# Engineering Solutions For Oil & Gas Industries

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Engineering Solutions For Oil & Gas Industries Company Profile & a Brief Introduction On the Standards, Practices & General Services Offered V.2 - Jan 2012
• early involvement of expert personnel for crisis man- agement
• maintenance vendors qa youssef a. tahan chief operations officerproduction fabrication
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Published : Tuesday, March 27, 2012
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INDIANET @IITJEE, Where technology meets education! Visual Physics, Maths & Chemistry; Classroom & Online Courses.IITJEE 2003 Mains Questions & Solutions  Maths (The questions are based on memory) Breakup of questions: Algebra Trigonometry Coordinate Geometry Calculus Vector/3D 8 1 1 8 2  1z z 1 2 1.that Prove <1 if |z1| < 1 < |z2| zz 1 2
Sol.
2.
Sol.
T.P.T.
 1z z 1 2 <1 zz 1 2
| 1z z|<|zz| or, T.P.T.1 2 2 1 (1z z)(1z z)<(zz)(zz) or, T.P.T.1 2 1 2 2 1 2 1 2 2 2 2 1+|z| |z||z||z|<0 or, T.P.T. 1 2 1 2 2 2 2 (1|z|)+(|z|1)|z|<0 or, T.P.T. 1 1 2 2 2 (1|z|)(1|z|)<0 or, T.P.T. 1 2 Which is true because of |z1| < 1 < |z2|
[2]
P(x) is a polynomial function such that P(1) = 0, P(x) > P(x),x > 1. Prove that P(x) > 0,x > 1 [2]  P(x) – P(x) > 0,x > 1 xx e.P(x)e P(x)>0 x  ,x > 1 (multiplying byewhich is +ve) d x (e.p(x))>0,x>1 dx–x  e .P(x) is an increasing function of x, x [1, ) (as P(x) being polynomial function is a continuous function). Thus for x > 1 –x –1 .P(1) e .P(x) > e –x .P(x) > 0, as P(1) = 0 e x  P(x) > 0. (ase+ve)
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3.
Sol.4.
Sol.
5.
Sol.
INDIANET @IITJEE, Where technology meets education! Visual Physics, Maths & Chemistry; Classroom & Online Courses.
a  IfA=b  c 
b c a
cT3 3 3 a, abc =1,A A=I+ c + b , then find the value of a .  b 
T A A=IT 2  (det ) = 1A) (det A = 1(det A) det A =±1. 3 3 3 3 3 3 Now det A = –(a + b + c – 3abc) = –(a + b + c ) + 3 3 3 3 Thus(a+b+c)+3= ±1=k(say) (say) 3 3 3 = 3 – k = 2 or 4.+ b + c  a
2 2 Find the point on x + 2y = 6, which is nearest to the line x + y = 7.
63 Let Pcos ( , sin) be the required point Tangent at P should be of slope = – 1 (slope of the line x + y = 7) 2 2 Now x + 2y = 6 dy  x + 2y=0 dx dy x =  dx2y 
y
x + y = 7
2 2 x y + =1 6 3
dy6 cos1 =  = cot = 1   dxPsin2 3 2  cot=21 2  sincos= , P lies in the I’st quadrant)= (as 3 3   2 1   P6. , 3.(2, 1) 3 3   Then nn nn1knnk n k k1 22  ........(1)  =  0k1k1k0k            Prove that where n n  =C m m   .
n n1 k  k1nn knnk  2  2  ....(1)   0k1k1k0  k n n n n1 fficient of x in = coe C(1+2x)C x(1+2x)+.............  0 1
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[2]
[2]
x
[2]
6.
Sol.
7.
Sol.
INDIANET @IITJEE, Where technology meets education! Visual Physics, Maths & Chemistry; Classroom & Online Courses.
kn = coefficient of x in+2x)(x) [(1] k n = coefficient of x in (1 + x) n . = Ck
n i a z=1 1 i |a|<2, If I{1, 2, 3……., n}. Prove that for no z, |z| < and can i i=1 3 occur simultaneously. [2] i i 1= a Z |a Z| i i 2 3n 1|a Z|+|a Z|+|a Z|+.......+|a Z| 1 2 3n 2 3n 2(|Z|+|Z|+|Z|+......+|Z|)  < 2 3 n + |Z| + |Z| + |Z| > 3/2+….. + |Z|  1 Case I
|Z| < 1 2  1 + …..+ |Z| + |Z| > 3/2 1 3 > 1|Z| 2  2 > 3 – 3 |Z| > 1/3 |Z| Case II
|Z|1, then obviously, |Z| < 1/3 is not possible n i i a z=1 Hence |Z| < 1/3 and can not occur simultaneously for any ai, |ai| < 2. i=1
f:[2a, 2a]!R If be an odd function such that left hand derivative at x = a is zero and f(x) = f(2a – x), x(a, 2a), then find left hand derivative offat x = –a.
f(a+h)f(a) Lt=0  f(a) =h, x(0, 2a) h!0 f(a+h)f(a) Lt  Nowf(–a) =hh!0 f(ah)+f(a) =Lt  h,fis an odd function. h!0
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[2]
8.
Sol.
9.
Sol.
INDIANET @IITJEE, Where technology meets education! Visual Physics, Maths & Chemistry; Classroom & Online Courses.
f(a+h)+f(a) Lt  =h,f(x) =f(2a – x), x(a, 2a)h!0 f(a+h)f(a) Lt=0  =hh!0
Iff(x) is an even function, then prove that #/ 2#/ 4 f(cos 2x)cosxdx=2f(sin 2x)cosxdx " " 0 0
#/ 2 I=f(cos 2x)cosx dx " 0 #/ 2 I=f(cos 2(#/ 2x))cos(#/ 2x)dx " 0 #/ 2 I=f(cos 2x)sinxdx " 0 , asfis even #/ 2 2I=f(cos 2x)(cosx+sinx)dx " 0 #/ 2 2f(cos 2x)sin(x+ #/ 4)dx " 0  = #/ 4 # # 2f(cos(#/ 2+2t))cos(t)dt "x+ = +t  = ,4 2#/ 4 #/ 4 2f(sin 2t)costdt "  = #4 #/ 4 I=2f(sin 2t)costdt " 0 #/ 4 I=2f(sin 2x)cosxdx " 0 .
[2]
A person has to go through three successive tests. Probability of his passing first exam is P. Probability of passing successive tests is P or P/2 according as he passed the last test or not. He is selected if he passes at least two tests. Find the probability of his selection. [2]
Person is selected if either he passes all the tests or exactly two of the tests. 3 P (passing all the tests) = P.P.P = P Probability of passing two tests  = P(first two tests) + P(first and third tests) + P(second and third tests)
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10.
Sol.
11.
Sol.
INDIANET @IITJEE, Where technology meets education! Visual Physics, Maths & Chemistry; Classroom & Online Courses.
P P  = P.P.(1–P) + P.(1 – P).+(1P)P2 2 1 1 22 2  = P (1 – P) +P(1P)+P(1P) 2 2 2  = 2P (1 – P) 3 2 2 3 Thus required probability = P + 2 P (1 – P) = 2P – P .
In a combat betweenA,BandC,Atries to hitBandC,andBandCtry to hitA. Probability of A, B and C hitting the targets are 2/3, 1/2, and 1/3 respectively. If A is hit, find the probability thatBhitsAandCdoes not. [2] The required probability is given by
P(A|BC).P(BC) P(BC|A)= P(A|BC)P(BC)+P(A|BC).P(BC)+P(A|BC).P(BC)+P(A|BC).P(BC) 1 2 1.× 2 3 1 2 1 1 1 1 1 2 1.× +1.× +1.× +0× ×  = 2 3 2 3 2 3 2 3 1 3 1 1 11 + +  = 3 6 6 =2
Three normals with slopes m1, m2and m3are drawn from a point P not on the axis 2 of the parabola y = 4x. If m1m2=%, results in the locus of P being a part of the parabola, find the value of%. [4] Any normal of slope m to the m1 yparabola 2 y = 4x is 3 y = mx – 2m – m (1) P(h, k) If it passes through (h, k), then 3 x k = mh – 2m – m 3  m + (2 – h)m + k = 0 (2) m3 m2 Thus m1m2m3= –k. k % Now m1m2=%m3= Now m3satisfies (2), so 3 k k  (2h)+k=0 3 % %

3 2 3 k + (2– h)k%– k%= 0
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INDIANET @IITJEE, Where technology meets education! Visual Physics, Maths & Chemistry; Classroom & Online Courses. Thus locus of P is 3 2 3  y + (2 – x)y%– y%= 0 2 2 3 + (2 – x) y %%= 0, as y'0 (P does not lie on the axis of the parabola) 2 2 2 3 = y %x – 2%+%2 2 2 3  If it is a part of the parabola y = 4x, then%= 4 and –2%+%= 0 %= 2 12. Letf : [0, 4]!R be a differentiable function 2 2  (i) For some a, b(0, 4), show thatf(4) –f(0) = 8f(a).f(b) 4 2 2  (ii) Show thatf(t)dt=2 %f(%)+(f((), for some 0 <%;(< 2. "  0 [4] Sol.(i) Using mean value theorem, there exists b(0, 4) such that f(4)f(0) f(b) =4 (1) 2 2(f(4)f(0)) (f(4))(f(0))=(f(4)+f(0))×4  Now4 From (1) 2 2 (f(4))(f(0))=f(b)(f(4)+f(0))×4  Hence it is sufficient to prove that f(0)+f(4) =f(a) 2[f(0),f(4)] [f(4),f(0)]  Range of functionf must contain the interval or according as f(0)f(4) orf(0)f(4) f(0)+f(4)  Now2is the mean value off(0) andf(4) f(0)+f(4)   2 range of the function f(0)+f(4) f(a)=  a[0, 4] for which2. Hence proved. 2 t=xt=xdt=2x dx (ii) Let . 4 2 2 f(t)dt=2xf(x)dx=2(20)f()) " " Thus for some)(0, 2), (using mean value 0 0 theorem for definite integral of a differentiable function). 4 f(t)dt=2(f())+f())) "  Thus 0 2 2 2(%f(%)+ (f(())  = , where%=(=).
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13.
Sol.
14.
Sol.
INDIANET @IITJEE, Where technology meets education! Visual Physics, Maths & Chemistry; Classroom & Online Courses.
If Inarea of n–sided regular polygon inscribed in a unit circle and O represents nthe area of the n–sided regular polygon circumscribing it, prove that 2 O2I n n   I=1+1  n 2 n    [4] I=2n n ×area of*OA1I11 2n× ×A I×OI 1 1 1  =2 O# # n×sin×cos +A1OI1=#/n  =n nA1 n2#I1 sin  =2n. B1 O1 O=2n n ×area of*OB1O1
1 2n× ×B O×O 1 1 1  =2O # # n×tan×1ntan  =n =n2 O n2In 1+1    2 n    Now R.H.S. =
On22#O # n21+1sin  1+cos   2n  2n  = = O## n22 ×2 cosO.cos n  =2n =n# #n2# 2 ntan .cos sin  =n n =2nI = n. Hence proved
Find the equation of the plane passing through (2, 1, 0); (4, 1, 1); (5, 0, 1). Find the point Q such that its distance from the plane is equal to the distance of point P(2, 1, 6) from the plane and the line joining P and Q is perpendicular to the plane. [4] Let equation of the plane be  ax + by + cz + d = 0 (1) (1) passes through the points (2, 1, 0); (4, 1, 1); (5, 0, 1)
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15.
Sol.
16.
Sol.
INDIANET @IITJEE, Where technology meets education! Visual Physics, Maths & Chemistry; Classroom & Online Courses.2  a = –d/3; b = –d/3; c =d3  x + y – 2z – 3 = 0 (2) which is the required equation of the plane obviously Q is the image of P in the plane. It is easy to see that Q(6, 5, –2) uˆ,vˆ, ˆwˆ dvˆ%, If be three non–coplanar unit vectors with angles betweenuan is G G G ˆ ˆ ˆa,b,c envˆ dwis(and betweenwanduis,. If betwe an are the unit vectors along angle bisectors of%,(,,respectively, then G G G G G G12 2% 2( 2, [a×b,b×c,c×a]=[uˆvˆ ˆw]sec sec sec  162 2 2 prove that . [4] G(uˆ+vˆ) a= 2 | cos%/ 2 | G (vˆ+wˆ) b= 2 | cos(/ 2 | G(ˆw+uˆ) c= 2 | cos,/ 2 | 2 ((uˆ+vˆ)×(vˆ+ˆw)).(ˆw+uˆ)G G G2= G G G G G G  × × × =8 | cos%/ 2.cos(/ 2.cos,/ 2 | [a b,b c,c a] [a,b,c]  -2 [uˆvˆ ˆw]2 2 2 sec(%/ 2).sec((/ 2).sec(,/ 2) =16
2 2 2 If a, b and c are in arithmetic progression and a , b and c are in Harmonic progression, then prove that either a = b = c or a, b and –c/2 are in Geometric Progression. [4]
Given that 2b = a + c (1) 2 2 2 a , b , c are in H.P. 2 2 22a c b= 2 2 anda+c (2) 2 2 22a c b= 2 From (2) 4b2ac, using (1) 2 2 ) = 0) (ac + 2b – b  (ac 2 2  b = ac or 2b = –ac. 2 Case I: b = ac 2 a+c   =ac 2  , using (1)
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