Question:
The least positive value of \(a\) for which the equation
\[
\int_{0}^{x}(t^{2}-8t+13)dt=x\sin\frac{a}{x}
\]
has a solution is:
The least positive value of \(a\) for which the equation
\[
\int_{0}^{x}(t^{2}-8t+13)dt=x\sin\frac{a}{x}
\]
has a solution is:
Show Hint
Whenever a trigonometric equation forces a condition like $\text{quadratic polynomial} = \sin\theta$, check the vertex of the quadratic first. Often, the vertex value is exactly 1 or -1, which collapses the entire range of solutions down to a single contact point!
Updated On: May 25, 2026
- $3\pi$
- $4\pi$
- $\pi$
- $2\pi$
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
To find the solutions of an equation containing a definite integral variable, we first evaluate the integral analytically to find its polynomial form. We can then use the mathematical boundaries of trigonometric functions (like knowing $-1 \le \sin\theta \le 1$) to find the valid range for the parameter $a$.
Step 1: Evaluate the definite integral on the left side.
Integrate the polynomial expression with respect to $t$: $$\int_{0}^{x}(t^{2}-8t+13)\,dt = \left[ \frac{t^3}{3} - \frac{8t^2}{2} + 13t \right]_{0}^{x} = \frac{x^3}{3} - 4x^2 + 13x$$ Substitute this back into the primary equation: $$\frac{x^3}{3} - 4x^2 + 13x = x \sin\frac{a}{x}$$
Step 2: Isolate the sine function term.
Since $x \neq 0$ for the expression $\frac{a}{x}$ to be well-defined, divide the entire equation by $x$: $$\frac{x^2}{3} - 4x + 13 = \sin\frac{a}{x} \quad \cdots (1)$$
Step 3: Find the minimum value of the quadratic expression.
For equation (1) to have a valid real root solution, the value of the quadratic polynomial on the left side must fall within the natural output boundaries of the sine function: $$-1 \le \frac{x^2}{3} - 4x + 13 \le 1$$ Let us find the absolute minimum point of this quadratic function, $\phi(x) = \frac{x^2}{3} - 4x + 13$, by setting its derivative to zero: $$\phi'(x) = \frac{2x}{3} - 4 = 0 \quad \Rightarrow \quad \frac{2x}{3} = 4 \quad \Rightarrow \quad x = 6$$ Calculate the minimum value at this vertex point: $$\phi(6) = \frac{6^2}{3} - 4(6) + 13 = \frac{36}{3} - 24 + 13 = 12 - 24 + 13 = 1$$ This shows that the minimum value of the quadratic function is exactly 1.
Step 4: Solve for the parameter $a$.
Since the left side's minimum value is 1 and the right side's maximum value is 1, the two curves can only meet at this exact peak touchpoint: $$\sin\frac{a}{x} = 1 \quad \text{at } x = 6$$ Substitute $x = 6$ into the equation: $$\sin\left(\frac{a}{6}\right) = 1$$ The smallest positive angle where the sine function reaches 1 is $\frac{\pi}{2}$: $$\frac{a}{6} = \frac{\pi}{2} \quad \Rightarrow \quad a = \frac{6\pi}{2} = 3\pi$$ Let's re-verify the option values and the quadratic coefficients from the image text. The equation reads: $\int_{0}^{x}(t^{2}-8t+13)dt=x \sin\frac{a}{x}$. If the minimum value is exactly 1, then $a/6 = \pi/2 \implies a = 3\pi$, which matches option (A). % Final Correct Answer Selection Update Correct Answer: (A) $3\pi$
Integrate the polynomial expression with respect to $t$: $$\int_{0}^{x}(t^{2}-8t+13)\,dt = \left[ \frac{t^3}{3} - \frac{8t^2}{2} + 13t \right]_{0}^{x} = \frac{x^3}{3} - 4x^2 + 13x$$ Substitute this back into the primary equation: $$\frac{x^3}{3} - 4x^2 + 13x = x \sin\frac{a}{x}$$
Step 2: Isolate the sine function term.
Since $x \neq 0$ for the expression $\frac{a}{x}$ to be well-defined, divide the entire equation by $x$: $$\frac{x^2}{3} - 4x + 13 = \sin\frac{a}{x} \quad \cdots (1)$$
Step 3: Find the minimum value of the quadratic expression.
For equation (1) to have a valid real root solution, the value of the quadratic polynomial on the left side must fall within the natural output boundaries of the sine function: $$-1 \le \frac{x^2}{3} - 4x + 13 \le 1$$ Let us find the absolute minimum point of this quadratic function, $\phi(x) = \frac{x^2}{3} - 4x + 13$, by setting its derivative to zero: $$\phi'(x) = \frac{2x}{3} - 4 = 0 \quad \Rightarrow \quad \frac{2x}{3} = 4 \quad \Rightarrow \quad x = 6$$ Calculate the minimum value at this vertex point: $$\phi(6) = \frac{6^2}{3} - 4(6) + 13 = \frac{36}{3} - 24 + 13 = 12 - 24 + 13 = 1$$ This shows that the minimum value of the quadratic function is exactly 1.
Step 4: Solve for the parameter $a$.
Since the left side's minimum value is 1 and the right side's maximum value is 1, the two curves can only meet at this exact peak touchpoint: $$\sin\frac{a}{x} = 1 \quad \text{at } x = 6$$ Substitute $x = 6$ into the equation: $$\sin\left(\frac{a}{6}\right) = 1$$ The smallest positive angle where the sine function reaches 1 is $\frac{\pi}{2}$: $$\frac{a}{6} = \frac{\pi}{2} \quad \Rightarrow \quad a = \frac{6\pi}{2} = 3\pi$$ Let's re-verify the option values and the quadratic coefficients from the image text. The equation reads: $\int_{0}^{x}(t^{2}-8t+13)dt=x \sin\frac{a}{x}$. If the minimum value is exactly 1, then $a/6 = \pi/2 \implies a = 3\pi$, which matches option (A). % Final Correct Answer Selection Update Correct Answer: (A) $3\pi$
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