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Class 12 RD Sharma Solutions – Chapter 9 Continuity – Exercise 9.1 | Set 3

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  • Last Updated : 26 May, 2021
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Question 31. If f(x)=\begin{cases}\frac{2^{x+2}-16}{4^x-16},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases}     is continuous at x = 2, find k.

Solution: 

Given that,

f(x)=\begin{cases}\frac{2^{x+2}-16}{4^x-16},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases}

Also, f(x) is continuous at x = 2

So, LHL = RHL = f(2)        …..(i)

Now,

f(2) = k  ……(ii)

Let us consider LHL,

\lim_{x\to2^-}f(x) =\lim_{h\to0}f(2-h)

=\lim_{h\to0}\frac{2^{(2-h)+2}-16}{4^{(2-h)}-16}

=\lim_{h\to0}\frac{2^{4-h}-16}{4^{(2-h)}-16}

=\lim_{h\to0}\frac{2^4.2^{-h}-16}{4^2.4^{-h}-16}

=\lim_{h\to0}\frac{16.2^{-h}-16}{16.4^{-h}-16}

=\lim_{h\to0}\frac{16(2^{-h}-1)}{16(4^{-h}-1)}

=\lim_{h\to0}\frac{2^{-h}-1}{(2^{-h})^2-1^2}

=\lim_{h\to0}\frac{2^{-h}-1}{(2^{-h}-1)(2^{-h}+1)}=1/2          ……(iii)

Using eq(i), (ii) and (iii), we get

k = 1/2

Question 32. If f(x)=\begin{cases}\frac{cos^2x-sin^2x-1}{\sqrt{x^2+1}-1},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}     is continuous at x = 0, find k.

Solution:

Given that, 

f(x)=\begin{cases}\frac{cos^2x-sin^2x-1}{\sqrt{x^2+1}-1},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}

Also, f(x) is continuous at x = 2

So, LHL = RHL

Now,

\lim_{x\to0}f(x)=f(0)

⇒ \lim_{x\to0}\frac{cos^2x-sin^2x-1}{\sqrt{x^2+1}-1}=k

⇒ \lim_{x\to0}\frac{1-sin^2x-sin^2x-1}{\sqrt{x^2+1}-1}=k

⇒ \lim_{x\to0}\frac{-2sin^2x}{\sqrt{x^2+1}-1}=k

⇒ \lim_{x\to0}\frac{-2(sin^2x)(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)}=k

⇒ \lim_{x\to0}\frac{-2(sin^2x)(\sqrt{x^2+1}+1)}{x^2}=k

⇒ -2\lim_{x\to0}\frac{(sin^2x)(\sqrt{x^2+1}+1)}{x^2}=k

⇒ -2\lim_{x\to0}(\frac{sinx}{x})^2\lim_{x\to0}(\sqrt{x^2+1}+1)=k

⇒ -2 × 1 × (1 + 1) = k

⇒ k = -4

Question 33. Extend the definition of the following by continuity f(x) = \frac{1-cos7(x-π)}{5(x-π)^2}       at the point x = π.

Solution: 

Given that, 

\frac{1-cos7(x-π)}{5(x-π)^2}

As we know that a f(x) is continuous at x = π if,

LHL = RHL = f(π)  ……(i)

Let us consider LHL,

\lim_{x\toπ^-}f(x) =\lim_{h\to0}f(π-h)

=\lim_{h\to0}\frac{1-cos7(π-h-π)}{5((π-h)-π)^2}

=\lim_{h\to0}\frac{2sin^2(7/2)h}{5h^2}

=\lim_{h\to0}(2/5)(\frac{sin(7/2)h}{(7/2)h})^2×(7/2)^2

= (2/5) × (49/4) = 49/10

 Thus, from eq(i) we get,

f(π) = 49/10

Hence, f(x) is continuous at x = π

Question 34. If f(x) = \frac{2x+3sinx}{3x+2sinx}      , x ≠ 0 is continuous at x = 0, then find f(0).

Solution:

Given that, 

f(x) = \frac{2x+3sinx}{3x+2sinx}

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)     ……(i)

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)

=\lim_{h\to0}\frac{2(-h)+3sin(-h)}{3(-h)+2sin(-h)}

=\lim_{h\to0}\frac{-2h-3sinh}{-3h-2sinh}

=\lim_{h\to0}\frac{\frac{2h+3sinh}{h}}{\frac{3h+2sinh}{h}}

=\lim_{h\to0}\frac{2+3\frac{sinh}{h}}{3+2\frac{sinh}{h}}=\frac{2+3}{3+2}=1

From eq(i) we get,

f(0) = 1

Question 35. Find the value of k for which f(x)=\begin{cases}\frac{1-cos4x}{8x^2}  ,& \text{when }x\neq0 \\k,& \text{when }x=0\end{cases}     is continuous at x = 0

Solution:

Given that, 

f(x)=\begin{cases}\frac{1-cos4x}{8x^2}  ,& \text{when }x\neq0 \\k,& \text{when }x=0\end{cases}

Also, f(x) is continuous at x = 0

LHL = RHL = f(0)     …..(i)

f(0) = k

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)

=\lim_{h\to0}\frac{1-cos4(-h)}{8(-h)^2}

=\lim_{h\to0}\frac{1-cos4h}{8h^2}

=\lim_{h\to0}\frac{2sin^22h}{8h^2}

=\lim_{h\to0}(\frac{sin2h}{2h})^2=1

Thus, from eq(i) we get,

k = 1

Question 36. In each of the following, find the value of the constant k so that the given function is continuous at the indicated point:

(i) f(x)=\begin{cases}\frac{1-cos2kx}{x^2},& \text{if }x\neq0 \\8,& \text{if }x=0\end{cases}     at x = 0

Solution:

Given that, 

f(x)=\begin{cases}\frac{1-cos2kx}{x^2},& \text{if }x\neq0 \\8,& \text{if }x=0\end{cases}

Also, f(x) is continuous at x = 0

\lim_{x\to0}f(x) =f(0)

⇒ \lim_{x\to0}\frac{1-cos2kx}{x^2}=8

⇒ \lim_{x\to0}\frac{2k^2sin^2kx}{k^2x^2}=8

⇒ 2k^2\lim_{x\to0}(\frac{sinkx}{kx})^2=8

⇒ 2k2 × 1 = 8

⇒ k2 = 4

⇒ k = ±2

(ii) f(x)=\begin{cases}(x-1)\frac{tanπx}{2},& \text{if }x\neq1 \\k,& \text{if }x=1\end{cases}   at x = 1

Solution:

Given that, 

f(x)=\begin{cases}(x-1)\frac{tanπx}{2},& \text{if }x\neq1 \\k,& \text{if }x=1\end{cases}

Also, f(x) is continuous at x = 1

\lim_{x\to1}f(x) =f(1)

⇒ \lim_{x\to1}(x-1)tan(πx/2)=k

Now, on putting x – 1 = y, we get

\lim_{y\to0}ytan\frac{π(y+1)}{2}=k

⇒ \lim_{y\to0}ytan(\frac{πy}{2}+π/2)=k

⇒ \lim_{y\to0}ytan(\frac{π}{2}+\frac{πy}{2})=k

⇒ -\lim_{y\to0}ycot(\frac{π}{2})=k

⇒ \frac{-2}{π}\lim_{y\to0}\frac{\frac{πy}{2}cos(\frac{πy}{2})}{sin\frac{πy}{2}}=k

⇒ \frac{-2}{π}\frac{\lim_{y\to0}cos(\frac{πy}{2})}{\lim_{y\to0}(\frac{sin(\frac{πy}{2})}{\frac{πy}{2}})}=k

⇒ (-2/π) × (1/1) = k

⇒ k = (-2/π)

(iii) f(x)=\begin{cases}k(x^2-2x),& \text{if }x<0 \\cosx,& \text{if }x\geq0\end{cases}    at x = 0

Solution:

Given that, 

f(x)=\begin{cases}k(x^2-2x),& \text{if }x<0 \\cosx,& \text{if }x\geq0\end{cases}

Also, f(x) is continuous at x = 0

Let us consider LHL, at x = 0

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)

=\lim_{h\to0}f(-h)

=\lim_{h\to0}k(h^2+2h)=0

Let us consider RHL at x = 0

\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)

=\lim_{h\to0}f(h)

=\lim_{h\to0}cosh=1

\lim_{x\to0^-}f(x)≠\lim_{x\to0^+}f(x)

Hence, no value of k exists for which function is continuous at x = 0.

(iv) f(x)=\begin{cases}kx+1,& \text{if }x\leqπ \\cosx,& \text{if }x>π\end{cases}   at x = π

Solution:

Given that, 

f(x)=\begin{cases}kx+1,& \text{if }x\leqπ \\cosx,& \text{if }x>π\end{cases}

Also, f(x) is continuous at x = π

Let us consider LHL 

\lim_{x\toπ^-}f(x) =\lim_{h\to0}f(π-h)

=\lim_{h\to0}k(π-h)+1=kπ+1

Let us consider RHL 

\lim_{x\toπ^+}f(x) =\lim_{h\to0}f(π+h)

=\lim_{h\to0}cos(π+h)

cosπ = -1

As we know that f(x) is continuous at x = π, so

\lim_{x\toπ^-}f(x)=\lim_{x\toπ^+}f(x)

⇒ kπ + 1 = -1

⇒ k = (-2/π)

(v) f(x)=\begin{cases}kx+1,& \text{if }x\leq5 \\3x-5,& \text{if }x>5\end{cases}   at x = 5

Solution:

Given that, 

f(x)=\begin{cases}kx+1,& \text{if }x\leq5 \\3x-5,& \text{if }x>5\end{cases}

Also, f(x) is continuous at x = 5

Let us consider LHL 

\lim_{x\to5^-}f(x)=\lim_{h\to0}f(5-h)

=\lim_{h\to0}k(5-h)+1

= 5k + 1

Let us consider RHL 

\lim_{x\to5^+}f(x)=\lim_{h\to0}f(5+h)

=\lim_{h\to0}3(5+h)-5

= 10

As we know that f(x) is continuous at x = 5, so

\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)

⇒ 5k + 1 = 10

⇒ k = 9/5

(vi) f(x)=\begin{cases}\frac{x^2-25}{x-5},& \text{if }x\neq5 \\k,& \text{if }x=5\end{cases}   at x = 5 

Solution:

Given that, 

f(x)=\begin{cases}\frac{x^2-25}{x-5},& \text{if }x\neq5 \\k,& \text{if }x=5\end{cases}

Also, f(x) is continuous at x = 5

So, 

f(x) = (x2 – 25)/(x – 5), if x ≠ 5 & f(x) = k, if x = 5

⇒ f(x)= {(x – 5)(x+5)/(x-5)}, if x ≠ 5 & f(x) = k, if x = 5

⇒ f(x)= (x + 5), if x ≠ 5 & f(x) = k, if x = 5

As we know that f(x) is continuous at x = 5, so

\lim_{x\to5}f(x)=f(5)

⇒ \lim_{x\to5}(x+5)=k

⇒ k = 5 + 5 = 10

(vii) f(x)=\begin{cases}kx^2,& \text{if }x\geq1 \\4,& \text{if }x<1\end{cases}   at x = 1

Solution:

Given that, 

f(x)=\begin{cases}kx^2,& \text{if }x\geq1 \\4,& \text{if }x<1\end{cases}

Also, f(x) is continuous at x = 1

Let us consider LHL 

\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)

=\lim_{h\to0}4=4

Let us consider RHL 

\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)

=\lim_{h\to0}k(1+h)^2

= k

As we know that f(x) is continuous at x = 1, so

\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)

⇒ k = 4

(viii) f(x)=\begin{cases}k(x^2+2),& \text{if }x\leq0 \\3x+1,& \text{if }x>0\end{cases}   at x = 0

Solution:

Given that, 

f(x)=\begin{cases}k(x^2+2),& \text{if }x\leq0 \\3x+1,& \text{if }x>0\end{cases}

Also, f(x) is continuous at x = 0

Let us consider LHL 

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)

=\lim_{h\to0}k((-h)^2+2)

= 2k

Let us consider RHL 

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)

=\lim_{h\to0}3h+1

= 1

As we know that f(x) is continuous at x = 0, so

\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)

⇒ 2k = 1

⇒ k = 1/2

(ix) f(x)=\begin{cases}\frac{x^3+x^2-16x+20}{(x-2)^2},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases}   at x = 2

Solution:

Given that, 

f(x)=\begin{cases}\frac{x^3+x^2-16x+20}{(x-2)^2},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases}

Also, f(x) is continuous at x = 2

f(x)= \frac{x^3+x^2-16x+20}{(x-2)^2}        , if x ≠ 2 & f(x) = k, if x = 2

⇒ f(x)= \frac{x^3+x^2-16x+20}{x^2-4x+4}         , if x ≠ 2 & f(x) = k, if x = 2

⇒ f(x)= \frac{(x+5)(x^2-4x+4)}{x^2-4x+4}         , if x ≠ 2 & f(x) = k, if x = 2

⇒ f(x)= (x + 5), if x ≠ 2 & f(x) = k, if x = 2

As we know that f(x) is continuous at x = 2, so

\lim_{x\to2}f(x)=f(2)

⇒ \lim_{x\to2}(x+5)=f(2)

⇒ k = 2 + 5 = 7

Question 37. Find the values of a and b so that the function f given by 

f(x)=\begin{cases}1,& \text{if }x\leq3 \\ax+b,& \text{if }3<x<5\\7, &\text{if }x\geq5\end{cases}    is continuous at x = 3 and x = 5.

Solution:

Given that, 

f(x)=\begin{cases}1,& \text{if }x\leq3 \\ax+b,& \text{if }3<x<5\\7, &\text{if }x\geq5\end{cases}

Let us consider LHL at x = 3,

\lim_{x\to3^-}f(x)=\lim_{h\to0}f(3-h)

=\lim_{h\to0}(1)

= 1

Let us consider RHL at x = 3,

\lim_{x\to3^+}f(x)=\lim_{h\to0}f(3+h)

=\lim_{h\to0}a(3+h)+b

= 3a + b

Let us consider LHL at x = 5,

\lim_{x\to5^-}f(x)=\lim_{h\to0}f(5-h)

=\lim_{h\to0}(a(5-h)+b)

= 5a + b

Let us consider RHL at x = 5,

\lim_{x\to5^+}f(x)=\lim_{h\to0}f(5+h)

=\lim_{h\to0}7

= 7

It is given that f(x) is continuous at x = 3 and x = 5, then

\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)         and \lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)

⇒ 1 = 3a + b …..(i) 

and 5a + b = 7 …….(ii)

On solving eq(i) and (ii), we get

a = 3 and b = -8

Question 38. If f(x)=\begin{cases}\frac{x^2}{2},& \text{if }0\leq x \leq 1 \\2x^2-3x+(\frac{3}{2}),& \text{if }1<x\leq2\end{cases}   . Show that f is continuous at x = 1.

Solution:

Given that, 

f(x)=\begin{cases}\frac{x^2}{2},& \text{if }0\leq x \leq 1 \\2x^2-3x+(\frac{3}{2}),& \text{if }1<x\leq2\end{cases}

So, 

Let us consider LHL at x = 1,

\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)

=\lim_{h\to0}\frac{(1-h)^2}{2}

= 1/2

Let us consider RHL at x = 1,

\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)

=\lim_{h\to0}[2(1+h)^2-3(1+h)+3/2]

= 2 – 3 + 3/2 = 1/2

Also,

f(1) = (1)2/2 = 1/2

\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)

LHL = RHL = f(1)

Hence, the f(x) is continuous at x = 1

Question 39. Discuss the continuity of the f(x) at the indicated points:

(i) f(x) = |x| + |x – 1| at x = 0, 1.

Solution:

Given that, 

f(x) = |x| + |x – 1|

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL at x = 0,

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)

=\lim_{h\to0}[|0-h|+|0-h-1|]=1

Let us consider RHL at x = 0,

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)

=\lim_{h\to0}[|0+h|+|0+h-1|]=1

Also,

f(0) = |0| + |0 – 1| = 0 + 1 = 1

LHL = RHL = f(0)

Now, we check the continuity of the given f(x) at x = 1,

Let us consider LHL at x = 1,

\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)

=\lim_{h\to0}f(|1-h|+|1-h-1|)=1+0

= 1

Let us consider RHL at x = 1

\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)

=\lim_{h\to0}f(|1+h|+|1+h-1|)=1+0

= 1

Also,

f(1) = |1| + |1 – 1| = 1 + 0 = 1

LHL = RHL = f(1)

Hence, f(x) is continuous at x = 0, 1.

(ii) f(x) = |x – 1| + |x + 1| at x = -1, 1.

Solution:

Given that, 

f(x) = |x – 1| + |x + 1| at x = -1, 1.

So, here we check the continuity of the given f(x) at x = -1,

Let us consider LHL at x = -1,

\lim_{x\to-1^-}f(x)=\lim_{h\to0}f(-1-h)

=\lim_{h\to0}[|-1-h-1|+|-1-h+1|]=2+0=2

Let us consider RHL at x = -1,

\lim_{x\to-1^+}f(x)=\lim_{h\to0}f(-1+h)

=\lim_{h\to0}[|-1+h-1|+|-1+h+1|]=2+0=2

Also,

f(-1) = |-1 – 1| + |-1 + 1| = |-2| = 2

LHL = RHL = f(-1)

Now, we check the continuity of the given f(x) at x = 1,

Let us consider LHL at x = 1,

\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)

=\lim_{h\to0}f(|1-h-1|+|1-h+1|)=0+2

= 2

\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)

=\lim_{h\to0}f(|1+h-1|+|1+h+1|)=0+2

= 2

Also,

f(1) = |1 + 1| + |1 – 1| = 2

LHL = RHL = f(1)

Hence, f(x) is continuous at x = -1, 1.

Question 40. Prove that f(x)=\begin{cases}\frac{x-|x|}{x},& \text{if }x\neq0 \\2,& \text{if }x=0\end{cases}   is discontinuous at x = 0.

Solution:

Prove that f(x)=\begin{cases}\frac{x-|x|}{x},& \text{if }x\neq0 \\2,& \text{if }x=0\end{cases}   is discontinuous at x = 0.

Proof:

Let us consider LHL at x = 0,

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)

=\lim_{h\to0}f(-h)

=\lim_{h\to0}2=2

Let us consider RHL at x = 0,

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)

=\lim_{h\to0}f(h)

=\lim_{h\to0}0=0

LHL ≠ RHL

Hence, f(x) is discontinuous at x = 0.

Question 41. If f(x)=\begin{cases}2x^2+k,& \text{if }x\geq0 \\-2x^2+k,& \text{if }x<0\end{cases}    then what should be the value of k so that f(x) is continuous at x = 0.

Solution:

Given that,

f(x)=\begin{cases}2x^2+k,& \text{if }x\geq0 \\-2x^2+k,& \text{if }x<0\end{cases}

Let us consider LHL at x = 0,

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)

=\lim_{h\to0}f(-h)

=\lim_{h\to0}-2(-h)^2+k

= k

Let us consider RHL at x = 0,

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)

=\lim_{h\to0}f(h)

=\lim_{h\to0}f(2h^2+k)

= k

It is given that f(x) is continuous at x = 0.

LHL = RHL = f(0)

⇒ \lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=k

k can be any real number.

Question 42. For what value λ of is the function

f(x)=\begin{cases}λ(x^2-2x),& \text{if }x\leq0 \\4x+1,& \text{if }x>0\end{cases}   continuous at x = 0 ? What about continuity at x = ±1?

Solution:

Given that,

f(x)=\begin{cases}λ(x^2-2x),& \text{if }x\leq0 \\4x+1,& \text{if }x>0\end{cases}

Check for x = 0, 

Hence, there is no value of λ for which f(x) is continuous at x = 0.

Now for x = 1,

f(1) = 4x + 1 = 4 × 1 + 1 = 5

Hence, for any values of λ, f is continuous at x = 1.

Now for x = -1, 

f(-1) = λ(1 + 2)= 3λ

=\lim_{x\to-1}λ(1+2)=3λ

 =\lim_{x\to-1}f(x)=f(-1)

Hence, for any values of λ, f is continuous at x=-1.

Question 43. For what values of k is the following function continuous at x = 2? 

 f(x)=\begin{cases}2x+1,& \text{if }x<2 \\k,& \text{if }x=2\\3x-1,& \text{if }x>2\end{cases}

Solution:

Given that,

f(x)=\begin{cases}2x+1,& \text{if }x<2 \\k,& \text{if }x=2\\3x-1,& \text{if }x>2\end{cases}

We have,

Let us consider LHL at x = 2,

=\lim_{x\to2^-}f(x)=\lim_{h\to0}f(2-h)

=\lim_{h\to0}(2(2-h)+1)

= 5

Let us consider RHL at x = 2,

\lim_{x\to2^+}f(x)=\lim_{h\to0}f(2+h)

=\lim_{h\to0}3(2+h)-1

= 5

Also,

f(2) = k

It is given that f(x) is continuous at x = 2.

LHL = RHL = f(2)

⇒ 5 = 5 = k

Hence, for k = 5, f(x) is continuous at x = 2.

Question 44. Let f(x)=\begin{cases}1-sin^3x3cos^2x,& \text{if }x<(\frac{π}{2}) \\a,& \text{if }x=(\frac{π}{2})\\\frac{b(1-sinx)}{(π-2x)^2},& \text{if }x>(\frac{π}{2})\end{cases}    If f(x) is continuous at x = (π/2), find a and b.

Solution:

Given that,

f(x)=\begin{cases}1-sin^3x3cos^2x,& \text{if }x<(\frac{π}{2}) \\a,& \text{if }x=(\frac{π}{2})\\\frac{b(1-sinx)}{(π-2x)^2},& \text{if }x>(\frac{π}{2})\end{cases}

Let us consider LHL at x = π/2

=\lim_{x\to(\frac{π}{2})^-}f(x)=\lim_{h\to0}f(\frac{π}{2}-h)

=\lim_{h\to0}\frac{1-sin^3(\frac{π}{2}-h)}{3cos^2(\frac{π}{2}-h)}

=\lim_{h\to0}\frac{1-cos^3h}{3sin^2h}

=\frac{1}{3}\lim_{h\to0}(\frac{(1-cosh)(1+cos^2h+cosh)}{(1-cosh)(1+cosh)})

=\frac{1}{3}\lim_{h\to0}(\frac{(1+cos^2h+cosh)}{(1+cosh)})

=\frac{1}{3}(\frac{1+1+1}{1+1})

= 1/2

Let us consider RHL at x = π/2

=\lim_{x\to(\frac{π}{2})^+}f(x)=\lim_{h\to0}f(\frac{π}{2}+h)

=\lim_{h\to0}(\frac{b[1-sin(\frac{π}{2}+h)]}{[π-2(\frac{π}{2}+h)]^2})

=\lim_{h\to0}(\frac{b(1-cosh)}{[-2h]^2})

=\lim_{h\to0}(\frac{2bsin^2(\frac{h}{2})}{4h^2})

=\lim_{h\to0}(\frac{2bsin^2(\frac{h}{2})}{\frac{16h^2}{4}})

=(\frac{b}{8})\lim_{h\to0}(\frac{sin(\frac{h}{2})}{\frac{h}{2}})^2

= b/8 × 1

= b/8

Also,

f(π/2) = a

It is given that f(x) is continuous at x = π/2.

LHL = RHL = f(π/2)

So, 

⇒ 1/2 = b/8 = a

⇒ a = 1/2 and b = 4

Question 45. If the functions f(x), defined below is continuous at x = 0, find the value of k,

f(x)=\begin{cases}\frac{1-cos2x}{2x^2},& \text{if }x<0 \\k,& \text{if }x=0\\\frac{x}{|x|},& \text{if }x>0\end{cases}

Solution:

Given that,

f(x)=\begin{cases}\frac{1-cos2x}{2x^2},& \text{if }x<0 \\k,& \text{if }x=0\\\frac{x}{|x|},& \text{if }x>0\end{cases}

Let us consider LHL at x = 0,

=\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)

=\lim_{h\to0}(\frac{1-cos2(-h)}{2(-h)^2})

=\lim_{h\to0}(\frac{1-cos2h}{2h^2})

=\frac{1}{2}\lim_{h\to0}(\frac{2sin^2h}{h^2})

=\frac{2}{2}\lim_{h\to0}(\frac{sin^2h}{h^2})

=\frac{2}{2}\lim_{h\to0}(\frac{sinh}{h})^2

= 1 × 1

Let us consider RHL at x = 0,

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)

=\lim_{h\to0}f(h)

=\lim_{h\to0}1=1

Also,

f(0) = k

It is given that f(x) is continuous at x = 0,

LHL = RHL = f(0)

So, 

⇒ 1 = 1 = k

Hence, the required value of k is 1.

Question 46. Find the relationship between ‘a’ and ‘b’ so that function ‘f’ defined by

f(x)=\begin{cases}ax+1,& \text{if }x\leq3 \\bx+3,& \text{if }x>3\\\end{cases}   is continuous at x = 3.

Solution:

Given that,

f(x)=\begin{cases}ax+1,& \text{if }x\leq3 \\bx+3,& \text{if }x>3\\\end{cases}

Let us consider LHL at x = 3,

\lim_{x\to3^-}f(x)=\lim_{h\to0}f(3-h)

=\lim_{h\to0}a(3-h)+1

= 3a + 1

Let us consider RHL at x = 3,

\lim_{x\to3^+}f(x)=\lim_{h\to0}f(3+h)

=\lim_{h\to0}b(3+h)+3

= 3b + 3

It is given that f(x) is continuous at x = 3,

LHL = RHL = f(3)

So, 

⇒ 3a + 1 = 3b + 3

⇒ 3a – 3b = 2

Hence, the required relationship between a and b is 3a – 3b = 2.


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