Limits, Continuity & Differentiability
Differentiability + Definite Integration
nta_pyq_2024_jan
Grade 12
Question:
Let $a$ and $b$ be real constants such that the function $f$ defined by $f(x)=\begin{cases}x^2+3x+a & ,\; x\le 1\\bx+2 & ,\; x>1\end{cases}$ be differentiable on $\mathbb{R}$. Then the value of $\displaystyle\int_{-2}^{2}f(x)\,dx$ equals
$\frac{15}{6}$
$\frac{19}{6}$
21
17
Step-by-Step Solution
Key Concept: Continuity at $x=1$: $1+3+a=b+2\Rightarrow a=b-2$. Differentiability at $x=1$: $f'(1^-)=2(1)+3=5$; $f'(1^+)=b$. So $b=5$, $a=3$. Then compute $\int_{-2}^2 f(x)dx=\int_{-2}^1(x^2+3x+3)dx+\int_1^2(5x+2)dx$.
$b=5$, $a=3$.
$\int_{-2}^1(x^2+3x+3)dx=\left[\frac{x^3}{3}+\frac{3x^2}{2}+3x\right]_{-2}^1=\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(-\frac{8}{3}+6-6\right)=\frac{1}{3}+\frac{3}{2}+3+\frac{8}{3}=6+\frac{3}{2}=\frac{15}{2}$...
From solution: total $=6+\frac{3}{2}+12-\frac{5}{2}=17$.
Correct Answer: 4