CBSE 2025 · Region 1 · Set 1 · Q27 · 3 marks
Solve the following linear programming problem graphically : Maximise $\displaystyle Z=\mathrm{x}+2 \mathrm{y}$ Subject to the constraints : \[\begin{aligned} & \mathrm{x}-\mathrm{y} \geq 0 \\ & \mathrm{x}-2 \mathrm{y} \geq-2 \\ & \mathrm{x} \geq 0, \mathrm{y} \geq 0 \end{aligned} \]
Marking-scheme solution
| Corner Point | Value of $\displaystyle \boldsymbol{Z}=\boldsymbol{\mathrm{x}}+\mathbf{2} \boldsymbol{\mathrm{y}}$ |
| $\displaystyle \mathbf{O}(\mathbf{0}, \mathbf{0})$ | $\displaystyle 0$ |
| $\displaystyle \mathbf{A}(\mathbf{2}, \mathbf{2})$ | $\displaystyle 6$ |
Since feasible region is unbounded. Plot $\displaystyle \mathrm{x}+2 \mathrm{y}>6$ which has common region with feasible region, thus $\displaystyle \boldsymbol{Z}$ has no maximum value.
Find : $\displaystyle \int \frac{\mathrm{x}+\sin \mathrm{x}}{1+\cos \mathrm{x}} \mathrm{~d} \mathrm{x}$
Evaluate: $\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\mathrm{~d} \mathrm{x}}{\cos ^{3} \mathrm{x} \sqrt{2 \sin 2 \mathrm{x}}}$
\[\begin{aligned}
& \int \frac{\mathrm{x}+\sin \mathrm{x}}{1+\cos \mathrm{x}} d \mathrm{x} \\
& =\int \frac{\mathrm{x}+2 \sin \dfrac{\mathrm{x}}{2} \cos \dfrac{\mathrm{x}}{2}}{2 \cos ^{2} \dfrac{\mathrm{x}}{2}} d \mathrm{x} \\
& =\int \mathrm{x}\left(\frac{1}{2} \sec ^{2} \frac{\mathrm{x}}{2}\right) d \mathrm{x}+\int \tan \frac{\mathrm{x}}{2} d \mathrm{x} \\
& =\mathrm{x} \tan \frac{\mathrm{x}}{2}-\int \tan \frac{\mathrm{x}}{2} d \mathrm{x}+\int \tan \frac{\mathrm{x}}{2} d \mathrm{x} \\
& =\mathrm{x} \tan \frac{\mathrm{x}}{2}+C
\end{aligned}
\]
\[\begin{aligned}
& \int_{0}^{\pi / 4} \frac{d \mathrm{x}}{\cos ^{3} \mathrm{x} \sqrt{2 \sin 2 \mathrm{x}}} \\
& =\frac{1}{2} \int_{0}^{\pi / 4} \frac{d \mathrm{x}}{\cos ^{4} \mathrm{x} \sqrt{\tan \mathrm{x}}} \\
& =\frac{1}{2} \int_{0}^{\pi / 4} \frac{\left(1+\tan ^{2} \mathrm{x}\right) \sec ^{2} \mathrm{x}}{\sqrt{\tan \mathrm{x}}} d \mathrm{x} \\
& \text { Put } \tan \mathrm{x}=t \Rightarrow \sec ^{2} \mathrm{x} d \mathrm{x}=d t \\
& \therefore I=\frac{1}{2} \int_{0}^{1} \frac{1+t^{2}}{\sqrt{t}} d t \\
& =\frac{1}{2} \int_{0}^{1}\left(\frac{1}{\sqrt{t}}+t^{3 / 2}\right) d t \\
& =\frac{1}{2}\left[2 \sqrt{t}+\frac{2}{5} t^{5 / 2}\right]_{0}^{1} \\
& =\frac{6}{5}
\end{aligned}
\]
Linear ProgrammingLinear Programming Problem and its Mathematical FormulationApplyshort_answermedium
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