Question 1

If n is a positive integer, then 2.4^ 2 n+1 +3^ 3 n+1 is divisible by:

Accepted Answer

Let P(n)=2 4^ 2 n+1 +3^ 3 n+1 Then P(1)=2.4^3+3^4=209, which is divisible by 11 but not divisible by 2,7 or 27 . Further, let P(k)=2.4^ 2 k+1 +3^ 3 k+1 is divisible by 11 , that is, 2.4^ 2 k+1 +3^ 3 k+1 =11 q for some integer q. Now aligned P ( k +1) & =2 4^ 2 k +3 +3^ 3 k +4 \ & =2 4^ 2 k +1 4^2+3^ 3 k +1 3^3 \ & =16 2 4^ 2 k +1 +27 3^ 3 k +1 \ & =16 2 4^ 2 k +1 +(16+11) 3^ 3 k +1 \ & =16 [2 4^ 2 k +1 +3^ 3 k +1 ]+11 3^ 3 k +1 \ & =16 11 q +11 3^ 3 k +1 \ & =11 (16 q +3^ 3 k +1 )=11 ~m al

Question 2

The solution set of (x-2)^ x^3-6 x+8 >1 is

Accepted Answer

Clearly x>2. Write the given inequality as aligned & (x^2-6 x+8 ) (x-2)>0 (x-2)(x-4) (x-2)> \ & 0 \ & (x-4) (x-2)>0 [ x>2] aligned

Question 3

If x, y and z are real numbers, the x^ 2 +4 y^ 2 +9 z^ 2 -6 y z-3 z x-2 x y is always

Accepted Answer

x^ 2 +4 y^ 2 +9 z^ 2 -6 y z-3 z x-2 x y =x^ 2 +(2 y)^ 2 +(3 z)^ 2 -(2 y)(3 z)-(3 z)(x)- x(2 y) 0 x, y, z [ a ^ 2 + b ^ 2 + c ^ 2 - ab - bc - ca 0 ]

Basic of Mathematics — Previous Year Questions & Solutions