Solution - Career Point

75. If the lines represented by x 2 + 2λx + 2y 2 = 0 & linesrepresented by (1 + λ)x 2 – 8xy + y 2 = 0 are equallyinclined then λ equals :(A) – 2 (B) + 2(C) ± 2 (D) ± 484. In a class of 100 students there are 70 boys whoseaverage marks in a subject are 75.If the averagemarks of the complete class is 72, then what is theaverage of the girls.(A) 73 (B) 65(C) 68 (D) 7485. A letter is taken at random from the letters of word'STATISTICS' and a another letter is taken at randomfrom letters of word 'ASSISTANT'. The probabilitythat they are the same letter is -4(A) 1/45 (B) 13/90(C) 19/0 (D) 5/1886. If A = {x : x ∈ I ; – 2 ≤ x ≤ 2}2Β = { x : x ∈ I ; 0 ≤ x ≤ 3}(A) 2 (B) – 3C = {x : x ∈ N ; 1 ≤ x ≤ 2} and(C) – 5(D) NoneD = {(x, y) ∈ N × N; x + y = 8} then -(A) n(A ∪ (B ∪ C)) = 5 (B) n(D) = 6(C) n(B ∪ C) = 5 (D) None of these87. Solution of sec 2 dyy + 2x tan y = x 3 is -dx2x(A) tan y = ce − + (x 2 – 1)∫ 22–1xf (g(x)) f 'g(x). g'(x) dx where (B) tan y = ce − + (x 2 – 1)12xg(1) = g(2) is equal to -(C) tan y = ce − – (x 2 – 1)(A) 1 (B) 2(D) None of these(C) 0(D) None88. The equation of common tangent to the curves2 2x + sin xy 2 = 8x and xy = –1 is -sec 2 x dx and f(0) = 0 then2(A) 3y = 9x + 2 (B) y = 2x + 11 xf(1) =(C) 2y = x + 8 (D) y = x + 2(A) 1 – π/4 (B) π/4 – 1(C) tan1 – π/4 (D) None of these89. Let f be twice differentiable function such thatf "(x) = – f(x) and f '(x) = g(x)sec 1 xh(x) = (f(x)) 2 + (g(x)) 2 . If h(5) = 11is -x −[x]then h(10) is equal to(A) R(B) R – {(– 1, 1)I}(A) 22 (B) 11(C) R – I (D) R – [0, 1)(C) 0(D) None90. We are required to from different words with the help(A) D f = [– 1, 1] (B) R f = {–π/2, π/2}of letter of the word INTEGER. Let m 1 , be the(C) R f = {π/2} (D) None of thesenumber of words in which I and N are never togetherand m 2 be the number of words which begin withm1I and end with R. Thenm2is given by -76. locus of centre of a variable circletx 2 + ty 2 + 2(t 2 + 1)x – 2(t 2 – 1)y + t = 0 is a :(A) Straight line (B) Parabola(C) Ellipse(D) Hyperbola477. If∫f (x) dx = 4 and (3 f (x))− 1 ∫− dx = 7 then the2value of∫ 4f (x) dx is -78. Bisector of angle between lines 2x + y – 6 = 0 &4x – 2y + 7 = 0 which contains origin is -(A) acute angle bisector ; x = 5/8(B) acute angle bisector ; y = 19/4(C) obtuse angle bisector ; x = 5/8(D) obtuse angle bisector ; y = 19/479. The value of [ ]80. If f(x) =∫+81. The domain of the function f(x) =82. Let f(x) = sin –1 x + sec –1 x, then -83. If z 1 , z 2 , z 3 represents the vertices of an equilateraltriangle such that|z 1 | = |z 2 | = |z 3 | then -(A) z 1 + z 2 = z 3 (B) z 1 + z 2 + z 3 = 0(A) 30 (B) 1/30(C) 6 (D) 42XtraEdge for IIT-JEE 75APRIL 2010