Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If \mathop lim\limits_x \to 0 ( f( x ) \over x^2 + 1 ) = 3 then f( - 1) is equal to :

Limits Continuity and Differentiability PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q41 with Solution

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Correct answer: A

Since $f(x)$ is a polynomial of degree $4$ and has extreme values at $x=1$ and $x=2$ , we must have $f'(1)=0,\qquad f'(2)=0.$

Because $f'(x)$ is a cubic polynomial, let $f'(x)=k(x-1)(x-2)(x-a)$ for some constants $k\neq 0$ and $a$ .

Now use the given limit: $\lim_{x\to 0}\left(\frac{f(x)}{x^2}+1\right)=3.$ This gives $\lim_{x\to 0}\frac{f(x)}{x^2}=2.$ For this limit to be finite, we must have $f(0)=0 \quad \text{and} \quad f'(0)=0.$ Also, $\frac{f(x)}{x^2}\to 2 \implies f''(0)=4,$ because near $x=0$ , $f(x)=\frac{f''(0)}{2}x^2+\cdots$ .

Thus $x=0$ is a double root of $f(x)$ , so write $f(x)=x^2(ax^2+bx+c).$ Since $\lim_{x\to 0}\frac{f(x)}{x^2}=c=2,$ we get $f(x)=x^2(ax^2+bx+2).$

Differentiate: $f'(x)=4ax^3+3bx^2+4x.$ Given $f'(1)=0$ and $f'(2)=0$ , $4a+3b+4=0 \qquad (1)$ $32a+12b+8=0 \qquad (2)$

From (2), dividing by $4$ : $8a+3b+2=0 \qquad (3)$ Subtract (1) from (3): $4a-2=0 \implies a=\frac12.$ Then from (1):

\implies 2+3b+4=0 \implies 3b=-6 \implies b=-2.$$ Hence $$f(x)=x^2\left(\frac12 x^2-2x+2\right).$$ 4. Now compute $f(-1)$: $$f(-1)=(-1)^2\left(\frac12(-1)^2-2(-1)+2\right)$$ $$=1\left(\frac12+2+2\right)=\frac12+4=\frac92.$$ 5. Therefore, $$\boxed{f(-1)=\frac92}.$$ So the correct option is **A**.

View written solutionFree

Correct answer: A

Since $f(x)$ is a polynomial of degree $4$ and has extreme values at $x=1$ and $x=2$ , we must have $f'(1)=0,\qquad f'(2)=0.$

Because $f'(x)$ is a cubic polynomial, let $f'(x)=k(x-1)(x-2)(x-a)$ for some constants $k\neq 0$ and $a$ .

Now use the given limit: $\lim_{x\to 0}\left(\frac{f(x)}{x^2}+1\right)=3.$ This gives $\lim_{x\to 0}\frac{f(x)}{x^2}=2.$ For this limit to be finite, we must have $f(0)=0 \quad \text{and} \quad f'(0)=0.$ Also, $\frac{f(x)}{x^2}\to 2 \implies f''(0)=4,$ because near $x=0$ , $f(x)=\frac{f''(0)}{2}x^2+\cdots$ .

Thus $x=0$ is a double root of $f(x)$ , so write $f(x)=x^2(ax^2+bx+c).$ Since $\lim_{x\to 0}\frac{f(x)}{x^2}=c=2,$ we get $f(x)=x^2(ax^2+bx+2).$

Differentiate: $f'(x)=4ax^3+3bx^2+4x.$ Given $f'(1)=0$ and $f'(2)=0$ , $4a+3b+4=0 \qquad (1)$ $32a+12b+8=0 \qquad (2)$

From (2), dividing by $4$ : $8a+3b+2=0 \qquad (3)$ Subtract (1) from (3): $4a-2=0 \implies a=\frac12.$ Then from (1):

\implies 2+3b+4=0 \implies 3b=-6 \implies b=-2.$$ Hence $$f(x)=x^2\left(\frac12 x^2-2x+2\right).$$ 4. Now compute $f(-1)$: $$f(-1)=(-1)^2\left(\frac12(-1)^2-2(-1)+2\right)$$ $$=1\left(\frac12+2+2\right)=\frac12+4=\frac92.$$ 5. Therefore, $$\boxed{f(-1)=\frac92}.$$ So the correct option is **A**.