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Maths Question 17 – JEE-MAIN 2025

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Let x=−1 and x=2 be the critical points of the function f(x)=x3+ax2+bloge⁡|x|+1,x≠0. Let m and M respectively be the absolute minimum and the absolute maximum values of f in the interval [−2,−12]. Then |M+m| is equal to (Take loge⁡2=0.7) :

To find the absolute maximum and minimum of a function on a closed interval, evaluate the function at its critical points within the interval and at the endpoints of the interval.

Step 1: Determine coefficients a and b✦ Active

The derivative of f(x)=x3+ax2+bloge⁡|x|+1 is f′(x)=3x2+2ax+bx. Since x=−1 and x=2 are critical points, f′(−1)=0 and f′(2)=0.

f′(−1)=3(−1)2+2a(−1)+b−1=3−2a−b=0⇒2a+b=3(1) f′(2)=3(2)2+2a(2)+b2=12+4a+b2=0⇒8a+b=−24(2)

Solving equations (1) and (2) simultaneously yields a=−92 and b=12. Thus, the function is f(x)=x3−92x2+12loge⁡|x|+1.

Step 2: Evaluate function at relevant points○ Expand

The given interval is [−2,−12]. For x in this interval, |x|=−x. The critical point x=−1 lies within this interval. We evaluate f(x) at the endpoints and the critical point within the interval, using loge⁡2=0.7.

f(−2)=(−2)3−92(−2)2+12loge⁡(2)+1=−8−18+12(0.7)+1=−26+8.4+1=−16.6 f(−12)=(−12)3−92(−12)2+12loge⁡(12)+1=−18−98−12loge⁡(2)+1=−1.25−8.4+1=−8.65 f(−1)=(−1)3−92(−1)2+12loge⁡(1)+1=−1−4.5+0+1=−4.5
💡 Teacher's Secret Hint

Remember that loge⁡1=0 and loge⁡(1/x)=−loge⁡x.

Step 3: Determine absolute maximum and minimum and calculate |M+m|○ Expand

Comparing the values f(−2)=−16.6, f(−12)=−8.65, and f(−1)=−4.5:

The absolute maximum value is M=−4.5.

The absolute minimum value is m=−16.6.

Therefore, |M+m|=|−4.5+(−16.6)|=|−4.5−16.6|=|−21.1|=21.1.

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