MHT CET · Maths · Definite Integration
If \(\int_{1}^{k}\left(3 x^{2}+2 x+1\right) d x=11\), then \(\mathrm{k}=\)
- A \(\frac{1}{2}\)
- B \(-2\)
- C \(-\frac{1}{2}\)
- D 2
Answer & Solution
Correct Answer
(D) 2
Step-by-step Solution
Detailed explanation
(C)
We have \(\left[3\left(\frac{x^{3}}{3}\right)+2\left(\frac{x^{2}}{2}\right)+x\right]_{1}^{k}=11\)
\(\therefore\left[x^{3}+x^{2}+x\right]_{1}^{k}=11\)
\(\left(k^{3}+k^{2}+k\right)-(1+1+1)=11 \Rightarrow k^{3}+k^{2}+k=14\)
\(\therefore \mathrm{k}\left(\mathrm{k}^{2}+\mathrm{k}+1\right)=2 \times 7 \Rightarrow \mathrm{k}=2\)
We have \(\left[3\left(\frac{x^{3}}{3}\right)+2\left(\frac{x^{2}}{2}\right)+x\right]_{1}^{k}=11\)
\(\therefore\left[x^{3}+x^{2}+x\right]_{1}^{k}=11\)
\(\left(k^{3}+k^{2}+k\right)-(1+1+1)=11 \Rightarrow k^{3}+k^{2}+k=14\)
\(\therefore \mathrm{k}\left(\mathrm{k}^{2}+\mathrm{k}+1\right)=2 \times 7 \Rightarrow \mathrm{k}=2\)
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