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Application of Integrals: Coordinate Plane Region Consists Points Satisfying Inequalit

JEE Maths question with a full step-by-step solution.

Question
In the coordinate plane, the region MM consists of all the points (x,y)\left(x,y\right) satisfying the inequalities y≥0y \ge 0, y≤xy \le x and y≤2−xy \le 2-x simultaneously. The region NN, which varies with the parameter tt, consists of all the points (x,y)\left(x,y\right) satisfying the inequalities t≤x≤t+1t \le x \le t+1 and 0≤t≤10 \le t \le 1 simultaneously. The area of M∩NM \cap N is −k1t2+k2t+12-k_1t^2+k_2t+\dfrac12, then k1+k2k_1+k_2 is
Solution
Answer: 2
Question attachment Step 1: y≥0y \ge 0, y≤xy \le x and y≤2−xy \le 2-x meet in the triangle with vertices
(0,0),(1,1),(2,0)\left(0,0\right),\qquad \left(1,1\right),\qquad \left(2,0\right)
 whose base is 22 and height 11, so its area is 11. Step 2: NN is the vertical strip t≤x≤t+1t \le x \le t+1 of width 11, sliding as tt takes the values in [0,1]\left[0,1\right]. Because 0≤t0 \le t and t+1≤2t+1 \le 2, the strip always lies inside [0,2]\left[0,2\right], the range of MM. Step 3: For 0≤x≤20 \le x \le 2 the cross-section of MM is 0≤y≤h(x)0 \le y \le h\left(x\right), where
h(x)={x,0≤x≤1,2−x,1≤x≤2.h\left(x\right) = \begin{cases} x , & 0 \le x \le 1 ,\\ 2-x , & 1 \le x \le 2 .\end{cases}
 So M∩NM\cap N has area ∫tt+1h(x)dx\displaystyle\int_t^{t+1}h\left(x\right)dx, and since t≤1≤t+1t \le 1 \le t+1 the integral splits at x=1x = 1. Step 4:
∫t1x dx=[x22]t1=12−t22\int_t^{1}x\,dx = \left[\frac{x^2}{2}\right]_t^1 = \frac12-\frac{t^2}{2}
∫1t+1(2−x)dx=[2x−x22]1t+1=(2t+2−(t+1)22)−32\int_1^{t+1}\left(2-x\right)dx = \left[2x-\frac{x^2}{2}\right]_1^{t+1} = \left(2t+2-\frac{\left(t+1\right)^2}{2}\right)-\frac32
 Step 5: On adding,
Area=12−t22+2t+2−t2+2t+12−32=−t2+t+12\text{Area} = \frac12-\frac{t^2}{2}+2t+2-\frac{t^2+2t+1}{2}-\frac32 = -t^2+t+\frac12
 The same expression comes from the area of MM minus the two corner triangles cut off, 1−t22−(1−t)221-\dfrac{t^2}{2}-\dfrac{\left(1-t\right)^2}{2}. Step 6: Both sides are quadratics in tt agreeing for every tt in [0,1]\left[0,1\right], so their coefficients are equal:
−k1t2+k2t+12=−t2+t+12⟹k1=1,k2=1-k_1t^2+k_2t+\frac12 = -t^2+t+\frac12 \quad\Longrightarrow\quad k_1 = 1 ,\quad k_2 = 1
k1+k2=2k_1+k_2 = 2
 Answer: 22
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