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UPSC NDA I 2023 Mathematics Solved Paper

Question Numbers: 71-73

Consider the following for the next three (03) items that follow :

Let

f(x) = Pe^{x} + Qe^{2x} + Re^{3x}

where

P, Q, R

are real numbers. Further

f(0) = 6

f'(\ln 3) = 282

and

\int\limits_{0}^{\ln 2}f(x)dx=11

Question : 71 of 120

Marks: +1, -0

‘What is the value of Q ?

Solution:

Concept:

Use differentiation and integration of exponentials, then solve a system of equations using Cramer's rule.

Explanation:

Given

f(x)=Pe^{x}+Qe^{2x}+Re^{3x}

f(0)=P+Q+R=6

(equation 1).

f'(x)=Pe^{x}+2Qe^{2x}+3Re^{3x}

x=\ln3

f'(\ln3)=3P+18Q+81R=282

Divide by 3:

P+6Q+27R=94

(equation 2).

Integrate from 0 to

\ln2

\int_{0}^{\ln2}f(x)dx = \left[Pe^{x}+\frac{Q}{2}e^{2x}+\frac{R}{3}e^{3x}\right]_{0}^{\ln2}

Evaluate:

P(2-1)+\frac{Q}{2}(4-1)+\frac{R}{3}(8-1)=P+\frac{3Q}{2}+\frac{7R}{3}=11

Multiply by 6:

6P+9Q+14R=66

(equation 3).

System:

P+Q+R=6

P+6Q+27R=94

6P+9Q+14R=66

Using Cramer's rule:

\Delta = \begin{vmatrix}1&1&1\\1&6&27\\6&9&14\end{vmatrix} = -38

\Delta_{2} = \begin{vmatrix}1&6&1\\1&94&27\\6&66&14\end{vmatrix} = -76

Q = \frac{\Delta_{2}}{\Delta} = \frac{-76}{-38}=2

Answer:

2

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