CBSE 2024 · Region 3 · Set 1 · Q33 · 5 marks
It is given that function $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}^{4}-62 \mathrm{x}^{2}+a \mathrm{x}+9$ attains local maximum value at $\displaystyle \mathrm{x}=1$. Find the value of ' $\displaystyle a$ ', hence obtain all other points where the given function $\displaystyle \mathrm{f}(\mathrm{x})$ attains local maximum or local minimum values.The perimeter of a rectangular metallic sheet is $\displaystyle 300$ cm . It is rolled along one of its sides to form a cylinder. Find the dimensions of the rectangular sheet so that volume of cylinder so formed is maximum.
It is given that function $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}^{4}-62 \mathrm{x}^{2}+a \mathrm{x}+9$ attains local maximum value at $\displaystyle \mathrm{x}=1$. Find the value of ' $\displaystyle a$ ', hence obtain all other points where the given function $\displaystyle \mathrm{f}(\mathrm{x})$ attains local maximum or local minimum values.
The perimeter of a rectangular metallic sheet is $\displaystyle 300$ cm . It is rolled along one of its sides to form a cylinder. Find the dimensions of the rectangular sheet so that volume of cylinder so formed is maximum.
Marking-scheme solution
Let length of rectangle be $\displaystyle \mathrm{x}$ cm and breadth be $\displaystyle (150-\mathrm{x}) \mathrm{cm}$.
Let $\displaystyle r$ be the radius of cylinder $\displaystyle \Rightarrow 2 \pi r=\mathrm{x} \Rightarrow r=\frac{\mathrm{x}}{2 \pi}$V=\pi r^{2} h=\pi\left(\frac{\mathrm{x}^{2}}{4 \pi^{2}}\right)(150-\mathrm{x})=\frac{75 \mathrm{x}^{2}}{2 \pi}-\frac{\mathrm{x}^{3}}{4 \pi}Length of rectangle is $\displaystyle 100$ cm and breadth of rectangle is $\displaystyle 50$ cm .
Application of DerivativesMaxima and MinimaApplynumerichard
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.