Question

If a function \[ f:[1,7]\rightarrow\mathbb{R} \] is defined as \[ f(x)= \begin{cases} 2x^2-1, & x\le\sqrt7,[2mm] \sqrt{7x}+6, & \sqrt7[2mm] \sin\pi x, & 5\le x\le6,[2mm] x-[x], & 6<x\le7, \end{cases} \] then the number of points of discontinuity in \([1,7]\) is

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(1\)

Hint:

For piecewise functions, check continuity only at the junction points by comparing the left-hand limit, right-hand limit, and the function value.

Answer 1

The Correct Option is D

Step 1: Check the transition at \(x=\sqrt7\). Left limit: \[ \lim_{x\to\sqrt7^-}(2x^2-1) = 2(7)-1 = 13. \] Right limit: \[ \lim_{x\to\sqrt7^+}(\sqrt{7x}+6) = \sqrt{49}+6 = 13. \] Also, \[ f(\sqrt7)=13. \] Hence, \[ f(x) \] is continuous at \[ x=\sqrt7. \]

Step 2: Check the transition at \(x=5\). Left limit: \[ \lim_{x\to5^-}(\sqrt{7x}+6) = \sqrt{35}+6. \] Right limit: \[ \lim_{x\to5^+}\sin\pi x = \sin5\pi = 0. \] Since \[ \sqrt{35}+6\ne0, \] the function is discontinuous at \[ x=5. \]

Step 3: Check the transition at \(x=6\). Left limit: \[ \lim_{x\to6^-}\sin\pi x = 0. \] Right limit: \[ \lim_{x\to6^+}(x-[x]) = 0. \] Also, \[ f(6)=0. \] Hence, \[ f(x) \] is continuous at \[ x=6. \] Therefore, the only point of discontinuity is \[ x=5. \] Hence, the number of discontinuities is \[ \boxed{1}. \] Thus, \[ \boxed{(D)} \] is the correct answer.