Question

ax + b
1<x<5
1. If f(x)= 7x-5 5 5x<10 is continuous, then (a, b) is equal to :
bx + 3a x 10
(D) (0,0)
(C) (10,5)
(B) (5,5)
(A) (5, 10)​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

$\sf f(x) = \begin{cases} & \sf{ (ax + b) \: \: \: \: when \: 1 \leqslant x < 5} \\ \\ & \sf{ (7x - 5) \: \: \: \: \: when \: 5 \leqslant x < 10} \\ \\ & \sf{ (bx + 3a) \: \: \: \: \: when \: x \geqslant 10} \end{cases}$

f(x) is continuous (a, b) =

(A) (5, 10)

(B) (5, 5)

(C) (10,5)

(D) (0,0)

EVALUATION

Here by the given :

$\sf f(x) = \begin{cases} & \sf{ (ax + b) \: \: \: \: when \: 1 \leqslant x < 5} \\ \\ & \sf{ (7x - 5) \: \: \: \: \: when \: 5 \leqslant x < 10} \\ \\ & \sf{ (bx + 3a) \: \: \: \: \: when \: x \geqslant 10} \end{cases}$

By the given condition f(x) is continuous

So f(x) is continuous at x = 5

$\displaystyle \sf{\lim_{x \to 5 + } f(x) = \lim_{x \to 5 - } f(x)}$

$\displaystyle \sf{ \implies \: \lim_{x \to 5 + } (7x - 5) = \lim_{x \to 5 - }(ax + b)}$

$\displaystyle \sf{ \implies \: (35 - 5) =(5a + b)}$

$\displaystyle \sf{ \implies \: (5a + b) = 35 \: \: \: - - - (1)}$

Again f(x) is continuous at x = 10

$\displaystyle \sf{\lim_{x \to 10 + } f(x) = \lim_{x \to 10 - } f(x)}$

$\displaystyle \sf{ \implies \: \lim_{x \to 10 + } (bx + 3a) = \lim_{x \to 10- }(7x - 5)}$

$\displaystyle \sf{ \implies \: (10b + 3a) = 70 - 5}$

$\displaystyle \sf{ \implies \: (10b + 3a) = 65 \: \: \: - - - (2)}$

Solving Equation 1 and Equation 2 we get

a = 5 & b = 5

FINAL ANSWER

Hence the correct option is (B) (5,5)

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