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Application of Derivatives question

2021 · 31 Aug · Shift 1 · Q39
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  5. /2021 · 31 Aug · Shift 1 · Q39

Application of Derivatives question

2021 · 31 Aug · Shift 1 · Q39

JEE MainMathematicsApplication of DerivativesNumerical+4 / −1
If 'R' is the least value of 'a' such that the function f(x) = x2 + ax + 1 is increasing on [1, 2] and 'S' is the greatest value of 'a' such that the function f(x) = x2 + ax + 1 is decreasing on [1, 2], then the value of |R −-− S| is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. We are given f(x)=x2+ax+1f(x)=x^2+ax+1f(x)=x2+ax+1 and need conditions on aaa so that fff is increasing or decreasing on [1,2][1,2][1,2].

  2. Compute the derivative: f′(x)=2x+af'(x)=2x+af′(x)=2x+a

  3. For f(x)f(x)f(x) to be increasing on [1,2][1,2][1,2], we need f'(x) 0 \quad \text{for all } x\in[1,2]. Since f′(x)=2x+af'(x)=2x+af′(x)=2x+a is a linear function increasing in xxx, its minimum on [1,2][1,2][1,2] occurs at x=1x=1x=1. So it is enough to require f′(1)=2+a≥0f'(1)=2+a\ge 0f′(1)=2+a≥0 a≥−2a\ge -2a≥−2 Hence the least value of aaa is R=−2R=-2R=−2

  4. For f(x)f(x)f(x) to be decreasing on [1,2][1,2][1,2], we need f′(x)≤0for all x∈[1,2].f'(x)\le 0 \quad \text{for all } x\in[1,2].f′(x)≤0for all x∈[1,2]. Again, since f′(x)=2x+af'(x)=2x+af′(x)=2x+a increases with xxx, its maximum on [1,2][1,2][1,2] occurs at x=2x=2x=2. So it is enough to require f′(2)=4+a≤0f'(2)=4+a\le 0f′(2)=4+a≤0 a≤−4a\le -4a≤−4 Hence the greatest value of aaa is S=−4S=-4S=−4

  5. Therefore, ∣R−S∣=∣−2−(−4)∣=∣2∣=2|R-S|=|-2-(-4)|=|2|=2∣R−S∣=∣−2−(−4)∣=∣2∣=2

So the required integer is 2\boxed{2}2​

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