Question:
The triangle formed by the tangent to the curve $f(x)=x^2 + bx - b$ at the point $(1,1)$ and the coordinate axes lies in the first quadrant. If its area is 2, then the value of $b$ is
The triangle formed by the tangent to the curve $f(x)=x^2 + bx - b$ at the point $(1,1)$ and the coordinate axes lies in the first quadrant. If its area is 2, then the value of $b$ is
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Area with axes = $\frac{1}{2} \times x\text{-intercept} \times y\text{-intercept}$.
Updated On: Apr 23, 2026
- $-1$
- $3$
- $1$
- $-3$
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The Correct Option is C
Solution and Explanation
Concept:
Equation of tangent + intercept form → area of triangle.
Step 1: Differentiate function.
\[ f'(x) = 2x + b \]
Step 2: Slope at $x=1$.
\[ m = 2(1) + b = 2 + b \]
Step 3: Equation of tangent.
\[ y - 1 = (2+b)(x - 1) \] \[ y = (2+b)x - (2+b) + 1 \] \[ y = (2+b)x - (1+b) \]
Step 4: Find intercepts.
$x$-intercept: \[ 0 = (2+b)x - (1+b) \Rightarrow x = \frac{1+b}{2+b} \] $y$-intercept: \[ y = -(1+b) \]
Step 5: Area of triangle.
\[ \frac{1}{2} \times \left|\frac{1+b}{2+b}\right| \times |1+b| = 2 \]
Step 6: Solve equation.
Simplifying gives: \[ b = 1 \] Conclusion:
$b = 1$
Step 1: Differentiate function.
\[ f'(x) = 2x + b \]
Step 2: Slope at $x=1$.
\[ m = 2(1) + b = 2 + b \]
Step 3: Equation of tangent.
\[ y - 1 = (2+b)(x - 1) \] \[ y = (2+b)x - (2+b) + 1 \] \[ y = (2+b)x - (1+b) \]
Step 4: Find intercepts.
$x$-intercept: \[ 0 = (2+b)x - (1+b) \Rightarrow x = \frac{1+b}{2+b} \] $y$-intercept: \[ y = -(1+b) \]
Step 5: Area of triangle.
\[ \frac{1}{2} \times \left|\frac{1+b}{2+b}\right| \times |1+b| = 2 \]
Step 6: Solve equation.
Simplifying gives: \[ b = 1 \] Conclusion:
$b = 1$
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