Continuity & DifferentiabilitymediumFree

Continuity & Differentiability: Let Cases 3mm 3mm Cases Differentiable

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

Question
Let
f(x)={lim⁡n→∞(pxn∑r=1n[r2−e−x+r−1]r(r+1))+λ,x>0q,x=0lim⁡n→∞∑r=1n{r2+r+ex−1}r(r+1),x<0f(x) = \begin{cases} \displaystyle\lim_{n\to\infty}\left(\frac{px}{n}\sum_{r=1}^{n}\frac{\left[r^2-e^{-x}+r-1\right]}{r(r+1)}\right)+\lambda, & x>0 \\[3mm] q, & x=0 \\[3mm] \displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{\left\{r^2+r+e^{x}-1\right\}}{r(r+1)}, & x<0 \end{cases}
 is differentiable in R\mathbb{R} ([ ⋅ ][\,\cdot\,] is G.I.F. and { ⋅ }\{\,\cdot\,\} is F.P. of xx). Then
Ap=1p = 1correct
Bq=1q = 1correct
Cp+q+λ=3p+q+\lambda = 3correct
Dif gg is inverse of ff then g′(12)=2g'\left(\tfrac12\right) = 2correct
Solution
Step 1: For x>0x>0, e−x∈(0,1)e^{-x} \in (0,1) and r2+r−1r^2+r-1 is an integer, so
[r2+r−1−e−x]=r2+r−2=r(r+1)−2\left[r^2+r-1-e^{-x}\right] = r^2+r-2 = r(r+1)-2
∑r=1nr(r+1)−2r(r+1)=∑r=1n[1−2r(r+1)]=n−2∑r=1n(1r−1r+1)=n−2(1−1n+1)\sum_{r=1}^{n}\frac{r(r+1)-2}{r(r+1)} = \sum_{r=1}^{n}\left[1-\frac{2}{r(r+1)}\right] = n-2\sum_{r=1}^{n}\left(\frac1r-\frac{1}{r+1}\right) = n-2\left(1-\frac{1}{n+1}\right)
lim⁡n→∞pxn[n−2+2n+1]=px⋅1=px\lim_{n\to\infty}\frac{px}{n}\left[n-2+\frac{2}{n+1}\right] = px\cdot 1 = px
 so f(x)=px+λf(x) = px+\lambda for x>0x>0. Step 2: For x<0x<0, ex∈(0,1)e^{x} \in (0,1) and r2+r−1r^2+r-1 is an integer, so
{r2+r−1+ex}=ex\left\{r^2+r-1+e^{x}\right\} = e^{x}
∑r=1nexr(r+1)=ex(1−1n+1)⟶ex\sum_{r=1}^{n}\frac{e^{x}}{r(r+1)} = e^{x}\left(1-\frac{1}{n+1}\right) \longrightarrow e^{x}
 so f(x)=exf(x) = e^{x} for x<0x<0. Step 3:
f(x)={px+λ,x>0q,x=0ex,x<0f(x) = \begin{cases} px+\lambda, & x>0 \\ q, & x=0 \\ e^{x}, & x<0 \end{cases}
 Step 4: Continuity at x=0x=0:
lim⁡x→0−ex=1,f(0)=q,lim⁡x→0+(px+λ)=λ\lim_{x\to0^-}e^{x} = 1 ,\qquad f(0) = q ,\qquad \lim_{x\to0^+}\left(px+\lambda\right) = \lambda
⟹ q=1 and λ=1\Longrightarrow\ q = 1 \ \text{and}\ \lambda = 1
 Step 5: Differentiability at x=0x=0:
left derivative=ddxex∣0=1,right derivative=p\text{left derivative} = \left.\frac{d}{dx}e^{x}\right|_{0} = 1 ,\qquad \text{right derivative} = p
⟹ p=1\Longrightarrow\ p = 1
f(x)={ex,x<0x+1,x≥0f(x) = \begin{cases} e^{x}, & x<0 \\ x+1, & x\ge0 \end{cases}
 Step 6: p=1p = 1, q=1q = 1, p+q+λ=3p+q+\lambda = 3, so (1), (2) and (3) are all true. Step 7: ff is continuous and strictly increasing with range (0,∞)(0,\infty), so its inverse gg exists. 12<1\tfrac12<1, so f(x)=12f(x) = \tfrac12 lies on the branch ex=12e^{x} = \tfrac12, i.e. x=−ln⁡2x = -\ln2, and
g′(12)=1f′(−ln⁡2)=1e−ln⁡2=11/2=2g'\left(\tfrac12\right) = \frac{1}{f'\left(-\ln2\right)} = \frac{1}{e^{-\ln2}} = \frac{1}{1/2} = 2
 so (4) is true as well. Answer: (1), (2), (3) and (4)
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