Algebra Exercises

May 29th, 2015
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1. The number of real roots of the equation ecosx - e-cosx 4 = 0 is(A)0(B) 1(C) 2(D) None of these 1. (A) ecosx = t Þ t2 –4t – 1 = 0 Þ t = ecosx = 2 ± Ö5 Since ecosx Î [1/e, e], so number of real roots is 0. 2. The number of real

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QEEGroup I1.The number of real roots of the equation ecosx - e-cosx4 = 0 is(A)0(B)1(C)2(D)None of these1.(A)ecosx = tt2 4t 1 = 0t = ecosx = 2 5Since ecosx [1/e, e], so number of real roots is 0.2.The number of real roots of the quadratic equation (n > 1) is (A) 1 (B) 2(C) n(D) 02.(D) Each term of is non-negative, so no real root.3.The set of values of a for which the equation x3 3x + a has three distinct real roots, is(A) ( -, ) (B) (-2, 2)(C) ( -1, 1) (D) none of these 3.(B) Let f(x) = x3 3x +a f(x) = 3x2 3.For three distinct real roots (i) f(x) = 0 should have two distinct real roots and and (ii) f() f() < 0 Here = 1, = 1. Now f() f() < 0 (1 3 + a) (1+ 3 + a) < 0 (a 2) (a + 2) < 0 2 < a < 2.4.If a1, a2, a3 (a1 > 0) are in G. P. with common ratio r, then the value of r, for which the inequality 9a1 + 5 a3 > 14 a2 holds, can not lie in the interval(A) [1, ) (B) [1, 9/5](C) [4/5, 1](D) [5/ 9, 1]4.(B) Since a1, a2, a3 (a1 > 0) are in G.P. So, a2 = a1 r ; a3 = a1 r2 Given inequality 9a1 + 5 a3 > 14 a29 a1+ 5 a1 r2 > 14 a1r 5r2 14 r +9 > 0 (r 1) ( r 9/5) > 0 r > 9/ 5 and r < 1 r [1, 9/ 5]. 5.Consider the graph of f(x) = ax2 +bx +c in the adjacent figure. We can conclude that (A) c >

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