Question:
\( f(x)=
\begin{cases}
|x^3 + x^2 + 3x + \sin x|\left(3 + \sin\frac{1}{x}\right), & x\neq 0 \\
0, & x=0
\end{cases}
\)
The number of points, where \( f(x) \) attains its minimum value, is
\( f(x)=
\begin{cases}
|x^3 + x^2 + 3x + \sin x|\left(3 + \sin\frac{1}{x}\right), & x\neq 0 \\
0, & x=0
\end{cases}
\)
The number of points, where \( f(x) \) attains its minimum value, is
Show Hint
For absolute value expressions, minimum occurs when inside becomes zero.
Updated On: Apr 23, 2026
- 1
- 2
- 3
- infinitely many
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
Minimum occurs when expression becomes smallest possible value.
Step 1: Analyze components.
\[ |x^3 + x^2 + 3x + \sin x| \ge 0 \] \[ 3 + \sin\frac{1}{x} \ge 2 \]
Step 2: Minimum possible value.
\[ f(x) \ge 0 \]
Step 3: Check when $f(x)=0$.
Occurs when: \[ |x^3 + x^2 + 3x + \sin x| = 0 \] \[ \Rightarrow x^3 + x^2 + 3x + \sin x = 0 \]
Step 4: Check solution.
At $x=0$: \[ f(0)=0 \] For $x\neq 0$, equation has no real solution due to dominance of polynomial.
Step 5: Conclusion.
Minimum occurs only at $x=0$ Conclusion:
Number of points = 1
Step 1: Analyze components.
\[ |x^3 + x^2 + 3x + \sin x| \ge 0 \] \[ 3 + \sin\frac{1}{x} \ge 2 \]
Step 2: Minimum possible value.
\[ f(x) \ge 0 \]
Step 3: Check when $f(x)=0$.
Occurs when: \[ |x^3 + x^2 + 3x + \sin x| = 0 \] \[ \Rightarrow x^3 + x^2 + 3x + \sin x = 0 \]
Step 4: Check solution.
At $x=0$: \[ f(0)=0 \] For $x\neq 0$, equation has no real solution due to dominance of polynomial.
Step 5: Conclusion.
Minimum occurs only at $x=0$ Conclusion:
Number of points = 1
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