If the derivative function fx- b x 2 -ax-4- x- -1 a x 2 -b- x- -1 is everywhere continuous- then

The correct option is B a=2,b=3 b(1)^2+a ( 1)+4=a( 1)^2+b b a+4=+a+b ⇒ 2a= 4 ⇒ a=2 Derivative is continuous #160; 2b( 1)+a=2a( 1) 2b ...

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Standard XII

Mathematics

Theorems for Continuity

If the deriva...

Question

If the derivative function $f (x) = {\begin{matrix} b x^{2} + a x + 4, x \geq - 1 a x^{2} + b, x < - 1 \end{matrix}$ is everywhere continuous, then

a = 2, b = 3

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a = 3, b = 2

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a = - 2, b = - 3

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a = - 3, b = - 2

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Solution

The correct option is B $a = 2, b = 3$
$b (1)^{2} + a (- 1) + 4 = a (- 1)^{2} + b$
$b - a + 4 = + a + b \Rightarrow 2 a = 4 \Rightarrow a = 2$
Derivative is continuous
$2 b (- 1) + a = 2 a (- 1)$
$- 2 b + a = - 2 a$
$+ 2 b = + 3 a$
$b = 3 \frac{a}{2}$
$b = \frac{3 \times 2}{2} \Rightarrow b = 3$

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If the derivative function f(x)={bx2+ax+4,x≥−1ax2+b,x<−1 is everywhere continuous, then

The correct option is B a=2,b=3b(1)2+a(−1)+4=a(−1)2+bb−a+4=+a+b⇒2a=4⇒a=2Derivative is continuous 2b(−1)+a=2a(−1)−2b+a=−2a+2b=+3ab=3a2b=3×22⇒b=3

If the derivative function $f (x) = {\begin{matrix} b x^{2} + a x + 4, x \geq - 1 a x^{2} + b, x < - 1 \end{matrix}$ is everywhere continuous, then

The correct option is B $a = 2, b = 3$
$b (1)^{2} + a (- 1) + 4 = a (- 1)^{2} + b$
$b - a + 4 = + a + b \Rightarrow 2 a = 4 \Rightarrow a = 2$
Derivative is continuous
$2 b (- 1) + a = 2 a (- 1)$
$- 2 b + a = - 2 a$
$+ 2 b = + 3 a$
$b = 3 \frac{a}{2}$
$b = \frac{3 \times 2}{2} \Rightarrow b = 3$