3x=7y=(21)2
Find the relation b/w x, y and z where x, y and z ≠ 0.
A. z=xy
B. z=x+y
C. z=(xy/x+y)
D. None of these
Answer : Option C
Let 3x = 7y = (21)z = k
3x = k, 7y = k, 21z = k
3 = k1/x, 7 = k1/y, 21 = k1/z
Now, 21 = 3 × 7
K1/z=k1/x ×k1/y
1/z=1/x+1/y z=xy/x+y
Find the value of √6+√6+√6+---∞
A. 3
B. 10
C. -2
D. None of these
Answer : Option A
√6+√6+√6+---∞ = x
x =√6+x
x2 = 6+x
x2 - x - 6 = 0
(x - 3) (x + 2) = 0
x = 3 or -2
But the given expression is positive, so rejecting x = - 2, we get the answer to be x = 3.
Among the 21/3, 31/4, 51/6, 61/12 which one is greater.
A. 21/3
B. 31/4
C. 51/6,
D. 61/12
Answer : Option B
(212/3)1/12, (312/4)1/12, (512/6)1/12, (212/12)1/12
(24)1/12, (33)1/12, (52)1/12, (6)1/12
So 31/4 is greatest.
-icon.jpg?null&itok=2FjKP8-O "Chain Rule Practice Problems"){kind=icon}
