The area bounded by fx= sin⁡2x x≥0 cos⁡2x x<0 with the x-axis, x=-(π)/4 and x=(π)/4 is k square units. Then, the value of 4k is equal to

Correct answer: 4. Required area is A=∫-(π)/40cos⁡2xdx+∫0(π)/4sin⁡2xdx =(sin⁡2x)/2-(π)/40-(cos⁡2x)/20(π)/4 =(0--1)/2-(0-1)/2=1 sq. units ⇒k=1 ∴4k=4

Q27 / 90 · NTA Abhyas (JEE Main) 2020 · Mathematics · Area Under Curves

The area bounded by $f (x) = \{\begin{cases} \sin (2 x) & x \geq 0 \\ \cos (2 x) & x < 0 \end{cases}$ with the $x$ -axis, $x = - \frac{π}{4}$ and $x = \frac{π}{4}$ is $k$ square units. Then, the value of $4 k$ is equal to

Correct answer: 4

Required area is

A = \int_{- \frac{π}{4}}^{0} \cos 2 x d x + \int_{0}^{\frac{π}{4}} \sin 2 x d x

= {[\frac{\sin 2 x}{2}]}_{- \frac{π}{4}}^{0} - {[\frac{\cos 2 x}{2}]}_{0}^{\frac{π}{4}}

= \frac{0 - (- 1)}{2} - \frac{(0 - 1)}{2} = 1

sq. units

\Rightarrow k = 1

∴ 4 k = 4

Hence, the correct answer is 4.