NIMCET 2026 — Mathematics PYQ
NIMCET | Mathematics | 2026Let be a function defined by . Then which of the following statement is true?
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Let f:R→R be a function defined by f(x)=∣x+1∣e−x2. Then which of the following statement is true?
f has a point of global maxima in (1,2)
f has a point of maxima in (−2,−1)
(Correct Answer)f has a point of global minima in (0,1)
f has a point of local minima in (0,1)
f has a point of maxima in (−2,−1)
To solve this, we first analyze the function by removing the modulus sign. The definition of ∣x+1∣ is:
∣x+1∣={x+1−(x+1)amp;if x≥−1amp;if xlt;−1
Therefore, the function f(x) is defined as:
f(x)={(x+1)e−x2−(x+1)e−x2amp;if x≥−1amp;if xlt;−1
Case 1: x > -1
Using the product rule on (x+1)e−x2:
f′(x)=(1)e−x2+(x+1)e−x2(−2x)
f′(x)=e−x2[1−2x2−2x]
Setting f′(x)=0:
−2x2−2x+1=0⟹2x2+2x−1=0
Using the quadratic formula x=2a−b±b2−4ac:
x=4−2±4−4(2)(−1)=4−2±12=2−1±3
Since we are in the domain x > -1, we consider x=2−1+3≈0.366.
Case 2: x < -1
Using the product rule on −(x+1)e−x2:
f′(x)=−[e−x2+(x+1)e−x2(−2x)]=−e−x2[1−2x2−2x]=e−x2(2x2+2x−1)
Setting f′(x)=0 leads to the same quadratic equation 2x2+2x−1=0. In the domain x < -1, we consider x=2−1−3≈−1.366.
Point x=2−1−3≈−1.366:
This point lies in the interval (−2,−1). Testing points around −1.366:
For x < -1.366, f'(x) > 0 (the function is increasing).
For -1.366 < x < -1, f'(x) < 0 (the function is decreasing).
Therefore, this is a point of local maxima.
Point x=2−1+3≈0.366:
Testing points around 0.366:
For x < 0.366, f'(x) > 0.
For x > 0.366, f'(x) < 0.
Therefore, this is also a point of local maxima.
Comparing our findings with the given options:
Option 1 (1,2): Incorrect, the max is at 0.366.
Option 2 ((−2,−1)): Correct, because we found a maximum at x≈−1.366, which lies in the interval (−2,−1).
Option 3 (0,1): Incorrect, global minimum is at x=−1 where f(−1)=0.
Option 4 (0,1): Incorrect, there is no minimum in this interval.
Correct Option: 2
To solve this, we first analyze the function by removing the modulus sign. The definition of ∣x+1∣ is:
∣x+1∣={x+1−(x+1)amp;if x≥−1amp;if xlt;−1
Therefore, the function f(x) is defined as:
f(x)={(x+1)e−x2−(x+1)e−x2amp;if x≥−1amp;if xlt;−1
Case 1: x > -1
Using the product rule on (x+1)e−x2:
f′(x)=(1)e−x2+(x+1)e−x2(−2x)
f′(x)=e−x2[1−2x2−2x]
Setting f′(x)=0:
−2x2−2x+1=0⟹2x2+2x−1=0
Using the quadratic formula x=2a−b±b2−4ac:
x=4−2±4−4(2)(−1)=4−2±12=2−1±3
Since we are in the domain x > -1, we consider x=2−1+3≈0.366.
Case 2: x < -1
Using the product rule on −(x+1)e−x2:
f′(x)=−[e−x2+(x+1)e−x2(−2x)]=−e−x2[1−2x2−2x]=e−x2(2x2+2x−1)
Setting f′(x)=0 leads to the same quadratic equation 2x2+2x−1=0. In the domain x < -1, we consider x=2−1−3≈−1.366.
Point x=2−1−3≈−1.366:
This point lies in the interval (−2,−1). Testing points around −1.366:
For x < -1.366, f'(x) > 0 (the function is increasing).
For -1.366 < x < -1, f'(x) < 0 (the function is decreasing).
Therefore, this is a point of local maxima.
Point x=2−1+3≈0.366:
Testing points around 0.366:
For x < 0.366, f'(x) > 0.
For x > 0.366, f'(x) < 0.
Therefore, this is also a point of local maxima.
Comparing our findings with the given options:
Option 1 (1,2): Incorrect, the max is at 0.366.
Option 2 ((−2,−1)): Correct, because we found a maximum at x≈−1.366, which lies in the interval (−2,−1).
Option 3 (0,1): Incorrect, global minimum is at x=−1 where f(−1)=0.
Option 4 (0,1): Incorrect, there is no minimum in this interval.
Correct Option: 2