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Provide solution for RD Sharma maths class 12 chapter Continuity exercise 8.2 question 4 subquestion (vii)

Answers (1)

Answer:

 a = 2, b = 1

Hint:

 Put LHL = RHL at x = 2 &x = 10

Given:

 f(x)= \begin{cases}5 & x \leq 2 \\ a x+b & 2<x<10 \\ 21 & x \geq 10\end{cases}

Explanation:

At x = 2

\begin{aligned} &f(2)=5 \\ &\lim _{x \rightarrow 2^{+}} f(x)=i m_{x \rightarrow 2} a x+b \\ &=2 a+b \end{aligned}

As, f(x) is continuous at x = 2 when 5 + 2a +b - (1)

At x = 10

\begin{gathered} \lim _{x \rightarrow 10^{-}} f(x)=\lim _{x \rightarrow 10} a x+b \\ =10 a+b \\ f(10)=21 \end{gathered}

As, f(x) is continuous at x = 10

10a + b =21

From (1) & (2) we have

               8a = 16

               a = 2

Put in (1)

               b = 1

Hence, a = 2 & b = 1

Posted by

Gurleen Kaur

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