Definite Integration
Grade 12
Question:
<p>Let f(x) = <img alt="" data-imgur-src="nB4XHKP.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/nB4XHKP.png" style="width: 150px; height: 75px;" /><br />
If f(x) is continuous at x = 2, then the value of k is equal to:</p>
<p style="display:inline"><span class="math-tex">\(\frac{7}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{9}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{5}{2}\)</span></p>
Step-by-Step Solution
Key Concept: To ensure continuity of an accumulation function at a point, the value of the function must equal the limit of the definite integral as the upper bound approaches that point.
<p>When <span class="math-tex">\(x<2, f(x)=\int_{-1}^{x}(2-t) d t=\left(2 t-\frac{t^{2}}{2}\right)_{-1}^{x}\)</span><br />
<span class="math-tex">\(=\left(2 x-\frac{x^{2}}{2}\right)-\left(-2-\frac{1}{2}\right)\)</span><br />
<span class="math-tex">\(=\frac{-x^{2}}{2}+2 x+\frac{5}{2}\)</span><br />
<span class="math-tex">\(\therefore \lim \limits_{x \rightarrow 2^{-}} f(x)=\frac{9}{2}\)</span><br />
When x > 2<br />
<span class="math-tex">\(f(x)=\int_{-1}^{2}(2-t) d t+\int_{2}^{x}(t-2) d t\)</span><br />
<span class="math-tex">\(=\left(2 t-\frac{t^{2}}{2}\right)_{-1}^{2}+\left(\frac{t^{2}}{2}-2 t\right)_{2}^{x}=\frac{9}{2}+\frac{x^{2}}{2}-2 x+2\)</span><br />
<span class="math-tex">\(\therefore \lim \limits_{x \rightarrow 2^{-}} f(x)=\frac{9}{2}\)</span><br />
Hence, f(2) = k <span class="math-tex">\(=\lim \limits_{x \rightarrow 2^+} f(x)=\frac{9}{2}\)</span></p>
Correct Answer: C