Let mathrmP( mathrmx) be a polynomial of degree 4 such that mathrmP(1)=7 and attains its minimum value 3 at both x=2 and x=3. If local maximum value of P(x) is (m/n), where m n are relatively prime natural numbers then find the value of (m-3 n).

Question

Let mathrmP( mathrmx) be a polynomial of degree 4 such that mathrmP(1)=7 and attains its minimum value 3 at both x=2 and x=3. If local maximum value of P(x) is (m/n), where m n are relatively prime natural numbers then find the value of (m-3 n). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Θ P(x)=a(x-2)2(x-3)2+3 ∴ P(1)=4 a+3=7 ⇒ a=1 ∴ P(x)=(x-2)2(x-3)2+3 P prime(x)=2(x-2)(x-3)2+2(x-2)2(x-3) =2(x-2)(x-3)(x-3+x-2) =2(x-2)(x-3)(2 x-5) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/17b8ab14e5b3d2ae220244e5a4d5490c-.png" /> ∴ Maximum value of P ( x ) occurs at x =(5/2) ∴ Local maximum value =3+(1/16)=(49/16).

Q. Let P(x) be a polynomial of degree 4 such that P(1)=7 and attains its minimum value 3 at both x=2 and x=3. If local maximum value of P(x) is nm​, where m&n are relatively prime natural numbers then find the value of (m−3n).

ΘP(x)=a(x−2)2(x−3)2+3 ∴P(1)=4a+3=7⇒a=1 ∴P(x)=(x−2)2(x−3)2+3 P′(x)=2(x−2)(x−3)2+2(x−2)2(x−3) =2(x−2)(x−3)(x−3+x−2) =2(x−2)(x−3)(2x−5) ∴ Maximum value of P(x) occurs at x=25​ ∴ Local maximum value =3+161​=1649​.

Q. Let $P (x)$ be a polynomial of degree 4 such that $P (1) = 7$ and attains its minimum value 3 at both $x = 2$ and $x = 3$ . If local maximum value of $P (x)$ is $\frac{m}{n}$ , where $m & n$ are relatively prime natural numbers then find the value of $(m - 3 n)$ .