32. Show that the function defined byf (x) = |\cos x |  is a continuous function.

Answers (1)

Given function is 
f (x) = |\cos x |
given function is defined for all values of x
f = g o h ,  g(x) = |x| and h(x) = cos x
Now,
g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}
g(x) is defined for all real numbers k
case(i)  k < 0
g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)
Hence, g(x) is continuous when k < 0

case (ii) k > 0
g(k) = k\\ \lim_{x\rightarrow k}g(x) = k\\ \lim_{x\rightarrow k}g(x) = g(k)
Hence, g(x) is continuous when k > 0

case (iii) k = 0
g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now, 
h(x) = cos x
Let suppose  x = c + h
if  x \rightarrow c , \ then \ h \rightarrow 0
h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h
                                                                                                 =\cos c\cos 0 + \sin c \sin 0 = \cos c
\lim_{x\rightarrow c}h(x) = h(c)
Hence, function h(x) = \cos x is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, f(x) = g o h is also continuous

Most Viewed Questions

Preparation Products

Knockout NEET May 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET May 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions