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Differentiation question

2023 · 29 Jan · Shift 1 · Q46
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  5. /2023 · 29 Jan · Shift 1 · Q46

Differentiation question

2023 · 29 Jan · Shift 1 · Q46

JEE MainMathematicsDifferentiationNumerical+4 / −1
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)−1,∀x,y∈Rf(x+y)=f(x)+f(y)-1,\forall x,y\in\mathbb{R}f(x+y)=f(x)+f(y)−1,∀x,y∈R. If f′(0)=2f'(0)=2f′(0)=2, then ∣f(−2)∣|f(-2)|∣f(−2)∣ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 3

  1. We are given f(x+y)=f(x)+f(y)−1  ∀x,y∈Rf(x+y)=f(x)+f(y)-1 \,\,\forall x,y\in\mathbb Rf(x+y)=f(x)+f(y)−1∀x,y∈R and fff is differentiable.

We need to find ∣f(−2)∣|f(-2)|∣f(−2)∣ given that f′(0)=2.f'(0)=2.f′(0)=2.

  1. First, find f(0)f(0)f(0).

Put x=0,y=0x=0, y=0x=0,y=0 in the functional equation: f(0)=f(0)+f(0)−1f(0)=f(0)+f(0)-1f(0)=f(0)+f(0)−1 f(0)=2f(0)−1f(0)=2f(0)-1f(0)=2f(0)−1 f(0)=1.f(0)=1.f(0)=1.

  1. Remove the constant term by defining g(x)=f(x)−1.g(x)=f(x)-1.g(x)=f(x)−1. Then g(x+y)=f(x+y)−1=(f(x)+f(y)−1)−1=(f(x)−1)+(f(y)−1)=g(x)+g(y).g(x+y)=f(x+y)-1=(f(x)+f(y)-1)-1=(f(x)-1)+(f(y)-1)=g(x)+g(y).g(x+y)=f(x+y)−1=(f(x)+f(y)−1)−1=(f(x)−1)+(f(y)−1)=g(x)+g(y). So ggg satisfies Cauchy’s equation: g(x+y)=g(x)+g(y).g(x+y)=g(x)+g(y).g(x+y)=g(x)+g(y).

Since fff is differentiable, ggg is also differentiable. Hence ggg must be linear: g(x)=cxg(x)=cxg(x)=cx for some constant ccc.

Therefore, f(x)=cx+1.f(x)=cx+1.f(x)=cx+1.

  1. Use the derivative condition.

Differentiate: f′(x)=c.f'(x)=c.f′(x)=c. So f′(0)=c=2.f'(0)=c=2.f′(0)=c=2. Hence, f(x)=2x+1.f(x)=2x+1.f(x)=2x+1.

  1. Now compute f(−2)f(-2)f(−2): f(−2)=2(−2)+1=−4+1=−3.f(-2)=2(-2)+1=-4+1=-3.f(−2)=2(−2)+1=−4+1=−3. Therefore, ∣f(−2)∣=∣−3∣=3.|f(-2)|=|-3|=3.∣f(−2)∣=∣−3∣=3.

  2. Final answer: 3\boxed{3}3​

This matches the stored correct answer.

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