />
3 3<br />
=<br />
3<br />
1/<br />
3<br />
^ 1 ^ 1 ^<br />
i + j + k<br />
1/<br />
3 1/<br />
3<br />
= – ^ i +<br />
<strong>Xtra</strong><strong>Edge</strong> for IIT-JEE 12 FEBRUARY <strong>2012</strong><br />
^<br />
3 j +<br />
^<br />
3k<br />
and when λ = – 2<br />
Position vector of E is given by,<br />
2×<br />
(– 2)<br />
+ 1 ^ 1 ^ 1 ^ – 4 + 1 ^<br />
i + j + k = i –<br />
– 2 + 1 – 2 + 1 – 2 + 1 – 1<br />
^ j – ^ k<br />
^<br />
= 3i – ^ j – ^ k<br />
Therefore, – ^ ^ ^<br />
i + 3 j + 3k and + ^<br />
3i – ^ j – ^ k are<br />
the position vector of E.<br />
13. Evaluate ∫ π<br />
Sol. Let,<br />
−π<br />
/ 3<br />
/ 3<br />
3<br />
π + 4x<br />
dx [IIT-2004]<br />
⎛ π ⎞<br />
2 − cos⎜|<br />
x | + ⎟<br />
⎝ 3 ⎠<br />
π / 3 πdx<br />
π / 3 x dx<br />
I = ∫ + 4<br />
−π<br />
/ 3 ⎛ π ⎞ ∫− π/<br />
3 ⎛ π ⎞<br />
2 − cos⎜|<br />
x | + ⎟ 2 − cos⎜|<br />
x | + ⎟<br />
⎝ 3 ⎠<br />
⎝ 3 ⎠<br />
a ⎡ 0,<br />
f ( −x)<br />
= − f ( x)<br />
⎤<br />
Using f ( x)<br />
dx= ⎢ a<br />
⎥<br />
∫− a ⎢2<br />
− = ⎥<br />
⎣ ∫ f ( x)<br />
dx,<br />
f ( x)<br />
f ( x)<br />
0<br />
⎦<br />
∴ I = 2 ∫ π<br />
0<br />
/ 3<br />
πdx<br />
+ 0<br />
⎛ π ⎞<br />
2 − cos⎜|<br />
x | + ⎟<br />
⎝ 3 ⎠<br />
⎧<br />
⎪<br />
⎨as<br />
⎪<br />
⎪⎩<br />
∫ π/<br />
3<br />
−π<br />
/ 3<br />
π/<br />
3 dx<br />
I = 2π∫ 0 2 − cos( x + π / 3)<br />
2π<br />
= 2π∫ dt<br />
2 t<br />
/ 3<br />
π / 3 − cos<br />
⎫<br />
3<br />
x dx<br />
⎪<br />
is odd ⎬<br />
⎛ π ⎞<br />
2 − cos⎜|<br />
x | + ⎟ ⎪<br />
⎝ 3 ⎠ ⎪⎭<br />
π<br />
, where x + = t<br />
3<br />
3
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