Question

26. Determine the values of a, b, c for which the function
sin (a +1)x+ sinx, for x < 0
f(x)=
for x = 0 is continuous at x = 0.
C
√x + bx? – da
for x > 0
by 3/2​

Answer 1

EXPLANATION.

Let,

$\implies f(x)= \dfrac{sin(a + 1)x + sinx}{x} \ \ \ , if \ x < 0$

$\implies c \ , if \ x = 0$

$\implies \dfrac{\sqrt{x + bx^{2} }- \sqrt{x} }{bx\sqrt{x} } \ \ \ if x > 0$

As we know that,

$\implies f(x)= \dfrac{sin(a + 1)x + sinx}{x} \ \ \ , if \ x < 0$

$\sf \implies \displaystyle \lim_{x \to 0^{-} } f(x) = \lim_{x \to 0} \dfrac{sin(a + 1)x + sin x}{x}$

As we know that,

Formula of :

⇒ sin(A + B) = sin(A).cos(B) + cos(A).sin(B).

Using this formula in the equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{[sin(a).cos(1) + cos(a).sin(1)]x + sinx}{x}$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{sin(ax).cos(x) + cos(ax).sin(x) + sinx}{x}$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{sin(ax).cos(x)}{x} + \displaystyle \lim_{x \to 0} \dfrac{cos(ax) . sin(x)}{x} + \displaystyle \lim_{x \to 0} \dfrac{sinx }{x}$

As we know that,

Some standard expansions.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{sin x}{x} = 1$

$\sf \implies \displaystyle \lim_{x \to 0} cos x = 1$

Using this expansions in the equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{sin(ax)}{x} \times\displaystyle \lim_{x \to 0} cos(x) + \displaystyle \lim_{x \to 0}\dfrac{sin x}{x} \times \displaystyle \lim_{x \to 0} cos(ax) + 1$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{sin(ax)}{ax} \times a \times 1 + 1 \times 1 + 1$

$\sf \implies \displaystyle a + 2.$

$\sf \implies \displaystyle \lim_{x \to 0^{+} } f(x) = \displaystyle \lim_{x \to 0} \dfrac{\sqrt{x + bx^{2} } - \sqrt{x} }{bx\sqrt{x} } \ \ if x > 0$

As we know that,

Put the value of x = 0 in the equation and check their indeterminant form, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{\sqrt{(0) + b(0)^{2} } - \sqrt{0} }{b(0)\sqrt{0} } = \dfrac{0}{0}$

As we can see that it is in the form of 0/0 indeterminant form.

If roots can exists then rationalizes the equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{\sqrt{x + bx^{2} }- \sqrt{x} }{bx\sqrt{x} } \times \dfrac{\sqrt{x + bx^{2} } + \sqrt{x} }{\sqrt{x + bx^{2} } + \sqrt{x} }$

As we know that,

Formula of :

⇒ (x² - y²) = (x + y)(x - y).

Using this formula in the equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{(\sqrt{x + bx^{2} } )^{2} - (\sqrt{x} )^{2} }{bx\sqrt{x} (\sqrt{x + bx^{2} } + \sqrt{x} )}$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{x + bx^{2} - x}{bx\sqrt{x} (\sqrt{x + bx^{2} } + \sqrt{x} )}$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{x^{2} }{x\sqrt{x} (\sqrt{x + bx^{2} } + \sqrt{x} )}$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{x}{\sqrt{x} (\sqrt{x + bx^{2} } + \sqrt{x} )}$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{\sqrt{x} }{\sqrt{x + bx^{2} } + \sqrt{x} }$

Divide numerator and denominator by √x, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{\bigg(\dfrac{\sqrt{x} }{\sqrt{x} } \bigg)}{\bigg(\dfrac{\sqrt{x + bx^{2} } }{\sqrt{x} } + \dfrac{\sqrt{x} }{\sqrt{x} }\bigg) }$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{1}{\bigg(\dfrac{\sqrt{x + bx^{2} } }{\sqrt{x} } + 1 \bigg)}$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{1}{\sqrt{1 + bx} + 1}$

Put the value of x = 0 in the equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{1}{\sqrt{1 + b(0)} + 1} = \dfrac{1}{2}$

$\sf \implies \displaystyle f(0) = \lim_{x \to 0} = c = \dfrac{1}{2}$

$\sf \implies \displaystyle a + 2 = \dfrac{1}{2}$

$\sf \implies \displaystyle a = \dfrac{1}{2} - 2 = \dfrac{-3}{2}$

⇒ b = x ∈ R - {0}.

Answer 2

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