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← JEE Advanced 2014 · Paper 1
MathematicsInteger+3Question 58 of 60

JEE Advanced 2014 · Paper 1, Mathematics question 58

Let f:[0,4π]→[0,π]f:[0,4\pi]\to[0,\pi] be defined by f(x)=cos⁡−1(cos⁡x)f(x)=\cos^{-1}(\cos x). The number of points x∈[0,4π]x\in[0,4\pi] satisfying the equation
f(x)=10−x10f(x)=\frac{10-x}{10}
is

Answer and worked solution

Official answer: 3

The graph of ff. cos⁡−1(cos⁡x)\cos^{-1}(\cos x) is a triangle wave:
f(x)={x0≤x≤π2π−xπ≤x≤2πx−2π2π≤x≤3π4π−x3π≤x≤4πf(x)=\begin{cases}x&0\le x\le\pi\\2\pi-x&\pi\le x\le2\pi\\x-2\pi&2\pi\le x\le3\pi\\4\pi-x&3\pi\le x\le4\pi\end{cases}

Line. y=1−x10y=1-\frac x{10} falls from 1 to 1−0.4π≈−0.261-0.4\pi\approx-0.26.

Solve on each piece.

  • x=1−x10x=1-\frac x{10}: x=1011≈0.91x=\frac{10}{11}\approx0.91, in [0,π][0,\pi]. ✓
  • 2π−x=1−x102\pi-x=1-\frac x{10}: x=2π−10.9≈5.87x=\frac{2\pi-1}{0.9}\approx5.87, in [π,2π][\pi,2\pi]. ✓
  • x−2π=1−x10x-2\pi=1-\frac x{10}: x=2π+11.1≈6.62x=\frac{2\pi+1}{1.1}\approx6.62, in [2π,3π][2\pi,3\pi]. ✓
  • 4π−x=1−x104\pi-x=1-\frac x{10}: x=4π−10.9≈12.85>4πx=\frac{4\pi-1}{0.9}\approx12.85>4\pi. ✗

Answer: 3