Basics & LogarithmshardFree

Basics & Logarithms — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If x,y,z∈Rx, y, z \in \mathbb{R} and 121x2+4y2+9z2−22x+4y+6z+3=0121x^2 + 4y^2 + 9z^2 - 22x + 4y + 6z + 3 = 0, then the value of 1x−1y−1z\dfrac{1}{x}-\dfrac{1}{y}-\dfrac{1}{z} is equal to:
A1717
B1616correct
C1515
D1414
Solution
Step 1: Complete the square in each variable
121x2−22x=(11x−1)2−1121x^2 - 22x = (11x-1)^2 - 1
4y2+4y=(2y+1)2−14y^2 + 4y = (2y+1)^2 - 1
9z2+6z=(3z+1)2−19z^2 + 6z = (3z+1)^2 - 1
 Step 2: Rewrite the equation
(11x−1)2+(2y+1)2+(3z+1)2−3+3=0(11x-1)^2 + (2y+1)^2 + (3z+1)^2 - 3 + 3 = 0
(11x−1)2+(2y+1)2+(3z+1)2=0(11x-1)^2 + (2y+1)^2 + (3z+1)^2 = 0
 Step 3: Solve using non-negativity of squares Each square must vanish:
x=111,y=−12,z=−13x = \frac{1}{11}, \quad y = -\frac{1}{2}, \quad z = -\frac{1}{3}
 Step 4: Evaluate the required expression
1x−1y−1z=11−(−2)−(−3)=16\frac{1}{x}-\frac{1}{y}-\frac{1}{z} = 11 - (-2) - (-3) = 16
 Answer: (2)
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