Exams Joint Entrance Examination Mathematics Application of Derivatives Question 57 Question 57 of 212 Medium NUMERICAL
Let the set of all values of p p p , for which f ( x ) = ( p 2 − 6 p + 8 ) ( sin 2 2 x − cos 2 2 x ) + 2 ( 2 − p ) x + 7 f(x)=({p}^{2}-6p+8)({\sin }^{2}2x-{\cos }^{2}2x)+2(2-p)x+7 f ( x ) = ( p 2 − 6 p + 8 ) ( sin 2 2 x − cos 2 2 x ) + 2 ( 2 − p ) x + 7 does not have any critical point, be the interval ( a , b ) (a,b) ( a , b ) . Then 16 a b 16ab 16 ab is equal to \_\_\_\_\_\_\_\_\_.
Answer: 252
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f ( x ) = ( p 2 − 6 p + 8 ) ( sin 2 2 x − cos 2 2 x ) + 2 ( 2 − p ) x + 7 f ( x ) = − cos 4 x ( p 2 − 6 p + 8 ) + 2 ( 2 − p ) x + 7 f ′ ( x ) = 4 sin 4 x ( p 2 − 6 p + 8 ) + 2 ( 2 − p ) ≠ 0 2 ( 2 − p ) + [ − 4 ( p 2 − 6 p + 8 ) , 4 ( p 2 − 6 p + 8 ) ] ⇒ [ − 4 p 2 + 24 p − 32 , 4 p 2 − 24 p + 32 ] + ( 4 − 2 p ) [ − 4 p 2 + 22 p − 28 , 4 p 2 − 26 p + 36 ] [ ( p − 2 ) ( − 4 p + 14 ) , ( p − 2 ) ( 4 p − 18 ) ] ⇒ ( p − 2 ) [ ( − 4 p + 14 ) , 4 p − 18 ] ⇒ p ∈ ( 7 2 , 9 2 ) ⇒ a = 7 2 , b = 9 2 ⇒ 16 a b = 4 × 63 = 252 \begin{gathered} f(x)=({p}^{2}-6p+8)({\sin }^{2}2x-{\cos }^{2}2x)+2(2-p)x+7 \\ f(x)=-\cos 4x({p}^{2}-6p+8)+2(2-p)x+7 \\ {f}^{'}(x)=4\sin 4x({p}^{2}-6p+8)+2(2-p)\ne 0 \\ 2(2-p)+[-4({p}^{2}-6p+8),4({p}^{2}-6p+8)] \\ \Rightarrow [-4{p}^{2}+24p-32,4{p}^{2}-24p+32]+(4-2p) \\ [-4{p}^{2}+22p-28,4{p}^{2}-26p+36] \\ [(p-2)(-4p+14),(p-2)(4p-18)] \\ \Rightarrow (p-2)[(-4p+14),4p-18]\Rightarrow p\in (\frac{7}{2},\frac{9}{2}) \\ \Rightarrow a=\frac{7}{2},b=\frac{9}{2} \\ \Rightarrow 16ab=4\times 63=252 \end{gathered} f ( x ) = ( p 2 − 6 p + 8 ) ( sin 2 2 x − cos 2 2 x ) + 2 ( 2 − p ) x + 7 f ( x ) = − cos 4 x ( p 2 − 6 p + 8 ) + 2 ( 2 − p ) x + 7 f ′ ( x ) = 4 sin 4 x ( p 2 − 6 p + 8 ) + 2 ( 2 − p ) = 0 2 ( 2 − p ) + [ − 4 ( p 2 − 6 p + 8 ) , 4 ( p 2 − 6 p + 8 )] ⇒ [ − 4 p 2 + 24 p − 32 , 4 p 2 − 24 p + 32 ] + ( 4 − 2 p ) [ − 4 p 2 + 22 p − 28 , 4 p 2 − 26 p + 36 ] [( p − 2 ) ( − 4 p + 14 ) , ( p − 2 ) ( 4 p − 18 )] ⇒ ( p − 2 ) [( − 4 p + 14 ) , 4 p − 18 ] ⇒ p ∈ ( 2 7 , 2 9 ) ⇒ a = 2 7 , b = 2 9 ⇒ 16 ab = 4 × 63 = 252 Answer the question to unlock