Application of DerivativesmediumFree

Is the Piecewise Function Increasing at x = 0 | JEE Advanced

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Question
If f(x)={2x+3x<05x=0x2+7x>0f(x) = \begin{cases} 2x+3 & x < 0 \\ 5 & x = 0 \\ x^{2}+7 & x > 0 \end{cases} then at x=0x = 0, f(x)f(x)
Ais decreasing
Bis increasingcorrect
Cis neither increasing nor decreasing
Dhas local maxima
Solution
Step 1: Note that ff is not continuous at 00, so the usual derivative test does not apply. Use the definition of increasing at a point instead: ff is increasing at x=0x = 0 if
x<0  ⟹  f(x)<f(0)andx>0  ⟹  f(x)>f(0).x < 0 \implies f(x) < f(0) \qquad \text{and} \qquad x > 0 \implies f(x) > f(0) .
 Step 2: Look at the values just to the left. For x<0x < 0,
f(x)=2x+3<3,f(x) = 2x+3 < 3 ,
 since 2x<02x < 0 there. Step 3: Compare with f(0)f(0).
f(x)<3<5=f(0)for all x<0.f(x) < 3 < 5 = f(0) \quad \text{for all } x < 0 .
 Step 4: Look at the values just to the right. For x>0x > 0,
f(x)=x2+7>7.f(x) = x^{2}+7 > 7 .
 Step 5: Compare with f(0)f(0) again.
f(x)>7>5=f(0)for all x>0.f(x) > 7 > 5 = f(0) \quad \text{for all } x > 0 .
 Step 6: Both requirements hold, so ff is increasing at x=0x = 0. Step 7: Reject the others. It is not decreasing, so (A) fails; it does satisfy the increasing condition, so (3) fails; and f(0)=5f(0) = 5 is smaller than every value on the right, so it is not a local maximum and (4) fails. Answer: (2).
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