Solution - Career Point
Solution - Career Point Solution - Career Point
56.[D]OH Zndust ⎯⎯→H3C–Cl⎯⎯⎯→CH 3AlCl3alk. KMnO 4⎯ ⎯⎯ → COOH64.[C] f(x) = sin x – cos x – kx + bf '(x) = cos x + sin x – kf '(x) = 2 sin(π/4 + x) – kif f(x) is decreases for all x ∴ f '(x) is –vei. e. k > max. of 2 sin(π/4 + x)i. e. k > 257.[D]58.[A] 4 HCl + O 2 ⎯⎯→ 2 H 2 O + 2 Cl 2Chlorine is in the form of cloud.59.[A] Coordination no. = 6Oxidation no. = 3no. of d electron = 6no. of Unpairedd electron = 060.[C] Resonance structure should have same numberof electron pairs.61.[A]MATHEMATICS30º 60ºAOLet the height of the tower is h1000 = h cot 30º – h cot 60º1000⇒ h =cot 30º −cot60ºh = 500 3 m62.[D] since variance is independent of change inorigin. Hence variance of observations 101,102, ...... 200 is same as variance of 151, 152,.....250.∴ V A = V BV A⇒ = 1VB63.[C] centroid of triangle is given by⎛⎞⎜c 2 + 3 a + b − 3, ⎟⎝ 3 3 ⎠if centroid lie on y axis ⇒ abscissae = 0c 2 + 3 = 0 ⇒ no real c existif centroid lie on x axis ⇒ ordinate = 0⇒ a + b – 3 = 0⇒ a + b = 3Ph65.[A] x lim → 0= x lim → 0⎡⎢⎢⎣⎛⎜⎜⎝xe 2− cos x ⎤⎥2x ⎥⎦−11−cos x⎟ ⎟ ⎞+22x x⎠xe 221 + x lim 2sin x / 2 1 3→ 0= 1 +2=x2 266.[A] Second determinant has been obtainedfrom the first by the operationC 1 → C 1 + 2C 2 – 3C 3 . so its value remainsunchanged67.[C] given 3sin x – 4sin 3 x – k = 0⇒ 3sin x – 4sin 3 x = k⇒ sin3x = k.........(i)angle A and B satisfy the equation (i)∴ sin 3A = sin3B = k ⇒ sin3A = sin3BBut A > B ⇒ A BNow, sin3A = sin(π – 3B)3A = π – 3Bπ 2πA + B = ⇒ C = 3 368.[B]69.[C]given statement(p q) ∨ ~ r → (p ∧ r)(F ↔ F) ∨ F → (F ∧ T)T ∨ F → FT → F = FpTTFFqTFTFp ∨ qTTTF~ pFFTTq∧~ pFFTF(p ∨ q) ↔ q∧~ pHence neither tautology nor contradiction70.[D] Given sequence can be written as5 20 10 20 , , , , ..............2 13 9 2320 20 20 20or , , , ..........8 13 18 23FFTTXtraEdge for IIT-JEE 106APRIL 2010
71.[B]which is a H.P.n th term of corresponding A.P.8 5 5 n + 3a n = + (n – 1) = 20 20 20n th term of H.P. =1 20 =5n + 3a n20 5also =5n + 3 17175n + 3 = 20 × = 68 5n =65 = 135Let any point P divides line joiningA(–2, 4, 7) & (3, –5, 8) in ratio λ : 1⎛ − 2 + 3λ4 − 5λ8λ + 7 ⎞then P ⎜ , , ⎟⎝ λ + 1 λ + 1 λ −1⎠if P lies on plane 2x – k = 0k⇒ x = 2k −2+ 3λ=2 λ + 1given λ = 9⎛ 25 ⎞k = ⎜ ⎟ × 2 ⇒ k = 5⎝ 10 ⎠72.[C] |a × b| = |a| |b| sin θ8 4sin θ = =5.2 5⇒ cos θ = 53|a – b| 2 = |a| 2 + |b| 2 – 2a.b= 4 + 25 – 2|a| |b| cos θ= 29 – 2.2.5 53 = 29 – 12AP 1if = PB 2⎛ a 2b ⎞co-ordinates of point P given by ⎜ , ⎟⎝ 3 3 ⎠locus of point P, a 2 + b 2 = (3x) 2 ⎛ 3y⎞+ ⎜ ⎟⎠⎝ 2or22x y+ = 14 16b > a then eccentricity e =2a1− =2b74.[D] parabola y = x 2 + 2px + p 2 + 13 – p 2(y – (13 – p 2 )) = (x + p) 2vertex is given by (– p, 13 – p 2 )is 4 units above x-axis⇒ 13 – p 2 = 4 ⇒ p = ± 3also lies in I st quadrant ⇒ p < 0⇒ p = – 375.[C] If lines x 2 + 2λx + 2y 2 = 0& (1 + λ) x 2 – 8xy + y 2 = 0are equally inclined⇒ Their bisectors eq n must be same2⇒ eq n x − y1−2&222x − y=(1 + λ)−1⇒ x 2 – y 2 =−4⇒ – λ =λ⇒ λ 2 = 4⇒ λ = ± 2= λxyxy− 4are samexy & x 2 – y 2 =− λxy− 4 / λ232= 36are same73.[C]|a – b| = 17A(0, b)176.[D] given circle in standard form isx 2 + y 2 ⎛ 1⎞⎛ 1⎞+ 2 ⎜ t + ⎟ x – 2 ⎜ t − ⎟ y + 1 = 0⎝ t ⎠ ⎝ t ⎠⎛ 1 1⎞centre is given by ⎜−t − , t − ⎟⎝ t t ⎠given length AB = 6or a 2 + b 2 = 36P2B(a, 0)⎛ 1⎞1or h = – ⎜ t + ⎟ & k = t –⎝ t ⎠ t22h 2 – k 2 ⎛ 1⎞⎛ 1⎞= ⎜ t + ⎟ – ⎜ t − ⎟ = 4⎝ t ⎠ ⎝ t ⎠locus x 2 – y 2 = 4 which is a hyperbolaXtraEdge for IIT-JEE 107APRIL 2010
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71.[B]which is a H.P.n th term of corresponding A.P.8 5 5 n + 3a n = + (n – 1) = 20 20 20n th term of H.P. =1 20 =5n + 3a n20 5also =5n + 3 17175n + 3 = 20 × = 68 5n =65 = 135Let any point P divides line joiningA(–2, 4, 7) & (3, –5, 8) in ratio λ : 1⎛ − 2 + 3λ4 − 5λ8λ + 7 ⎞then P ⎜ , , ⎟⎝ λ + 1 λ + 1 λ −1⎠if P lies on plane 2x – k = 0k⇒ x = 2k −2+ 3λ=2 λ + 1given λ = 9⎛ 25 ⎞k = ⎜ ⎟ × 2 ⇒ k = 5⎝ 10 ⎠72.[C] |a × b| = |a| |b| sin θ8 4sin θ = =5.2 5⇒ cos θ = 53|a – b| 2 = |a| 2 + |b| 2 – 2a.b= 4 + 25 – 2|a| |b| cos θ= 29 – 2.2.5 53 = 29 – 12AP 1if = PB 2⎛ a 2b ⎞co-ordinates of point P given by ⎜ , ⎟⎝ 3 3 ⎠locus of point P, a 2 + b 2 = (3x) 2 ⎛ 3y⎞+ ⎜ ⎟⎠⎝ 2or22x y+ = 14 16b > a then eccentricity e =2a1− =2b74.[D] parabola y = x 2 + 2px + p 2 + 13 – p 2(y – (13 – p 2 )) = (x + p) 2vertex is given by (– p, 13 – p 2 )is 4 units above x-axis⇒ 13 – p 2 = 4 ⇒ p = ± 3also lies in I st quadrant ⇒ p < 0⇒ p = – 375.[C] If lines x 2 + 2λx + 2y 2 = 0& (1 + λ) x 2 – 8xy + y 2 = 0are equally inclined⇒ Their bisectors eq n must be same2⇒ eq n x − y1−2&222x − y=(1 + λ)−1⇒ x 2 – y 2 =−4⇒ – λ =λ⇒ λ 2 = 4⇒ λ = ± 2= λxyxy− 4are samexy & x 2 – y 2 =− λxy− 4 / λ232= 36are same73.[C]|a – b| = 17A(0, b)176.[D] given circle in standard form isx 2 + y 2 ⎛ 1⎞⎛ 1⎞+ 2 ⎜ t + ⎟ x – 2 ⎜ t − ⎟ y + 1 = 0⎝ t ⎠ ⎝ t ⎠⎛ 1 1⎞centre is given by ⎜−t − , t − ⎟⎝ t t ⎠given length AB = 6or a 2 + b 2 = 36P2B(a, 0)⎛ 1⎞1or h = – ⎜ t + ⎟ & k = t –⎝ t ⎠ t22h 2 – k 2 ⎛ 1⎞⎛ 1⎞= ⎜ t + ⎟ – ⎜ t − ⎟ = 4⎝ t ⎠ ⎝ t ⎠locus x 2 – y 2 = 4 which is a hyperbolaXtraEdge for IIT-JEE 107APRIL 2010
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