The sum of the absolute maximum and minimum values of the function f(x)=|x^2-5 x+6|-3 x+2 in the interval [-1,3] is equal to :

Application of Derivatives PYQ — JEE Main 2023 · 1 Feb · Shift 2 · Q26 with Solution

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Correct answer: C

We need the absolute maximum and minimum of $f(x)=|x^2-5x+6|-3x+2$ on the interval $[-1,3]$ .

First factor the quadratic inside modulus: $x^2-5x+6=(x-2)(x-3).$

So we must determine where this expression is positive/negative on $[-1,3]$ .

Since the roots are $x=2,3$ :

Hence,

x^2-5x+6-3x+2=x^2-8x+8, & -1\le x\le 2,\\[4pt] (-x^2+5x-6)-3x+2=-x^2+2x-4, & 2\le x\le 3. \end{cases}$$ --- 3. **Find extrema on each interval** ### On $[-1,2]$ $$f(x)=x^2-8x+8$$ This is an upward opening parabola. Its derivative is $$f'(x)=2x-8.$$ Setting $f'(x)=0$ gives $$2x-8=0 \Rightarrow x=4,$$ which is **not** in $[-1,2]$. So on $[-1,2]$, extrema occur at endpoints: $$f(-1)=(-1)^2-8(-1)+8=1+8+8=17,$$ $$f(2)=2^2-8(2)+8=4-16+8=-4.$$ Thus on $[-1,2]$: - maximum = $17$ - minimum = $-4$ --- ### On $[2,3]$ $$f(x)=-x^2+2x-4$$ This is a downward opening parabola. Its derivative is $$f'(x)=-2x+2.$$ Setting $f'(x)=0$ gives $$-2x+2=0 \Rightarrow x=1,$$ which is **not** in $[2,3]$. So on $[2,3]$, extrema occur at endpoints: $$f(2)=-4+4-4=-4,$$ $$f(3)=-9+6-4=-7.$$ Thus on $[2,3]$: - maximum = $-4$ - minimum = $-7$ --- 4. **Absolute maximum and minimum on $[-1,3]$** Collect all relevant values: $$f(-1)=17,\quad f(2)=-4,\quad f(3)=-7.$$ Therefore: - **Absolute maximum value** = $17$ - **Absolute minimum value** = $-7$ Their sum is $$17+(-7)=10.$$ --- 5. **Compare with options** The required sum is $$\boxed{10}.$$ So the correct option is **C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.

View written solutionFree

Correct answer: C

We need the absolute maximum and minimum of $f(x)=|x^2-5x+6|-3x+2$ on the interval $[-1,3]$ .

First factor the quadratic inside modulus: $x^2-5x+6=(x-2)(x-3).$

So we must determine where this expression is positive/negative on $[-1,3]$ .

Since the roots are $x=2,3$ :

Hence,

x^2-5x+6-3x+2=x^2-8x+8, & -1\le x\le 2,\\[4pt] (-x^2+5x-6)-3x+2=-x^2+2x-4, & 2\le x\le 3. \end{cases}$$ --- 3. **Find extrema on each interval** ### On $[-1,2]$ $$f(x)=x^2-8x+8$$ This is an upward opening parabola. Its derivative is $$f'(x)=2x-8.$$ Setting $f'(x)=0$ gives $$2x-8=0 \Rightarrow x=4,$$ which is **not** in $[-1,2]$. So on $[-1,2]$, extrema occur at endpoints: $$f(-1)=(-1)^2-8(-1)+8=1+8+8=17,$$ $$f(2)=2^2-8(2)+8=4-16+8=-4.$$ Thus on $[-1,2]$: - maximum = $17$ - minimum = $-4$ --- ### On $[2,3]$ $$f(x)=-x^2+2x-4$$ This is a downward opening parabola. Its derivative is $$f'(x)=-2x+2.$$ Setting $f'(x)=0$ gives $$-2x+2=0 \Rightarrow x=1,$$ which is **not** in $[2,3]$. So on $[2,3]$, extrema occur at endpoints: $$f(2)=-4+4-4=-4,$$ $$f(3)=-9+6-4=-7.$$ Thus on $[2,3]$: - maximum = $-4$ - minimum = $-7$ --- 4. **Absolute maximum and minimum on $[-1,3]$** Collect all relevant values: $$f(-1)=17,\quad f(2)=-4,\quad f(3)=-7.$$ Therefore: - **Absolute maximum value** = $17$ - **Absolute minimum value** = $-7$ Their sum is $$17+(-7)=10.$$ --- 5. **Compare with options** The required sum is $$\boxed{10}.$$ So the correct option is **C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.