JEE Main 2022 — Application Of Derivatives Question with Solution

Given,

$$f(x)=4x^3-11x^2+8x-5$$

$$\therefore$$ $$f^{\prime }(x)=12x^2-22x+8$$

$$=2(6x^2-11x+4)$$

$$=2(6x^2-8x-3x+4)$$

$$=2\left[2x(3x-4)-1(3x-4)\right]$$

$$=2\left[(3x-4)(2x-1)\right]$$

$$f^{\prime }(x)$$ is positive before $$\frac{1}{2}$$ means slope of f(x) is positive before $$\frac{1}{2}$$ and f'(x) is negative after $$\frac{1}{2}$$ means slope of f(x) is negative after $$\frac{1}{2}$$.

$$\therefore$$ At $$\frac{1}{2}$$, f(x) is local maximum

Also, $$f^{\prime }(x)$$ is negative before $$\frac{4}{3}$$ means slope of f(x) is negative before $$\frac{4}{3}$$ and f'(x) is positive after $$\frac{4}{3}$$ means slope of f(x) is positive after $$\frac{4}{3}$$.

$$\therefore$$ At $$\frac{4}{3},f(x)$$ is local minimum

From wavy curve method,

$$f^{\prime }(x)>0,\forall x\in \left(-\alpha ,\frac{1}{2}\right)\cup \left(\frac{4}{3},\alpha \right)$$

$$\therefore$$ f(x) increasing in $$x\in \left(-\alpha ,\frac{1}{2}\right)\cup \left(\frac{4}{3},\alpha \right)$$

$$f^{\prime }(x)<0\forall x\in \left(\frac{1}{2},\frac{4}{3}\right)$$

$$\therefore$$ f(x) decreasing in $$x\in \left(\frac{1}{2},\frac{4}{3}\right)$$