Class 12-commerce NCERT Solutions Maths Chapter 5: Continuity and Differentiability
Continuity and Differentiability Exercise Ex. 5.1
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The given function is f(x) = 5x - 3
At x = 0, f(0) = 5 × 0 - 3 = -3
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The given function is f(x) = x2 - sinx + 5
It is evident that f is defined at x = ∏
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The given function f is continuous at x = 
, if f is defined at x = and if the value
of f at x = equals the limit of f at x =
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Continuity and Differentiability Exercise Ex. 5.2
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Then, (v ο u)(x) = v(u(x)) = v(x2 + 5) = sin (x2 + 5) = f(x)
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Continuity and Differentiability Exercise Ex. 5.3
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Continuity and Differentiability Exercise Ex. 5.4
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Continuity and Differentiability Exercise Ex. 5.5
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Continuity and Differentiability Exercise Ex. 5.6
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Continuity and Differentiability Exercise Ex. 5.7
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Continuity and Differentiability Exercise Misc. Ex.
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where sin x > cosx
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Yes.
Consider the function f(x)=|x-1|+|x-2|
Since we know that the modulus function is continuous everywhere, so there sum is also continuous
Therefore, function f is continuous everywhere
Now, let us check the differentiability of f(x) at x=1,2
At x=1
LHD = 
[Take x=1-h, h>0 such that h→0 as x→1-]
Now,
RHD = 
[Take x=1+h, h>0 such that h→0 as x→1+]
≠ LHD
Therefore, f is not differentiable at x=1.
At x=2
LHD = 
[Take x=2-h, h>0 such that h→0 as x→2-]
Now,
RHD = 
[Take x=2+h, h>0 such that h→0 as x→2+]
≠ LHD
Therefore, f is not differentiable at x=2.
Hence, f is not differentiable at exactly two points.
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