If f(x) is cubic polynomial which has local maximum at x=-1 . If f(2)=18, f(1)=-1 and f^ (x) has local minimum at x=0 , then

Correct answer: B. f(x) is increasing for x ∈[1,2 5 ] Since f(x) has local maxima at x=-1 and f^ (x) has local minima at x=0 . aligned f (x) & =λ x f^ (x) & =λ (x^2)/(2)+c (λ)/(2)+c & =0 λ=-2 c aligned [ f^ (-1)=0] Again, On integrating both sides, we get aligned & f(x)=λ (x^3)/(6)+c x+d & f(2)=λ((8)/(6))+2 c+d=18 & f(1)=(λ)/(6)+c+d=-1 aligned and Using Eqs. (i), (ii) and (iii), we get aligned f(x) & =(1)/(4)(19 x³-57 x+34) f^ (x) & =(1)/(4)(57 x²-57) & =(57)/(4)(x-1)(x+1), using number line rule aligned f(x) is increasing for [1,2 5 ] and f(x) is increasing for [1,2 5 ] and f(x) has local maximum at x=-1 and f(x) has local minimum at C x=1 . Also, f(0)= (34)/(4) .

JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2006 · Mathematics · Application of Derivatives

If f(x) is cubic polynomial which has local maximum at x=-1 . If...

f(x)

is cubic polynomial which has local maximum at

x=-1

. If

f(2)=18, f(1)=-1

and

f^{\prime}(x)

has local minimum at

x=0

, then

Options

Athe distance between $(-1,2)$ and $(a, f(a))$ where $x=a$ is the point of local minima is $2 \sqrt{5}$
B $f(x)$ is increasing for $x \in[1,2 \sqrt{5}]$
C $f(x)$ has local minima at $x=1$
Dthe value of $f(0)=5$

Correct answer

B. $f(x)$ is increasing for $x \in[1,2 \sqrt{5}]$

Step-by-step solution

Since

f(x)

has local maxima at x=-1 and

f^{\prime}(x)

has local minima at

x=0

\begin{aligned}f \wedge(x) & =\lambda x \\ f^{\prime}(x) & =\lambda \frac{x^2}{2}+c \\ \frac{\lambda}{2}+c & =0 \Rightarrow \lambda=-2 c \end{aligned}

\left[

f^{\prime}(-1)=0\right] Again, On integrating both sides, we get

\begin{aligned}& f(x)=\lambda \frac{x^3}{6}+c x+d \\ & f(2)=\lambda\left(\frac{8}{6}\right)+2 c+d=18 \\ & f(1)=\frac{\lambda}{6}+c+d=-1 \end{aligned}

and \therefore

Using Eqs. (i), (ii) and (iii), we get

\begin{aligned}f(x) & =\frac{1}{4}\left(19 x³-57 x+34\right) \\ \therefore \quad f^{\prime}(x) & =\frac{1}{4}\left(57 x²-57\right) \\ & =\frac{57}{4}(x-1)(x+1), \text{ using number line rule } \end{aligned}

\therefore f(x)

is increasing for

[1,2 \sqrt{5}] and f(x)

is increasing for

[1,2 \sqrt{5}] and f(x)

has local maximum at

x=-1 and f(x)

has local minimum at

C x=1

. Also, f(0)=

\frac{34}{4} \text{. }$

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