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Functions — JEE Maths practice question

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

Question
The domain of the function y=sin⁡x+cos⁡x+7x−x2−6y=\sqrt{\sin x+\cos x}+\sqrt{7x-x^{2}-6} is:
A[1,6][1,6]
B[1,3π4]∪[7π4,6]\left[1,\dfrac{3\pi}{4}\right]\cup\left[\dfrac{7\pi}{4},6\right]correct
C[1,π]∪[7π4,6]\left[1,\pi\right]\cup\left[\dfrac{7\pi}{4},6\right]
DNone of these
Solution
Step 1: First factor: sin⁡x+cos⁡x≥0\sqrt{\sin x+\cos x}\ge 0 requires sin⁡x+cos⁡x≥0\sin x+\cos x\ge 0. Write sin⁡x+cos⁡x=2sin⁡ ⁣(x+π4)≥0  ⇒  sin⁡ ⁣(x+π4)≥0\sin x+\cos x=\sqrt{2}\sin\!\left(x+\dfrac{\pi}{4}\right)\ge 0\;\Rightarrow\;\sin\!\left(x+\dfrac{\pi}{4}\right)\ge 0. This gives (for general nn): 2nπ−π4≤x≤2nπ+π−π4=2nπ+3π42n\pi-\dfrac{\pi}{4}\le x\le 2n\pi+\pi-\dfrac{\pi}{4}=2n\pi+\dfrac{3\pi}{4}. For n=0n=0: −π4≤x≤3π4-\dfrac{\pi}{4}\le x\le\dfrac{3\pi}{4}. For n=1n=1: 7π4≤x≤11π4\dfrac{7\pi}{4}\le x\le\dfrac{11\pi}{4}, etc. Step 2: Second factor: 7x−x2−6≥0  ⇒  x2−7x+6≤0  ⇒  (x−1)(x−6)≤0  ⇒  x∈[1,6]7x-x^{2}-6\ge 0\;\Rightarrow\;x^{2}-7x+6\le 0\;\Rightarrow\;(x-1)(x-6)\le 0\;\Rightarrow\;x\in[1,6]. Step 3: Intersect with x∈[1,6]x\in[1,6]: From n=0n=0 condition −π4≤x≤3π4-\dfrac{\pi}{4}\le x\le\dfrac{3\pi}{4}: intersect with [1,6][1,6] gives [1,3π4]\left[1,\dfrac{3\pi}{4}\right] (since 3π4≈2.36>1\dfrac{3\pi}{4}\approx 2.36>1). From n=1n=1 condition 7π4≤x≤11π4\dfrac{7\pi}{4}\le x\le\dfrac{11\pi}{4}: 7π4≈5.50\dfrac{7\pi}{4}\approx 5.50 and 11π4≈8.64\dfrac{11\pi}{4}\approx 8.64. Intersect with [1,6][1,6] gives [7π4,6]\left[\dfrac{7\pi}{4},6\right]. Domain =[1,3π4]∪[7π4,6]=\left[1,\dfrac{3\pi}{4}\right]\cup\left[\dfrac{7\pi}{4},6\right]. Answer: (2) [1,3π4]∪[7π4,6]\left[1,\dfrac{3\pi}{4}\right]\cup\left[\dfrac{7\pi}{4},6\right]
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