Let the function f(x) = 2x^3 + (2p - 7)x^2 + 3(2p - 9)x - 6 have a maxima for some value of x 0 . Then, the set of all values of p is

Application of Derivatives PYQ — JEE Main 2023 · 25 Jan · Shift 2 · Q30 with Solution

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Correct answer: NO VALUE OF $P$; THE QUESTION STATEMENT LIKELY HAS A TYPO. IF INTERPRETED DIFFERENTLY, THE STORED ANSWER MAY CORRESPOND TO AN ALTERED CONDITION, BUT FOR $X>0$ IT IS IMPOSSIBLE.

We are given $f(x)=2x^3+(2p-7)x^2+3(2p-9)x-6$ and it is stated that the function has a maxima for some value of $x>0$ .

We need the set of all values of $p$ for which a local maximum occurs at some positive $x$ .

Find the critical points using the first derivative: $f'(x)=6x^2+2(2p-7)x+3(2p-9)$ So, $f'(x)=6x^2+(4p-14)x+(6p-27)$
Factor $f'(x)$ . Observe: $6x^2+(4p-14)x+(6p-27)=(2x+2p-9)(3x+3)$ Since

=6x^2+(6p-21)x+(6p-27)$$ This does not match. So let us factor carefully another way. Try dividing by $1$ and checking roots directly. We solve $$6x^2+(4p-14)x+(6p-27)=0$$ Discriminant: $$\Delta=(4p-14)^2-4\cdot 6\cdot (6p-27)$$ $$=16p^2-112p+196-144p+648$$ $$=16p^2-256p+844$$ $$=4(4p^2-64p+211)$$ This is not very useful directly. Let us instead inspect whether the quadratic factors in a simpler form. Take $$f'(x)=6x^2+(4p-14)x+(6p-27)$$ Check $x=-\frac{3}{2}$: $$6\left(\frac{9}{4}\right)+(4p-14)\left(-\frac{3}{2}\right)+(6p-27) =\frac{27}{2}-6p+21+6p-27=\frac{15}{2}\neq 0$$ So not a fixed root. Thus we proceed using the condition for local maximum. 4. Second derivative: $$f''(x)=12x+4p-14$$ At a critical point $x=c$, for it to be a local maximum we need $$f'(c)=0 \quad \text{and} \quad f''(c)<0$$ with also $c>0$. 5. Since $f'(x)$ is a quadratic with leading coefficient positive, the smaller critical point corresponds to local maximum and the larger to local minimum. So we need the smaller root of $$6x^2+(4p-14)x+(6p-27)=0$$ to be positive. Let the roots be $\alpha<\beta$. Then for a cubic with positive leading coefficient, $\alpha$ is the point of local maximum and $\beta$ is the point of local minimum. Hence we need $$\alpha>0$$ 6. For both roots to be positive, necessary and sufficient conditions are: - sum of roots $>0$ - product of roots $>0$ - discriminant $\ge 0$ Now, $$\alpha+\beta=-\frac{4p-14}{6}=\frac{7-2p}{3}$$ So, $$\alpha+\beta>0 \implies 7-2p>0 \implies p<\frac{7}{2}$$ Also, $$\alpha\beta=\frac{6p-27}{6}=p-\frac{9}{2}$$ So, $$\alpha\beta>0 \implies p>\frac{9}{2}$$ These two conditions cannot hold together. Hence both roots cannot be positive. But we only need the smaller root $\alpha$ to be positive. Since $\alpha<\beta$, if $\alpha>0$, then automatically both roots are positive. So this is impossible. Therefore, there is no value of $p$ for which the local maximum occurs at a positive critical point. 7. Since the statement in the question appears to be written as “have a maxima for some value of $x<0$” or “$x\ne 0$” in many standard forms, let us inspect the likely intended interpretation from the options. If the intended condition is that the function has a maximum for some value of $x<0$, then we need the smaller root of $f'(x)=0$ to be negative. For a quadratic with positive leading coefficient, the smaller root is negative unless both roots are positive. From above, both roots are positive is impossible. So the smaller root is always negative whenever real roots exist. Thus we only need real critical points: $$\Delta\ge 0$$ But the options suggest a simple interval in $p$, so let us re-check the derivative for possible factoring from the original expression. Actually, $$f(x)=2x^3+(2p-7)x^2+3(2p-9)x-6$$ Differentiate correctly: $$f'(x)=6x^2+2(2p-7)x+3(2p-9)$$ $$=6x^2+(4p-14)x+(6p-27)$$ Now test $x=-1$: $$f'(-1)=6-(4p-14)+(6p-27)=2p-7$$ Not a fixed root. 8. Let us use the second derivative test at a positive critical point more directly. At a local maximum point $c$, $$f''(c)=12c+4p-14<0$$ If $c>0$, then $$12c>0 \implies 12c+4p-14>4p-14$$ For this to be negative, it is necessary that $$4p-14<0 \implies p<\frac{7}{2}$$ But even then, as shown above, a positive maximum critical point would force both roots positive, impossible because product condition gives $p>\frac{9}{2}$. So indeed no such $p$ exists under the literal reading $x>0$. 9. Therefore, the stored answer $D$ does not agree with the literal text of the question. The most likely issue is a typo in the question statement. Under the literal statement, the answer should be: no value of $p$.

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View written solutionFree

Correct answer: NO VALUE OF $P$; THE QUESTION STATEMENT LIKELY HAS A TYPO. IF INTERPRETED DIFFERENTLY, THE STORED ANSWER MAY CORRESPOND TO AN ALTERED CONDITION, BUT FOR $X>0$ IT IS IMPOSSIBLE.

We are given $f(x)=2x^3+(2p-7)x^2+3(2p-9)x-6$ and it is stated that the function has a maxima for some value of $x>0$ .

We need the set of all values of $p$ for which a local maximum occurs at some positive $x$ .

Find the critical points using the first derivative: $f'(x)=6x^2+2(2p-7)x+3(2p-9)$ So, $f'(x)=6x^2+(4p-14)x+(6p-27)$
Factor $f'(x)$ . Observe: $6x^2+(4p-14)x+(6p-27)=(2x+2p-9)(3x+3)$ Since

=6x^2+(6p-21)x+(6p-27)$$ This does not match. So let us factor carefully another way. Try dividing by $1$ and checking roots directly. We solve $$6x^2+(4p-14)x+(6p-27)=0$$ Discriminant: $$\Delta=(4p-14)^2-4\cdot 6\cdot (6p-27)$$ $$=16p^2-112p+196-144p+648$$ $$=16p^2-256p+844$$ $$=4(4p^2-64p+211)$$ This is not very useful directly. Let us instead inspect whether the quadratic factors in a simpler form. Take $$f'(x)=6x^2+(4p-14)x+(6p-27)$$ Check $x=-\frac{3}{2}$: $$6\left(\frac{9}{4}\right)+(4p-14)\left(-\frac{3}{2}\right)+(6p-27) =\frac{27}{2}-6p+21+6p-27=\frac{15}{2}\neq 0$$ So not a fixed root. Thus we proceed using the condition for local maximum. 4. Second derivative: $$f''(x)=12x+4p-14$$ At a critical point $x=c$, for it to be a local maximum we need $$f'(c)=0 \quad \text{and} \quad f''(c)<0$$ with also $c>0$. 5. Since $f'(x)$ is a quadratic with leading coefficient positive, the smaller critical point corresponds to local maximum and the larger to local minimum. So we need the smaller root of $$6x^2+(4p-14)x+(6p-27)=0$$ to be positive. Let the roots be $\alpha<\beta$. Then for a cubic with positive leading coefficient, $\alpha$ is the point of local maximum and $\beta$ is the point of local minimum. Hence we need $$\alpha>0$$ 6. For both roots to be positive, necessary and sufficient conditions are: - sum of roots $>0$ - product of roots $>0$ - discriminant $\ge 0$ Now, $$\alpha+\beta=-\frac{4p-14}{6}=\frac{7-2p}{3}$$ So, $$\alpha+\beta>0 \implies 7-2p>0 \implies p<\frac{7}{2}$$ Also, $$\alpha\beta=\frac{6p-27}{6}=p-\frac{9}{2}$$ So, $$\alpha\beta>0 \implies p>\frac{9}{2}$$ These two conditions cannot hold together. Hence both roots cannot be positive. But we only need the smaller root $\alpha$ to be positive. Since $\alpha<\beta$, if $\alpha>0$, then automatically both roots are positive. So this is impossible. Therefore, there is no value of $p$ for which the local maximum occurs at a positive critical point. 7. Since the statement in the question appears to be written as “have a maxima for some value of $x<0$” or “$x\ne 0$” in many standard forms, let us inspect the likely intended interpretation from the options. If the intended condition is that the function has a maximum for some value of $x<0$, then we need the smaller root of $f'(x)=0$ to be negative. For a quadratic with positive leading coefficient, the smaller root is negative unless both roots are positive. From above, both roots are positive is impossible. So the smaller root is always negative whenever real roots exist. Thus we only need real critical points: $$\Delta\ge 0$$ But the options suggest a simple interval in $p$, so let us re-check the derivative for possible factoring from the original expression. Actually, $$f(x)=2x^3+(2p-7)x^2+3(2p-9)x-6$$ Differentiate correctly: $$f'(x)=6x^2+2(2p-7)x+3(2p-9)$$ $$=6x^2+(4p-14)x+(6p-27)$$ Now test $x=-1$: $$f'(-1)=6-(4p-14)+(6p-27)=2p-7$$ Not a fixed root. 8. Let us use the second derivative test at a positive critical point more directly. At a local maximum point $c$, $$f''(c)=12c+4p-14<0$$ If $c>0$, then $$12c>0 \implies 12c+4p-14>4p-14$$ For this to be negative, it is necessary that $$4p-14<0 \implies p<\frac{7}{2}$$ But even then, as shown above, a positive maximum critical point would force both roots positive, impossible because product condition gives $p>\frac{9}{2}$. So indeed no such $p$ exists under the literal reading $x>0$. 9. Therefore, the stored answer $D$ does not agree with the literal text of the question. The most likely issue is a typo in the question statement. Under the literal statement, the answer should be: no value of $p$.

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