The sum of absolute maximum and absolute minimum values of the function f(x) = |2x^2 + 3x - 2| + \sin x\cos x in the interval [0, 1] is :

3 + 1 \over 2(1 + 2\cos (1))\sin (1)

3 + \sin (1)\cos ^2( 1 \over 2 ) \over 2

5 + 1 \over 2(\sin (1) + \sin (2))

2 + \sin ( 1 \over 2 )\cos ( 1 \over 2 )

Application of Derivatives PYQ — JEE Main 2022 · 24 Jun · Shift 1 · Q30 with Solution

View written solutionFree

Correct answer: B

Simplify the absolute value part

We have $f(x)=|2x^2+3x-2|+\sin x\cos x, \qquad x\in[0,1].$

First study $q(x)=2x^2+3x-2=(2x-1)(x+2).$

Its roots are $x=\frac12$ and $x=-2$ . On the interval $[0,1]$ , only $x=\frac12$ matters.

Now check the sign:

For $0\le x<\frac12$ , $q(x)<0$
For $\frac12\le x\le 1$ , $q(x)\ge 0$

Hence

\begin{cases} -(2x^2+3x-2)= -2x^2-3x+2, & 0\le x<\frac12,\\[4pt] 2x^2+3x-2, & \frac12\le x\le 1. \end{cases}$$ Also, $$\sin x\cos x=\frac12\sin 2x.$$ So $$f(x)= \begin{cases} -2x^2-3x+2+\frac12\sin 2x, & 0\le x<\frac12,\\[4pt] 2x^2+3x-2+\frac12\sin 2x, & \frac12\le x\le 1. \end{cases}$$ --- 2. **Find extrema on $\left[0,\frac12\right)$** Let $$f_1(x)=-2x^2-3x+2+\frac12\sin 2x.$$ Then $$f_1'(x)=-4x-3+\cos 2x.$$ Since $\cos 2x\le 1$, $$f_1'(x)\le -4x-3+1=-4x-2<0 \quad \text{for } x\in[0,\tfrac12).$$ So $f_1$ is strictly decreasing on $[0,\tfrac12)$. Therefore on this interval: - maximum occurs at $x=0$ - minimum occurs as we approach $x=\frac12$ Compute: $$f(0)=|{-2}|+0=2.$$ And $$f\left(\frac12\right)=|0|+\sin\frac12\cos\frac12=\frac12\sin 1.$$ --- 3. **Find extrema on $\left[\frac12,1\right]$** Let $$f_2(x)=2x^2+3x-2+\frac12\sin 2x.$$ Then $$f_2'(x)=4x+3+\cos 2x.$$ Since $\cos 2x\ge -1$, $$f_2'(x)\ge 4x+3-1=4x+2>0 \quad \text{for } x\in\left[\tfrac12,1\right].$$ So $f_2$ is strictly increasing on $\left[\frac12,1\right]$. Therefore on this interval: - minimum occurs at $x=\frac12$ - maximum occurs at $x=1$ Compute: $$f(1)=|2+3-2|+\sin 1\cos 1=3+\frac12\sin 2.$$ --- 4. **Absolute maximum and minimum on $[0,1]$** From monotonicity: - On $[0,\frac12]$, function decreases from $2$ to $\frac12\sin 1$ - On $[\frac12,1]$, function increases from $\frac12\sin 1$ to $3+\frac12\sin 2$ Thus, $$\text{absolute minimum}=f\left(\frac12\right)=\frac12\sin 1,$$ $$\text{absolute maximum}=f(1)=3+\frac12\sin 2.$$ Their sum is $$\frac12\sin 1+3+\frac12\sin 2.$$ Using $$\sin 2=2\sin 1\cos 1,$$ we get $$3+\frac12\sin 1+\frac12\sin 2 =3+\frac12\sin 1+\sin 1\cos 1 =3+\frac12(1+2\cos 1)\sin 1.$$ This matches **Option B**. --- 5. **Compare with stored correct answer** Stored correct answer: **B** Derived answer: **B** So the derived answer agrees with the stored answer.

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View written solutionFree

Correct answer: B

Simplify the absolute value part

We have $f(x)=|2x^2+3x-2|+\sin x\cos x, \qquad x\in[0,1].$

First study $q(x)=2x^2+3x-2=(2x-1)(x+2).$

Its roots are $x=\frac12$ and $x=-2$ . On the interval $[0,1]$ , only $x=\frac12$ matters.

Now check the sign:

For $0\le x<\frac12$ , $q(x)<0$
For $\frac12\le x\le 1$ , $q(x)\ge 0$

Hence

\begin{cases} -(2x^2+3x-2)= -2x^2-3x+2, & 0\le x<\frac12,\\[4pt] 2x^2+3x-2, & \frac12\le x\le 1. \end{cases}$$ Also, $$\sin x\cos x=\frac12\sin 2x.$$ So $$f(x)= \begin{cases} -2x^2-3x+2+\frac12\sin 2x, & 0\le x<\frac12,\\[4pt] 2x^2+3x-2+\frac12\sin 2x, & \frac12\le x\le 1. \end{cases}$$ --- 2. **Find extrema on $\left[0,\frac12\right)$** Let $$f_1(x)=-2x^2-3x+2+\frac12\sin 2x.$$ Then $$f_1'(x)=-4x-3+\cos 2x.$$ Since $\cos 2x\le 1$, $$f_1'(x)\le -4x-3+1=-4x-2<0 \quad \text{for } x\in[0,\tfrac12).$$ So $f_1$ is strictly decreasing on $[0,\tfrac12)$. Therefore on this interval: - maximum occurs at $x=0$ - minimum occurs as we approach $x=\frac12$ Compute: $$f(0)=|{-2}|+0=2.$$ And $$f\left(\frac12\right)=|0|+\sin\frac12\cos\frac12=\frac12\sin 1.$$ --- 3. **Find extrema on $\left[\frac12,1\right]$** Let $$f_2(x)=2x^2+3x-2+\frac12\sin 2x.$$ Then $$f_2'(x)=4x+3+\cos 2x.$$ Since $\cos 2x\ge -1$, $$f_2'(x)\ge 4x+3-1=4x+2>0 \quad \text{for } x\in\left[\tfrac12,1\right].$$ So $f_2$ is strictly increasing on $\left[\frac12,1\right]$. Therefore on this interval: - minimum occurs at $x=\frac12$ - maximum occurs at $x=1$ Compute: $$f(1)=|2+3-2|+\sin 1\cos 1=3+\frac12\sin 2.$$ --- 4. **Absolute maximum and minimum on $[0,1]$** From monotonicity: - On $[0,\frac12]$, function decreases from $2$ to $\frac12\sin 1$ - On $[\frac12,1]$, function increases from $\frac12\sin 1$ to $3+\frac12\sin 2$ Thus, $$\text{absolute minimum}=f\left(\frac12\right)=\frac12\sin 1,$$ $$\text{absolute maximum}=f(1)=3+\frac12\sin 2.$$ Their sum is $$\frac12\sin 1+3+\frac12\sin 2.$$ Using $$\sin 2=2\sin 1\cos 1,$$ we get $$3+\frac12\sin 1+\frac12\sin 2 =3+\frac12\sin 1+\sin 1\cos 1 =3+\frac12(1+2\cos 1)\sin 1.$$ This matches **Option B**. --- 5. **Compare with stored correct answer** Stored correct answer: **B** Derived answer: **B** So the derived answer agrees with the stored answer.

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