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Application of Derivatives question

2003 · Shift 0 · Q103
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Application of Derivatives question

2003 · Shift 0 · Q103

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
The real number xxx when added to its inverse gives the minimum sum at xxx equal :
  1. A
    -2
  2. B
    2
  3. C
    1
  4. D
    -1
View written solutionFree

Correct answer: C

  1. Let the function be f(x)=x+1x,x≠0.f(x)=x+\frac{1}{x}, \qquad x\neq 0.f(x)=x+x1​,x=0.

  2. We need the value of xxx for which this sum is minimum.

  3. Differentiate: f′(x)=1−1x2.f'(x)=1-\frac{1}{x^2}.f′(x)=1−x21​.

  4. Set the derivative equal to zero: 1−1x2=01-\frac{1}{x^2}=01−x21​=0 1x2=1\frac{1}{x^2}=1x21​=1 x2=1x^2=1x2=1 x=±1.x=\pm 1.x=±1.

  5. Use the second derivative test: f′′(x)=2x3.f''(x)=\frac{2}{x^3}.f′′(x)=x32​.

    • At x=1x=1x=1: f′′(1)=2>0,f''(1)=2>0,f′′(1)=2>0, so f(x)f(x)f(x) has a minimum at x=1x=1x=1.

    • At x=−1x=-1x=−1: f′′(−1)=−2<0,f''(-1)=-2<0,f′′(−1)=−2<0, so f(x)f(x)f(x) has a maximum at x=−1x=-1x=−1.

  6. Therefore, the minimum value of x+1xx+\dfrac{1}{x}x+x1​ occurs at x=1.x=1.x=1.

  7. Checking options:

    • A: −2-2−2 ❌
    • B: 222 ❌
    • C: 111 ✅
    • D: −1-1−1 ❌

Hence, the correct option is C.

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