If every element of a group G is its own inverse, then G 1) Finite 2) Infinite 3) Cyclic 4) Abelian Solution: If every element of a group G is its own inverse, then the group is Abelian.... View Article
In any group, the number of improper subgroups is 1) 2 2) 3 3) Depends on the group 4) 1 Solution: In any group, the number of improper subgroups is 2. Hence option (1) is the answer.... View Article
loga x is defined for (a>0) 1) all x 2) all negative real x not equal to 1 3) all positive real x not equal to 0 4) none of these Solution: loga x is... View Article
The remainder obtained when the polynomial x64 + x27 + 1 is divided by (x+1) is 1) 1 2) -1 3) 2 4) -2 Solution: We use the remainder theorem. P(x) = x64 + x27 + 1 Here a = -1 P(-1) = -164 + (-1)27 + 1 = 1 - 1... View Article
If 3x – 3x-1 = 6, then xx is equal to 1) 2 2) 4 3) 9 4) 3 Solution: Given 3x - 3x-1 = 6 => 3x - 3x/3 = 6 => 3x(1 - â…“) = 6 => 3x(â…”) = 6 => 3x = 18/2 = 9... View Article
(x+1)/(x-1)(x-2)(x-3) = 1) 1/(x-1) - 3/(x-2) + 2/(x-3) 2) 1/(x-1) + 3/(x-2) + 1/(x-3) 3) -3/(x-1) + 1/(x-2) + 2/(x-3) 4) None of these Solution:... View Article
(3x2 + 1)/(x2 – 6x + 8) is equal to 1) 3 + [49/2(x-4)] - [13/2(x-2)] 2) [49/2(x-4)] - 13/2(x-2) 3) -[49/2(x-4)] + 13/(x-2) 4) None of these Solution: (3x2 + 1)/(x2 - 6x... View Article
(3x+4)/(x+1)2(x-1) = A/(x-1) + B/(x+1) + C/(x+1)2, then A = 1) -1/2 2) 15/4 3) 7/4 4) -1/4 Solution: Given (3x+4)/(x+1)2(x-1) = A/(x-1) + B/(x+1) + C/(x+1)2 (3x+4)/(x+1)2(x-1) = [A(x+1)2 +... View Article
If (x2 + x + 1)/(x2 + 2x + 1) = A + [B/(x + 1)] + [C/(x + 1)2], then A – B is equal to 1) 4C 2) 4C + 1 3) 3C 4) 2C Solution: Given (x2 + x + 1)/(x2 + 2x + 1) = A + [B/(x + 1)] + [C/(x + 1)2] => (x2 + x + 1)/(x2 + 2x... View Article
If p and q are prime numbers satisfying the condition p2 – 2q2 =1, then the value of p2 + 2q2 is 1) 5 2) 15 3) 16 4) 17 Solution: Given that p and q are prime numbers satisfying the condition p2 - 2q2 =1 Let us take the first... View Article
Solution of the equation √(x + 10) + √(x – 2) = 6 are 1) 0 2) 6 3) 4 4) none of these Solution: Given √(x+ 10) + √(x - 2) = 6 Squaring both sides of the equation, we get x... View Article
If x = √7 + √3 and xy = 4, then x4 + y4 = 1) 400 2) 368 3) 200 4) none of these Solution: Given xy = 4 => y = 4/x Given x = √7 + √3 So y = 4/(√7... View Article
If [latex]\frac{2x+3}{(x+1)(x-3)} = \frac{a}{x+1}+\frac{b}{x-3}[/latex], then a+ b = 1) 1 2) 2 3) 9/4 4) none of these Solution: Given (2x + 3)/(x + 1)(x - 3) = [a/(x+1)] + [b/(x - 3)] (2x + 3)/(x + 1)(x - 3) = [a(x... View Article
If ax = bc, by = ac, cz = ab, then xyz = 1) 0 2) 1 3) x + y + z + 2 4) x + y + z Solution: Given ax = bc, by = ac and cz = ab (ax)y = (bc)y = bycy = ac.cy (axy)z =... View Article
If ax = by = cz and b2 = ac, then the value of y is 1) 2xz/(x+z) 2) xz/2(x-z) 3) xz/2(z-x) 4) 2xz/(x-z) Solution: Let ax = by = cz = k So a = k1/x b = k1/y c = k1/z Given b2 = ac... View Article
If a1/x = b1/y = c1/z and b2 = ac, then x + z = 1) y 2) 2y 3) 2xyz 4) none of these Solution: Let a1/x = b1/y = c1/z = k So a = kx b = ky c = kz Given b2 = ac => k2y =... View Article
If (2/3)x+2 = (3/2)2-2x, then x 1) 0 2) 1 c) 2 4) 4 Solution: (2/3)x+2 = (3/2)2-2x (2/3)x+2 = (2/3)2x-2 => x + 2 = 2x - 2 => x = 4 Hence option (4) is the... View Article
If log3 x + log3 y = 2 + log3 2 and log3(x + y) = 2, then 1) x = 1, y = 8 2) x = 8, y = 1 3) x = 3, y = 6 4) x = 9, y = 3 Solution: Given log3 x + log3 y = 2 + log3 2 => log3 xy = log332 + log3... View Article
If x = logb a, y = logc b, z = loga c, then xyz is 1) 0 2) 1 c) 2 4) 4 Solution: Given x = logb a, y = logcb, z = loga c We use property logn m = log m/log n So x = log a/log b y =... View Article
If a, b, c are consecutive positive integers and log(1 + ac)= 2k, then the value of k is 1) log a 2) log b c) 2 4) 1 Solution: Let a = n-1, b = n, c = n+1, where n > 1. Given log (1 + ac) = 2k => log (1 +... View Article
