Limits, Continuity and Differentiability: Mathematics | JEE Main
Let the function f(x)=2x3+(2p−7)x2+3(2p−9)x−6 have a maxima for some value of x<0 and a minima for some value of x>0. Then, the set of all values of p is
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Hint 1 of 4
What is the derivative f′(x) of the cubic function f(x)=2x3+(2p−7)x2+3(2p−9)x−6?
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Step-by-step solutionView
Correct answer
The cubic has a local maximum at x<0 and a local minimum at x>0 if and only if f′(x)=0 has two real roots of opposite signs, which requires f′(0)<0, giving p∈(−∞,29).
Option analysis
Why each option works or fails
A · (0,29)
Assuming that the parameter p itself must be positive, or falsely concluding that the product of roots being negative implies a lower bound of 0 on p. The condition for opposite signs is simply that the product of roots ac<0, which yields 2p−9<0 with no lower bound restriction on p.
B · (−∞,29)
None. This is the correct option. For f′(x)=6x2+2(2p−7)x+3(2p−9) with leading coefficient 6>0, having roots on opposite sides of 0 is equivalent to f′(0)<0, which gives 3(2p−9)<0⟹p<29.
C · (−29,29)
Confusing the condition p<29 with a symmetric interval ∣p∣<29 by misapplying an absolute value inequality. Do not introduce absolute value bounds unless the inequality was quadratic in p of the form p2<(29)2. Here the inequality is strictly linear: 2p−9<0.
D · (29,∞)
Reversing the direction of the inequality when requiring the product of the roots to be negative. Since the roots have opposite signs, their product must be strictly negative, giving 3(2p−9)<0, which leads to p<29, not p>29.
Reviewed route
Solution
StepWorking
01given
Given the cubic function f(x)=2x3+(2p−7)x2+3(2p−9)x−6. It has a local maximum at some x1<0 and a local minimum at some x2>0.
02approach
Since f(x) is a cubic with positive leading coefficient (a=2>0), f′(x) is a quadratic opening upwards. For f(x) to have a maximum at x1 and a minimum at x2, f′(x)=0 must have two distinct real roots α<β such that α=x1<0 and β=x2>0. That is, the roots of f′(x)=0 must have opposite signs. For a quadratic Ax2+Bx+C=0 with A>0, the roots have opposite signs if and only if the product of roots is strictly negative, i.e., C/A<0 (which automatically ensures D>0).
03execute
Differentiating f(x) with respect to x:
f′(x)=6x2+2(2p−7)x+3(2p−9)
Here, the leading coefficient is A=6>0.
For roots to be of opposite signs, the product of roots must be negative:
Product of roots=AC=63(2p−9)=22p−9<0
Solving the inequality:
2p−9<0⟹p<29
Hence, p∈(−∞,29).
✓verify
Check discriminant: D=4(2p−7)2−4(6)(6p−27)=4[(2p−7)2−6(2p−9)]=4[4p2−28p+49−12p+54]=4[4p2−40p+103].
For p<9/2, 4p2−40p+103=(2p−10)2+3>0 for all real p, so real distinct roots are always guaranteed whenever C/A<0.
Hints that build this answer step by step
What is the derivative f′(x) of the cubic function f(x)=2x3+(2p−7)x2+3(2p−9)x−6?
f′(x)=6x2+2(2p−7)x+3(2p−9)
For f(x) with positive leading coefficient (a=2>0) to have a local maximum at x<0 and a local minimum at x>0, what condition must the roots of f′(x)=0 satisfy?
The quadratic f′(x)=0 must have two real roots of opposite signs (one negative and one positive).
For a quadratic Ax2+Bx+C=0 with A=6>0 to have roots of opposite signs, which single condition is necessary and sufficient?
f′(0)<0⟹3(2p−9)<0
Solving the inequality 3(2p−9)<0 gives which range for p?
For a quadratic Ax2+Bx+C=0 with A>0, if the product of roots C/A<0, then AC<0. Therefore, D=B2−4AC=B2+4∣AC∣>0 is automatically strictly positive for any real B.