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Limits, Continuity and Differentiability

80010 If the function $f$ defined by
$f (x) = {\begin{cases} \cos x, if x \leq 0 \\ 3 x + α, if 0 < x < 2 \\ β x + 3, if 2 \leq x \leq 4 \\ 11, if x > 4 \end{cases}$
Where, $α$ and $β$ are real constants, is continuous on $R$ , then $α^{2} + β^{2} =$

1 3

2 9

3 5

4 1

Explanation:

(C) : $f (x)$ is continuous on R, So
$at, x = 0$
$lim_{x \to 0^{+}} f (x) = f (0)$
$0 + α = \cos 0 = 1$
$α = 1$
At,
$x = 2, lim_{x \to 2^{-}} f (x) = f (2)$
$6 + α = 2 β + 3$
$7 = 2 β + 3$
$2 β = 4$
$β = 2$
So, $α^{2} + β^{2} = 1 + 4 = 5$

AP EAMCET-2018-24.04.2018

Limits, Continuity and Differentiability

80011 If $f : [- 2, 2] \to R$ is defined by $f (x) = | \begin{array}{cc} \frac{\sqrt{1 + c x} - \sqrt{1 - c x}}{\frac{x}{x + 3}} & for - 2 \leq x < 0 \\ \frac{x + 1}{x +} & for 0 \leq x \leq 2 \end{array}$
continuous on $[- 2, 2]$ then $c$ is equal to

1

\frac{2}{\sqrt{3}}

2 3

3

\frac{3}{2}

4

\frac{3}{\sqrt{2}}

Explanation:

(B) : Given, $f : [- 2, 2] \to R$
$f (x) = {\begin{array}{cc} \frac{\sqrt{1 + c x} - \sqrt{1 - c x}}{x}, & - 2 \leq x < 0 \\ \frac{x + 3}{x + 1}, & 0 \leq x \leq 2 \end{array}$
Now, LHL $= lim_{x \to 0^{-}} f (x)$
$= lim_{h \to 0} \frac{\sqrt{1 - ch} - \sqrt{1 + ch}}{- h} \times \frac{\sqrt{1 - ch} + \sqrt{1 + ch}}{\sqrt{1 - ch} + \sqrt{1 + ch}}$
$= lim_{h \to 0} \frac{(1 - ch) - (1 + ch)}{- h \sqrt{1 - 0} + \sqrt{1 + 0}} = lim_{h \to 0} \frac{- 2 ch}{- h (1 + 1)} = c$
And, $RHL = lim_{x \to 0^{+}} f (x)$
$= lim_{h \to 0} f (0 + h) = lim_{h \to 0} \frac{0 + h + 3}{0 + h + 1} = lim_{h \to 0} \frac{h + 3}{h + 1}$
$= \frac{0 + 3}{0 + 1} = 3$
Since, $f$ is continuous at $x = 0$ ,
$∴ LHL = RHL$
$c = 3$

AP EAMCET-2014

Limits, Continuity and Differentiability

80012 $f (x) = {\begin{cases} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}} x \neq 0 \\ k \log 2 \log 3 x = 0 \end{cases}$
Find the value of ' $k$ ' for which the function $f$ is continuous.

1

\sqrt{2}

2 24

3

18 \sqrt{3}

4

24 \sqrt{2}

Explanation:

(D): At, $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$k \log 2 \log 3 = lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}}$
$= lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{2 \cos^{2} \frac{x}{2}}} [1 + \cos x = 2 \cos^{2} \frac{x}{2}]$
$= lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{2} \cos \frac{x}{2}} (\frac{0}{0} form)$
On using L-Hospital Rule, we get -
$= lim_{x \to 0} \frac{72^{x} \log 72 - 9^{x} \log 9 - 8^{x} \log 8}{- \sqrt{2} (- \sin \frac{x}{2}) \frac{1}{2}}$
$= lim_{x \to 0} \frac{2 [72^{x} \log 72 - 9^{x} \log 9 - 8^{x} \log 8]}{\sqrt{2} \sin (\frac{x}{2})}$
Again using L-Hospital rule, we get -
$= lim_{x \to 0} \frac{\sqrt{2} [72^{x} (\log 72)^{2} - 9^{x} (\log 9)^{2} - 8^{x} (\log 8)^{2}]}{\frac{1}{2} (\cos \frac{x}{2})}$
$= 2 \sqrt{2} [(\log 72)^{2} - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9 + \log 8)^{2} (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9 + \log 8)^{2} - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9)^{2} + (\log 8)^{2} + 2 \log (9) \log (8) - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} (2 \log 9 \cdot \log 8) = 4 \sqrt{2} \log 9 \log 8$
$= 4 \sqrt{2} \log 3^{2} \log 2^{3} = 4 \sqrt{2} 2 (\log 3) \cdot 3 \cdot \log (2)$
$= 24 \sqrt{2} \log (3) \log (2)$
So, $k \log 2 \cdot \log 3 = 24 \sqrt{2} \log (2) \log (3)$
$k = 24 \sqrt{2}$

AP EAMCET-2021-19.08.2021

Limits, Continuity and Differentiability

80013 If the function $f (x)$ , defined below is continuous in the interval $[0, π]$ , then
$f (x) = {\begin{cases} x + a \sqrt{2} (\sin x), 0 \leq x < \frac{π}{4} \\ 2 x (\cot x) + b, \frac{π}{4} \leq x \leq \frac{π}{2} \\ a (\cos 2 x) - b (\sin x), \frac{π}{4} < x \leq π \end{cases}$

1

a = \frac{π}{6}, b = \frac{π}{12}

2

a = \frac{π}{6}, b = \frac{π}{12}

3

a = \frac{- π}{6}, b = \frac{- π}{12}

4

a = \frac{π}{6}, b = \frac{- π}{12}

Explanation:

(D) : We have,
$f (x) = {\begin{cases} x + a \sqrt{2} (\sin x), 0 \leq x < \frac{π}{4} \\ 2 x (\cot x) + b, \frac{π}{4} \leq x \leq \frac{π}{2} \\ a (\cos 2 x) - b (\sin x), \frac{π}{4} < x \leq π \end{cases}$
At, $x = \frac{π}{4}$ LHL $=$ RHL
$lim_{\begin{matrix} x \to \frac{π^{-}}{4} \end{matrix}} f (x) = lim_{x \to \frac{π^{+}}{4}} f (x)$
$lim_{x \to \frac{π^{-}}{4}} (x + a \sqrt{2} \sin x) = lim_{x \to \frac{π^{+}}{4}} (2 x \cdot \cot x + b)$
$\frac{π}{4} + a \sqrt{2} \sin \frac{π}{4} = 2 \cdot \frac{π}{4} \cdot \cot \frac{π}{4} + b$
$\frac{π}{4} + a = \frac{π}{2} + b$
$a - b = \frac{π}{4}$
$at, x = \frac{π}{2} \to L H L = R H L$
$lim_{x \to \frac{π^{-}}{2}} f (x) = lim_{x \to \frac{π^{+}}{2}} f (x)$
$lim_{x \to \frac{π}{2}} (2 x \cot x + b) = lim_{x \to \frac{π}{2}} a (\cos 2 x) - b \sin x$
$2 \cdot \frac{π}{2} \cdot \cot \frac{π}{2} + b = a \cos 2 \cdot \frac{π}{2} - b \sin \frac{π}{2}$
$0 + b = - a - b$
$2 b = - a$
Solving equation (i) \& (ii), we get,
$a = \frac{π}{6}, b = \frac{- π}{12}$

AP EAMCET-2021-19.08.2021

Limits, Continuity and Differentiability

80010 If the function $f$ defined by
$f (x) = {\begin{cases} \cos x, if x \leq 0 \\ 3 x + α, if 0 < x < 2 \\ β x + 3, if 2 \leq x \leq 4 \\ 11, if x > 4 \end{cases}$
Where, $α$ and $β$ are real constants, is continuous on $R$ , then $α^{2} + β^{2} =$

1 3

2 9

3 5

4 1

Explanation:

(C) : $f (x)$ is continuous on R, So
$at, x = 0$
$lim_{x \to 0^{+}} f (x) = f (0)$
$0 + α = \cos 0 = 1$
$α = 1$
At,
$x = 2, lim_{x \to 2^{-}} f (x) = f (2)$
$6 + α = 2 β + 3$
$7 = 2 β + 3$
$2 β = 4$
$β = 2$
So, $α^{2} + β^{2} = 1 + 4 = 5$

AP EAMCET-2018-24.04.2018

Limits, Continuity and Differentiability

80011 If $f : [- 2, 2] \to R$ is defined by $f (x) = | \begin{array}{cc} \frac{\sqrt{1 + c x} - \sqrt{1 - c x}}{\frac{x}{x + 3}} & for - 2 \leq x < 0 \\ \frac{x + 1}{x +} & for 0 \leq x \leq 2 \end{array}$
continuous on $[- 2, 2]$ then $c$ is equal to

1

\frac{2}{\sqrt{3}}

2 3

3

\frac{3}{2}

4

\frac{3}{\sqrt{2}}

Explanation:

(B) : Given, $f : [- 2, 2] \to R$
$f (x) = {\begin{array}{cc} \frac{\sqrt{1 + c x} - \sqrt{1 - c x}}{x}, & - 2 \leq x < 0 \\ \frac{x + 3}{x + 1}, & 0 \leq x \leq 2 \end{array}$
Now, LHL $= lim_{x \to 0^{-}} f (x)$
$= lim_{h \to 0} \frac{\sqrt{1 - ch} - \sqrt{1 + ch}}{- h} \times \frac{\sqrt{1 - ch} + \sqrt{1 + ch}}{\sqrt{1 - ch} + \sqrt{1 + ch}}$
$= lim_{h \to 0} \frac{(1 - ch) - (1 + ch)}{- h \sqrt{1 - 0} + \sqrt{1 + 0}} = lim_{h \to 0} \frac{- 2 ch}{- h (1 + 1)} = c$
And, $RHL = lim_{x \to 0^{+}} f (x)$
$= lim_{h \to 0} f (0 + h) = lim_{h \to 0} \frac{0 + h + 3}{0 + h + 1} = lim_{h \to 0} \frac{h + 3}{h + 1}$
$= \frac{0 + 3}{0 + 1} = 3$
Since, $f$ is continuous at $x = 0$ ,
$∴ LHL = RHL$
$c = 3$

AP EAMCET-2014

Limits, Continuity and Differentiability

80012 $f (x) = {\begin{cases} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}} x \neq 0 \\ k \log 2 \log 3 x = 0 \end{cases}$
Find the value of ' $k$ ' for which the function $f$ is continuous.

1

\sqrt{2}

2 24

3

18 \sqrt{3}

4

24 \sqrt{2}

Explanation:

(D): At, $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$k \log 2 \log 3 = lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}}$
$= lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{2 \cos^{2} \frac{x}{2}}} [1 + \cos x = 2 \cos^{2} \frac{x}{2}]$
$= lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{2} \cos \frac{x}{2}} (\frac{0}{0} form)$
On using L-Hospital Rule, we get -
$= lim_{x \to 0} \frac{72^{x} \log 72 - 9^{x} \log 9 - 8^{x} \log 8}{- \sqrt{2} (- \sin \frac{x}{2}) \frac{1}{2}}$
$= lim_{x \to 0} \frac{2 [72^{x} \log 72 - 9^{x} \log 9 - 8^{x} \log 8]}{\sqrt{2} \sin (\frac{x}{2})}$
Again using L-Hospital rule, we get -
$= lim_{x \to 0} \frac{\sqrt{2} [72^{x} (\log 72)^{2} - 9^{x} (\log 9)^{2} - 8^{x} (\log 8)^{2}]}{\frac{1}{2} (\cos \frac{x}{2})}$
$= 2 \sqrt{2} [(\log 72)^{2} - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9 + \log 8)^{2} (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9 + \log 8)^{2} - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9)^{2} + (\log 8)^{2} + 2 \log (9) \log (8) - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} (2 \log 9 \cdot \log 8) = 4 \sqrt{2} \log 9 \log 8$
$= 4 \sqrt{2} \log 3^{2} \log 2^{3} = 4 \sqrt{2} 2 (\log 3) \cdot 3 \cdot \log (2)$
$= 24 \sqrt{2} \log (3) \log (2)$
So, $k \log 2 \cdot \log 3 = 24 \sqrt{2} \log (2) \log (3)$
$k = 24 \sqrt{2}$

AP EAMCET-2021-19.08.2021

Limits, Continuity and Differentiability

80013 If the function $f (x)$ , defined below is continuous in the interval $[0, π]$ , then
$f (x) = {\begin{cases} x + a \sqrt{2} (\sin x), 0 \leq x < \frac{π}{4} \\ 2 x (\cot x) + b, \frac{π}{4} \leq x \leq \frac{π}{2} \\ a (\cos 2 x) - b (\sin x), \frac{π}{4} < x \leq π \end{cases}$

1

a = \frac{π}{6}, b = \frac{π}{12}

2

a = \frac{π}{6}, b = \frac{π}{12}

3

a = \frac{- π}{6}, b = \frac{- π}{12}

4

a = \frac{π}{6}, b = \frac{- π}{12}

Explanation:

(D) : We have,
$f (x) = {\begin{cases} x + a \sqrt{2} (\sin x), 0 \leq x < \frac{π}{4} \\ 2 x (\cot x) + b, \frac{π}{4} \leq x \leq \frac{π}{2} \\ a (\cos 2 x) - b (\sin x), \frac{π}{4} < x \leq π \end{cases}$
At, $x = \frac{π}{4}$ LHL $=$ RHL
$lim_{\begin{matrix} x \to \frac{π^{-}}{4} \end{matrix}} f (x) = lim_{x \to \frac{π^{+}}{4}} f (x)$
$lim_{x \to \frac{π^{-}}{4}} (x + a \sqrt{2} \sin x) = lim_{x \to \frac{π^{+}}{4}} (2 x \cdot \cot x + b)$
$\frac{π}{4} + a \sqrt{2} \sin \frac{π}{4} = 2 \cdot \frac{π}{4} \cdot \cot \frac{π}{4} + b$
$\frac{π}{4} + a = \frac{π}{2} + b$
$a - b = \frac{π}{4}$
$at, x = \frac{π}{2} \to L H L = R H L$
$lim_{x \to \frac{π^{-}}{2}} f (x) = lim_{x \to \frac{π^{+}}{2}} f (x)$
$lim_{x \to \frac{π}{2}} (2 x \cot x + b) = lim_{x \to \frac{π}{2}} a (\cos 2 x) - b \sin x$
$2 \cdot \frac{π}{2} \cdot \cot \frac{π}{2} + b = a \cos 2 \cdot \frac{π}{2} - b \sin \frac{π}{2}$
$0 + b = - a - b$
$2 b = - a$
Solving equation (i) \& (ii), we get,
$a = \frac{π}{6}, b = \frac{- π}{12}$

AP EAMCET-2021-19.08.2021

Limits, Continuity and Differentiability

80010 If the function $f$ defined by
$f (x) = {\begin{cases} \cos x, if x \leq 0 \\ 3 x + α, if 0 < x < 2 \\ β x + 3, if 2 \leq x \leq 4 \\ 11, if x > 4 \end{cases}$
Where, $α$ and $β$ are real constants, is continuous on $R$ , then $α^{2} + β^{2} =$

1 3

2 9

3 5

4 1

Explanation:

(C) : $f (x)$ is continuous on R, So
$at, x = 0$
$lim_{x \to 0^{+}} f (x) = f (0)$
$0 + α = \cos 0 = 1$
$α = 1$
At,
$x = 2, lim_{x \to 2^{-}} f (x) = f (2)$
$6 + α = 2 β + 3$
$7 = 2 β + 3$
$2 β = 4$
$β = 2$
So, $α^{2} + β^{2} = 1 + 4 = 5$

AP EAMCET-2018-24.04.2018

Limits, Continuity and Differentiability

80011 If $f : [- 2, 2] \to R$ is defined by $f (x) = | \begin{array}{cc} \frac{\sqrt{1 + c x} - \sqrt{1 - c x}}{\frac{x}{x + 3}} & for - 2 \leq x < 0 \\ \frac{x + 1}{x +} & for 0 \leq x \leq 2 \end{array}$
continuous on $[- 2, 2]$ then $c$ is equal to

1

\frac{2}{\sqrt{3}}

2 3

3

\frac{3}{2}

4

\frac{3}{\sqrt{2}}

Explanation:

(B) : Given, $f : [- 2, 2] \to R$
$f (x) = {\begin{array}{cc} \frac{\sqrt{1 + c x} - \sqrt{1 - c x}}{x}, & - 2 \leq x < 0 \\ \frac{x + 3}{x + 1}, & 0 \leq x \leq 2 \end{array}$
Now, LHL $= lim_{x \to 0^{-}} f (x)$
$= lim_{h \to 0} \frac{\sqrt{1 - ch} - \sqrt{1 + ch}}{- h} \times \frac{\sqrt{1 - ch} + \sqrt{1 + ch}}{\sqrt{1 - ch} + \sqrt{1 + ch}}$
$= lim_{h \to 0} \frac{(1 - ch) - (1 + ch)}{- h \sqrt{1 - 0} + \sqrt{1 + 0}} = lim_{h \to 0} \frac{- 2 ch}{- h (1 + 1)} = c$
And, $RHL = lim_{x \to 0^{+}} f (x)$
$= lim_{h \to 0} f (0 + h) = lim_{h \to 0} \frac{0 + h + 3}{0 + h + 1} = lim_{h \to 0} \frac{h + 3}{h + 1}$
$= \frac{0 + 3}{0 + 1} = 3$
Since, $f$ is continuous at $x = 0$ ,
$∴ LHL = RHL$
$c = 3$

AP EAMCET-2014

Limits, Continuity and Differentiability

80012 $f (x) = {\begin{cases} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}} x \neq 0 \\ k \log 2 \log 3 x = 0 \end{cases}$
Find the value of ' $k$ ' for which the function $f$ is continuous.

1

\sqrt{2}

2 24

3

18 \sqrt{3}

4

24 \sqrt{2}

Explanation:

(D): At, $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$k \log 2 \log 3 = lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}}$
$= lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{2 \cos^{2} \frac{x}{2}}} [1 + \cos x = 2 \cos^{2} \frac{x}{2}]$
$= lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{2} \cos \frac{x}{2}} (\frac{0}{0} form)$
On using L-Hospital Rule, we get -
$= lim_{x \to 0} \frac{72^{x} \log 72 - 9^{x} \log 9 - 8^{x} \log 8}{- \sqrt{2} (- \sin \frac{x}{2}) \frac{1}{2}}$
$= lim_{x \to 0} \frac{2 [72^{x} \log 72 - 9^{x} \log 9 - 8^{x} \log 8]}{\sqrt{2} \sin (\frac{x}{2})}$
Again using L-Hospital rule, we get -
$= lim_{x \to 0} \frac{\sqrt{2} [72^{x} (\log 72)^{2} - 9^{x} (\log 9)^{2} - 8^{x} (\log 8)^{2}]}{\frac{1}{2} (\cos \frac{x}{2})}$
$= 2 \sqrt{2} [(\log 72)^{2} - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9 + \log 8)^{2} (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9 + \log 8)^{2} - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9)^{2} + (\log 8)^{2} + 2 \log (9) \log (8) - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} (2 \log 9 \cdot \log 8) = 4 \sqrt{2} \log 9 \log 8$
$= 4 \sqrt{2} \log 3^{2} \log 2^{3} = 4 \sqrt{2} 2 (\log 3) \cdot 3 \cdot \log (2)$
$= 24 \sqrt{2} \log (3) \log (2)$
So, $k \log 2 \cdot \log 3 = 24 \sqrt{2} \log (2) \log (3)$
$k = 24 \sqrt{2}$

AP EAMCET-2021-19.08.2021

Limits, Continuity and Differentiability

80013 If the function $f (x)$ , defined below is continuous in the interval $[0, π]$ , then
$f (x) = {\begin{cases} x + a \sqrt{2} (\sin x), 0 \leq x < \frac{π}{4} \\ 2 x (\cot x) + b, \frac{π}{4} \leq x \leq \frac{π}{2} \\ a (\cos 2 x) - b (\sin x), \frac{π}{4} < x \leq π \end{cases}$

1

a = \frac{π}{6}, b = \frac{π}{12}

2

a = \frac{π}{6}, b = \frac{π}{12}

3

a = \frac{- π}{6}, b = \frac{- π}{12}

4

a = \frac{π}{6}, b = \frac{- π}{12}

Explanation:

(D) : We have,
$f (x) = {\begin{cases} x + a \sqrt{2} (\sin x), 0 \leq x < \frac{π}{4} \\ 2 x (\cot x) + b, \frac{π}{4} \leq x \leq \frac{π}{2} \\ a (\cos 2 x) - b (\sin x), \frac{π}{4} < x \leq π \end{cases}$
At, $x = \frac{π}{4}$ LHL $=$ RHL
$lim_{\begin{matrix} x \to \frac{π^{-}}{4} \end{matrix}} f (x) = lim_{x \to \frac{π^{+}}{4}} f (x)$
$lim_{x \to \frac{π^{-}}{4}} (x + a \sqrt{2} \sin x) = lim_{x \to \frac{π^{+}}{4}} (2 x \cdot \cot x + b)$
$\frac{π}{4} + a \sqrt{2} \sin \frac{π}{4} = 2 \cdot \frac{π}{4} \cdot \cot \frac{π}{4} + b$
$\frac{π}{4} + a = \frac{π}{2} + b$
$a - b = \frac{π}{4}$
$at, x = \frac{π}{2} \to L H L = R H L$
$lim_{x \to \frac{π^{-}}{2}} f (x) = lim_{x \to \frac{π^{+}}{2}} f (x)$
$lim_{x \to \frac{π}{2}} (2 x \cot x + b) = lim_{x \to \frac{π}{2}} a (\cos 2 x) - b \sin x$
$2 \cdot \frac{π}{2} \cdot \cot \frac{π}{2} + b = a \cos 2 \cdot \frac{π}{2} - b \sin \frac{π}{2}$
$0 + b = - a - b$
$2 b = - a$
Solving equation (i) \& (ii), we get,
$a = \frac{π}{6}, b = \frac{- π}{12}$

AP EAMCET-2021-19.08.2021

Limits, Continuity and Differentiability

80010 If the function $f$ defined by
$f (x) = {\begin{cases} \cos x, if x \leq 0 \\ 3 x + α, if 0 < x < 2 \\ β x + 3, if 2 \leq x \leq 4 \\ 11, if x > 4 \end{cases}$
Where, $α$ and $β$ are real constants, is continuous on $R$ , then $α^{2} + β^{2} =$

1 3

2 9

3 5

4 1

Explanation:

(C) : $f (x)$ is continuous on R, So
$at, x = 0$
$lim_{x \to 0^{+}} f (x) = f (0)$
$0 + α = \cos 0 = 1$
$α = 1$
At,
$x = 2, lim_{x \to 2^{-}} f (x) = f (2)$
$6 + α = 2 β + 3$
$7 = 2 β + 3$
$2 β = 4$
$β = 2$
So, $α^{2} + β^{2} = 1 + 4 = 5$

AP EAMCET-2018-24.04.2018

Limits, Continuity and Differentiability

80011 If $f : [- 2, 2] \to R$ is defined by $f (x) = | \begin{array}{cc} \frac{\sqrt{1 + c x} - \sqrt{1 - c x}}{\frac{x}{x + 3}} & for - 2 \leq x < 0 \\ \frac{x + 1}{x +} & for 0 \leq x \leq 2 \end{array}$
continuous on $[- 2, 2]$ then $c$ is equal to

1

\frac{2}{\sqrt{3}}

2 3

3

\frac{3}{2}

4

\frac{3}{\sqrt{2}}

Explanation:

(B) : Given, $f : [- 2, 2] \to R$
$f (x) = {\begin{array}{cc} \frac{\sqrt{1 + c x} - \sqrt{1 - c x}}{x}, & - 2 \leq x < 0 \\ \frac{x + 3}{x + 1}, & 0 \leq x \leq 2 \end{array}$
Now, LHL $= lim_{x \to 0^{-}} f (x)$
$= lim_{h \to 0} \frac{\sqrt{1 - ch} - \sqrt{1 + ch}}{- h} \times \frac{\sqrt{1 - ch} + \sqrt{1 + ch}}{\sqrt{1 - ch} + \sqrt{1 + ch}}$
$= lim_{h \to 0} \frac{(1 - ch) - (1 + ch)}{- h \sqrt{1 - 0} + \sqrt{1 + 0}} = lim_{h \to 0} \frac{- 2 ch}{- h (1 + 1)} = c$
And, $RHL = lim_{x \to 0^{+}} f (x)$
$= lim_{h \to 0} f (0 + h) = lim_{h \to 0} \frac{0 + h + 3}{0 + h + 1} = lim_{h \to 0} \frac{h + 3}{h + 1}$
$= \frac{0 + 3}{0 + 1} = 3$
Since, $f$ is continuous at $x = 0$ ,
$∴ LHL = RHL$
$c = 3$

AP EAMCET-2014

Limits, Continuity and Differentiability

80012 $f (x) = {\begin{cases} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}} x \neq 0 \\ k \log 2 \log 3 x = 0 \end{cases}$
Find the value of ' $k$ ' for which the function $f$ is continuous.

1

\sqrt{2}

2 24

3

18 \sqrt{3}

4

24 \sqrt{2}

Explanation:

(D): At, $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$k \log 2 \log 3 = lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}}$
$= lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{2 \cos^{2} \frac{x}{2}}} [1 + \cos x = 2 \cos^{2} \frac{x}{2}]$
$= lim_{x \to 0} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{2} \cos \frac{x}{2}} (\frac{0}{0} form)$
On using L-Hospital Rule, we get -
$= lim_{x \to 0} \frac{72^{x} \log 72 - 9^{x} \log 9 - 8^{x} \log 8}{- \sqrt{2} (- \sin \frac{x}{2}) \frac{1}{2}}$
$= lim_{x \to 0} \frac{2 [72^{x} \log 72 - 9^{x} \log 9 - 8^{x} \log 8]}{\sqrt{2} \sin (\frac{x}{2})}$
Again using L-Hospital rule, we get -
$= lim_{x \to 0} \frac{\sqrt{2} [72^{x} (\log 72)^{2} - 9^{x} (\log 9)^{2} - 8^{x} (\log 8)^{2}]}{\frac{1}{2} (\cos \frac{x}{2})}$
$= 2 \sqrt{2} [(\log 72)^{2} - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9 + \log 8)^{2} (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9 + \log 8)^{2} - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} [(\log 9)^{2} + (\log 8)^{2} + 2 \log (9) \log (8) - (\log 9)^{2} - (\log 8)^{2}]$
$= 2 \sqrt{2} (2 \log 9 \cdot \log 8) = 4 \sqrt{2} \log 9 \log 8$
$= 4 \sqrt{2} \log 3^{2} \log 2^{3} = 4 \sqrt{2} 2 (\log 3) \cdot 3 \cdot \log (2)$
$= 24 \sqrt{2} \log (3) \log (2)$
So, $k \log 2 \cdot \log 3 = 24 \sqrt{2} \log (2) \log (3)$
$k = 24 \sqrt{2}$

AP EAMCET-2021-19.08.2021

Limits, Continuity and Differentiability

80013 If the function $f (x)$ , defined below is continuous in the interval $[0, π]$ , then
$f (x) = {\begin{cases} x + a \sqrt{2} (\sin x), 0 \leq x < \frac{π}{4} \\ 2 x (\cot x) + b, \frac{π}{4} \leq x \leq \frac{π}{2} \\ a (\cos 2 x) - b (\sin x), \frac{π}{4} < x \leq π \end{cases}$

1

a = \frac{π}{6}, b = \frac{π}{12}

2

a = \frac{π}{6}, b = \frac{π}{12}

3

a = \frac{- π}{6}, b = \frac{- π}{12}

4

a = \frac{π}{6}, b = \frac{- π}{12}

Explanation:

(D) : We have,
$f (x) = {\begin{cases} x + a \sqrt{2} (\sin x), 0 \leq x < \frac{π}{4} \\ 2 x (\cot x) + b, \frac{π}{4} \leq x \leq \frac{π}{2} \\ a (\cos 2 x) - b (\sin x), \frac{π}{4} < x \leq π \end{cases}$
At, $x = \frac{π}{4}$ LHL $=$ RHL
$lim_{\begin{matrix} x \to \frac{π^{-}}{4} \end{matrix}} f (x) = lim_{x \to \frac{π^{+}}{4}} f (x)$
$lim_{x \to \frac{π^{-}}{4}} (x + a \sqrt{2} \sin x) = lim_{x \to \frac{π^{+}}{4}} (2 x \cdot \cot x + b)$
$\frac{π}{4} + a \sqrt{2} \sin \frac{π}{4} = 2 \cdot \frac{π}{4} \cdot \cot \frac{π}{4} + b$
$\frac{π}{4} + a = \frac{π}{2} + b$
$a - b = \frac{π}{4}$
$at, x = \frac{π}{2} \to L H L = R H L$
$lim_{x \to \frac{π^{-}}{2}} f (x) = lim_{x \to \frac{π^{+}}{2}} f (x)$
$lim_{x \to \frac{π}{2}} (2 x \cot x + b) = lim_{x \to \frac{π}{2}} a (\cos 2 x) - b \sin x$
$2 \cdot \frac{π}{2} \cdot \cot \frac{π}{2} + b = a \cos 2 \cdot \frac{π}{2} - b \sin \frac{π}{2}$
$0 + b = - a - b$
$2 b = - a$
Solving equation (i) \& (ii), we get,
$a = \frac{π}{6}, b = \frac{- π}{12}$

AP EAMCET-2021-19.08.2021

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