Inverse Trigonometry Question no. 31 The value of cot(cosec -1 5/3 + tan -1 2/3) looks_one 4/17 looks_two 5/17 looks_3 6/17 looks_4 3/17 Answer Option looks_3 6/17 Solution Solution : Question no. 32 The Value of is looks_one π/4 looks_two π/2 looks_3 π/3 looks_4 π/6 Answer option looks_4 π/6 Solution Solution : Question no. 33 If x,y ,z are in A.P. and tan -1 x , tan -1 y, tan -1 z are also in A.P. , then : looks_one x=y=z looks_two 2x=3y=6z looks_3 6x=3y=2z looks_4 6x=4y=3z Answer option looks_one x=y=z Solution Solution : x,y ,z are in A.P. 2y = x + z tan -1 x , tan -1 y, tan -1 z are also in A.P. 2tan -1 y = tan -1 x + tan -1 z only x = y = z satisfy it. Question no. 34 Let x ∈(0,1) , the set of all x such that sin -1 x > cos -1 x is the interval looks_one (1/2, 1/√ 2 ) looks_two (1/√ 2 , 1) looks...

Inverse Trigonometry Question no. 16 what is equal to ? looks_one 0 looks_two 2 looks_3 1 looks_4 1/2 Answer Option looks_3 1 Solution Solution : Question no. 17 If sin -1 (1) +sin -1 (4/5)=sin -1 (x) then what is x equal to ? looks_one 3/5 looks_two 4/5 looks_3 1 looks_4 0 Answer Option looks_one 3/5 Solution Solution : Question no. 18 tan -1 (1+x) +tan -1 (1-x)=π/2 is satisfied by looks_one x=1 looks_two x=-1 looks_3 x=0 looks_4 x=1/2 Answer Option looks_3 x=0 Solution Solution : Question no. 19 If , where x= looks_one 1/√ 2 looks_two -1/√ 2 looks_3 ∓ √ (5/2) looks_4 ∓ 1/2 Answer Option looks_4 ∓ 1/2 Solution Solution : Question no. 20 tan -1 1 + tan -1 2 + tan -1 3= looks_one π/2 looks_two π looks_3 π/4 looks_4 none of these Answer Option looks_two ...

Probability Question no. 31 A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is looks_one looks_two looks_3 looks_4 none of these Answer Option looks_3 Solution Solution : P(x=7) = P(x=9) P(r=2) = 16 C 2 (1/2) 16 = 15/2 13 Question no. 32 A rifleman is firing at a distant target and has only 10% chance of hitting it. The least number of rounds, he must fire in order to have more than 50% chance of hitting it at least one is looks_one 11 looks_two 9 looks_3 7 looks_4 5 Answer Option looks_3 7 Solution Solution : P(hit)= 0.1 P(at least one) = P(x=1) + P(x=2)....+P(x=n) = 1-P(x=0) P(at least one) > 0.5 1-P(x=0) > 0.5 0.5 < P(x=0) 0.5 < (0.9) n n = 7 Question no. 33 If in a binomial distribution n =4 , P(X=0) = 16/81, Then P (X = 4) equals looks_one ...