Test: Monotonic-Increasing and Decreasing Functions(14 Sep) - JEE MCQ

f(x)= {−(x−1)/x²   x<1,   x(x−1)/x²  x≥1}
​f'(x)= {(x−2)/x³   x<1,      −(x−2)/x³   x≥1}
​x<1, if f′(x)<0 (for f(x) to be monotonically decreasing
⇒ (x−2)/x³<0
⇒x∈(0,2)
But x<1 ⇒ x∈(0,1)
For x≥1, if f′(x)<0
⇒ −(x−2)]/x³<0
⇒(x−2)/x³ > 0
⇒x∈(−∞,0)∪(2,∞)
But, x≥1   ⇒x∈(2,∞)
Hence, x∈(0,1)∪(2,∞)

f(x)=xlnx−x+1
f(1)=0
On differentiating w.r.t x, we get
f′(x)=x*1/x+lnx−1
=lnx
Therefore,f′(x)>0 for allx∈(1,∞)
f′(x)<0 for allx∈(0,1)
lim x→0 f(x)=1
f(x)>0
for allx∈(0,1)∪(1,∞)

f(x) = ln(1+x) - x ≤ 0
f(x) = ln(1+x) - x
f’(x) = 1/(1+x) - 1
= (1 - 1 - x)/(1+x) 
= -x/(1+x)
f’(x) ≤ 0
-x/(1+x) ≤ 0
0 ≤ x /(1+x)
for(x = 2) (-2)/(1-2)
= 2 > 0, therefore x > - 1

f’(x) > 0 
=> f(x) is increasing
f’’(x) < 0  => f(x) is convex

f(x) = x (x + 3)e–x/2 is continuous in [–3, 0] 
and f′(x)=(x2 + 3x) (– 1/2)e-x/2 + (2x + 3)e–x/2 
= – 1/2(x2 – x – 6)e–x/2 
Therefore f′(x) exists (i.e., finite) for all x 
Also f(–3) = 0, f (0) = 0 
So that f(–3) = f(0) 
Hence all the three conditions of the theorem are satisfied. 
Now consider f′(c)=0 
i.e., –1/2(c2 – c – 6)e–c/2 = 0 
c2 – c – 6 = 0 
(c + 2) (c – 3) = 0 c = 3 or –2 
Hence there exists – 2 ∈ (–3, 0) such that 
f′(–2) = 0