Q17
jee_main_2024_31_jan_morning
Continuity Check
Let g(x) be a linear function and
f(x) = begincases g(x) & , x le 0 \\ left(frac1+x2+xright)^frac1x & , x > 0 endcases
is continuous at x = 0. If f'(1) = f(-1), then the value of g(3) is
Solution
### Core Logic
Let g(x) = ax + b.
Since f(x) is continuous at x = 0:
lim_x to 0^+ f(x) = f(0)
lim_x to 0 left(frac1+x2+xright)^frac1x = b
As x to 0, the base approaches frac12, and exponent approaches infty. Thus, left(frac12right)^infty = 0.
So, b = 0. Thus, g(x) = ax.
### Step 1: Calculate Derivative
For x > 0, f(x) = left(frac1+x2+xright)^frac1x. Let y = f(x).
ln y = frac1x lnleft(frac1+x2+xright)
Differentiating both sides w.r.t x:
frac1y y' = -frac1x^2 lnleft(frac1+x2+xright) + frac1x cdot frac2+x1+x cdot frac1(2+x) - (1+x)1(2+x)^2
y' = y left[ -frac1x^2 lnleft(frac1+x2+xright) + frac1x(1+x)(2+x) right]
### Step 2: Apply Condition
At x=1, y = f(1) = frac23.
f'(1) = frac23 left[ -1 lnleft(frac23right) + frac16 right] = -frac23 lnleft(frac23right) + frac19
Also f(-1) = g(-1) = -a.
Given f'(1) = f(-1) implies -a = -frac23 lnleft(frac23right) + frac19.
a = frac23 lnleft(frac23right) - frac19
### Step 3: Evaluate g(3)
g(3) = 3a = 2 lnleft(frac23right) - frac13
g(3) = lnleft(frac49right) - frac13 = lnleft(frac49right) - ln(e^1/3) = lnleft(frac49e^1/3right)
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Continuity and Differentiability
Class 12 Maths: Application of Derivatives
