(a) The elasticity of demand with respect to price p for a commodity is $\eta_d = \frac{p + 2p^2}{100 - p - p^2}$. Find demand function where price is ₹ 5 and the demand is 70. OR (b) A machine produces a component of a product with a standard deviation of 1.6 cm in length. A random sample of 64 components was selected from the output and this sample has a mean length of 90 cm. The customer will reject the part if it is either less than 88 cm or more than 92 cm. Does the 95% confidence interval for the true mean length of all the components produced ensure acceptance by the customer ? | $1\%$ | $2\%$ | $5\%$ | $10\%$ | | --- | --- | --- | --- | | $|Z_\alpha| = 2.58$ | $|Z_\alpha| = 2.33$ | $|Z_\alpha| = 1.96$ | $|Z_\alpha| = 1.645$ |

(a) The elasticity of demand with respect to price – Class 12 Business Mathematics Statistics

Question 20 of 42

The elasticity of demand with respect to price p for a commodity is $\eta_d = \frac{p + 2p^2}{100 - p - p^2}$ . Find demand function where price is ₹ 5 and the demand is 70. OR
A machine produces a component of a product with a standard deviation of 1.6 cm in length. A random sample of 64 components was selected from the output and this sample has a mean length of 90 cm. The customer will reject the part if it is either less than 88 cm or more than 92 cm. Does the 95% confidence interval for the true mean length of all the components produced ensure acceptance by the customer ?

$1\%$	$2\%$	$5\%$	$10\%$
$	Z_\alpha	= 2.58$	$

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2022Subjective· 5mImportance★★★★★

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(a) Separate the elasticity relation and integrate: $x=A(100-p-p^2)$ , and $p=5,x=70$ give $A=1$ , so $x=100-p-p^2$ . (b) 95% CI $=90\pm0.392=(89.608,90.392)\subset(88,92)$ $\Rightarrow$ accepted.

(a) Demand function from elasticity.

Elasticity of demand is $\eta_d=-\dfrac{p}{x}\dfrac{dx}{dp}$ . Given $\eta_d=\dfrac{p+2p^2}{100-p-p^2}$ ,

$-\frac{p}{x}\frac{dx}{dp}=\frac{p+2p^2}{100-p-p^2}\;\Rightarrow\;\frac{dx}{x}=-\frac{p+2p^2}{p\,(100-p-p^2)}\,dp=-\frac{1+2p}{100-p-p^2}\,dp.$

Since $\dfrac{d}{dp}(100-p-p^2)=-(1+2p)$ , the right side is $\dfrac{d(100-p-p^2)}{100-p-p^2}$ . Integrating,

$\ln x=\ln(100-p-p^2)+\ln A\;\Rightarrow\;x=A(100-p-p^2).$

At $p=5,\ x=70$ : $100-5-25=70$ , so $70=A(70)\Rightarrow A=1.$

Hence the demand function is $x=100-p-p^2.$

(b) Confidence interval and acceptance.

$\sigma=1.6,\ n=64,\ \bar x=90.$ Standard error $=\dfrac{\sigma}{\sqrt n}=\dfrac{1.6}{8}=0.2.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set MARCH1 markMCQ
$p(x)$
(a) $MR = 0$
(b) $MC - MR = 0$
(c) $MC + MR = 0$
(d) $MC = 0$
›Reveal solutionSolution

Profit is maximum where marginal revenue equals marginal cost: $MR=MC$ , i.e. $MC-MR=0$ .

Profit is $p(x)=R(x)-C(x)$ , where $R$ is total revenue and $C$ is total cost. For a maximum, the first derivative must vanish:

$p'(x) = R'(x) - C'(x) = MR - MC = 0 \;\Rightarrow\; MR = MC.$

…

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CBSE 2025Set MARCH1 markMCQ
$Q. If the marginal revenue of a firm is a constant, then the demand function is :$
(a) $C(x)$
(b) $MR$
(c) $AC$
(d) $MC$
›Reveal solutionSolution

A constant marginal revenue integrates to $R=kx$ (no constant, since $R(0)=0$ ); the demand price $p=R/x=k$ equals $MR$ , option (b).

Integrate the constant $MR$ . Let $MR=k$ . Then

$R=\int MR\,dx=\int k\,dx=kx+C.$

Fix the constant. Revenue is zero when $x=0$ , so $C=0$ and $R=kx$ .

…

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CBSE 2023Set MARCH1 markMCQ
$p(x)$
(a) $MR=0$
(b) $MC-MR=0$
(c) $MC+MR=0$
(d) $MC=0$
›Reveal solutionSolution

Profit is maximised where its derivative is zero: $MR=MC$ , i.e. $MC-MR=0$ .

Profit is defined as revenue minus cost:

$P(x)=R(x)-C(x).$

For a maximum, set the first derivative to zero:

$P'(x)=R'(x)-C'(x)=0\;\Rightarrow\;MR-MC=0\;\Rightarrow\;MC-MR=0,$ …

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CBSE 2023Set MARCH1 markMCQ
$MR=100-9x^2$
(a) $100x-9x^2$
(b) $100-3x^2$
(c) $100+9x^2$
(d) $100x-3x^2$
›Reveal solutionSolution

Total revenue is the integral of marginal revenue; dividing $R$ by $x$ gives the demand (price) function $p=100-3x^2$ .

Given the marginal revenue $MR=100-9x^2$ . Total revenue is:

$R=\int MR\,dx=\int (100-9x^2)\,dx=100x-3x^3+c.$

At $x=0$ revenue is $0$ , so $c=0$ :

$R=100x-3x^3.$

…

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CBSE 2022Set MARCH1 markMCQ
$p(x)$
(a) $MR = 0$
(b) $MC - MR = 0$
(c) $MC + MR = 0$
(d) $MC = 0$
›Reveal solutionSolution

Profit peaks when $MC = MR$ , i.e. $MC - MR = 0$ .

Profit is total revenue minus total cost: $p(x) = R(x) - C(x)$ . For a maximum, differentiate and set the derivative to zero:

$\frac{dp}{dx} = \frac{dR}{dx} - \frac{dC}{dx} = MR - MC = 0.$

…

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CBSE 2022Set MARCH1 markMCQ
$\int \frac{dp}{p} = k \int \frac{dx}{x}$
(a) $-\frac{1}{\eta_d}$
(b) $\eta_d$
(c) $\frac{1}{\eta_d}$
(d) $-\eta_d$
›Reveal solutionSolution

The constant $k = -\dfrac{1}{\eta_d}$ .

The elasticity of demand is defined as

$\eta_d = -\frac{p}{x}\cdot\frac{dx}{dp}.$

Integrate the given relation $\displaystyle\int \frac{dp}{p} = k\int \frac{dx}{x}$ :

$\ln p = k\ln x + \text{constant} \;\Rightarrow\; p = C\,x^{k}.$

Differentiate: $\dfrac{dp}{dx} = Ck\,x^{k-1} = k\cdot\dfrac{p}{x}$ , so $\dfrac{dx}{dp} = \dfrac{x}{k\,p}$ .

…

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CBSE 2020Set MARCH1 markMCQ
$MC = 100\sqrt{x}$
(a) $\dfrac{200}{3x^{1/2}}$
(b) $\dfrac{200}{3}x^{1/2}$
(c) $\dfrac{200}{3}x^{3/2}$
(d) $\dfrac{200}{3x^{3/2}}$
›Reveal solutionSolution

Integrate $MC$ to get $TC$ , fix the constant using $TC = 0$ at $x = 0$ , then divide by $x$ to obtain the average cost $AC = \frac{200}{3}x^{1/2}$ .

Step 1 — Total cost from marginal cost.

$TC = \int MC\,dx = \int 100\sqrt{x}\,dx = 100\int x^{1/2}\,dx = 100\cdot \frac{x^{3/2}}{3/2} = \frac{200}{3}x^{3/2} + c.$

…

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