A function f(x) is said to be continuous in an open interval (a, b), if it is continuous at every point in this interval. A function f(x) is said to be continuous in the closed interval [a, b), if f(x) is continuous in (a, b) and _ x 0 f(a+h)=f(a) and _ x 0 f(b-h)=f(b) If function f(x)= (a+1)x+ x &,x 0 is continuous at x = 0, then answer the following questions. 1. The value of a is: 1. -3 2 2. 0 3. 1 2 4. -1 2 2. The value of b is: 1. 1 2. -1 3. 0 4. Any real number. 3. The value of c is: 1. 1 2. 1 2 3. -1 4. -1 2 4. The value of a + c is: 1. 1 2. 0 3. -1 4. -2 5. The value of c - a is: 1. 1 2. 0 3. -1 4. 2

L.H.L.(at x)= _ x 0 (a+1)x+ x (0 0 form ) Using L' Hospital rule, we get L.H.L. (at x = 0) = _ x 0 (a+1) (a+1)x+ =a+2 R.H.L. (at x = 0)= _ x 0 x+bx^2 - x bx^3 2 = _ x 0 1+bx -1 bx = _ x 0 1 1+bx +1 =1 2 Since,f(x) is continuous at x = 0. From (i) and (ii), we get a+2=c=1 2 =-3 2 ,c=1 2 Also, value of b does not affect the continuity of f(x), so b can be any real number. 1. (a) -3 2 2. (d) Any real number. 3. (b) 1 2 4. (c) -1 Solution: a+c=-3 2 +1 2 =-1 5. (d) 2 Solution: c-a=1 2 +3 2 =2

A function f(x) is said to be continuous in an open interval (a, …

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A function f(x) is said to be continuous in an open interval (a, b), if it is continuous at every point in this interval.
A function f(x) is said to be continuous in the closed interval [a, b), if f(x) is continuous in (a, b) and $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{a}+\text{h})=\text{f}(\text{a})$ and $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{b}-\text{h})=\text{f}(\text{b})$
If function $\text{f}(\text{x})=\begin{cases}\frac{\sin(\text{a}+1)\text{x}+\sin\text{x}}{\text{x}}&,\text{x}<0\\\text{c}&,\text{x}=0\\\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^{\frac{3}{2}}}&,\text{x}>0\end{cases}$ is continuous at x = 0, then answer the following questions.

The value of a is:

$-\frac{3}{2}$
$0$
$\frac{1}{2}$
$-\frac{1}{2}$

The value of b is:

1
-1
0
Any real number.

The value of c is:

$1$
$\frac{1}{2}$
$-1$
$-\frac{1}{2}$

The value of a + c is:

1
0
-1
-2

The value of c - a is:

1
0
-1
2

✓

Answer

$\text{L.H.L.}(\text{at x})=\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+1)\text{x}+\sin\text{x}}{\text{x}}\Big(\frac{0}{0}\text{ form}\Big)$
Using L' Hospital rule, we get
$\text{L.H.L.} (\text{at x} = 0)$
$=\lim\limits_{\text{x}\rightarrow0}(\text{a}+1)\cos(\text{a}+1)\text{x}+\cos\text{x}=\text{a}+2$
$\text{R.H.L.} (\text{at x} = 0)=\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^\frac{3}{2}}=\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{bx}}-1}{\text{bx}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\sqrt{1+\text{bx}}+1}=\frac{1}{2}$
Since,f(x) is continuous at x = 0.
$\therefore$ From (i) and (ii), we get
$\text{a}+2=\text{c}=\frac{1}{2}\Rightarrow\text{a}=-\frac{3}{2},\text{c}=\frac{1}{2}$
Also, value of b does not affect the continuity of f(x), so b can be any real number.

(a) $-\frac{3}{2}$

(d) Any real number.

(b) $\frac{1}{2}$

Solution:

$\text{a}+\text{c}=-\frac{3}{2}+\frac{1}{2}=-1$

(d) 2

Solution:

$\text{c}-\text{a}=\frac{1}{2}+\frac{3}{2}=2$

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