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Class 12 Maths
Continuity And Differentiability
Quiz 1
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If f (x) = 2x and g (x) = \(\frac{x^2}{2}\) + 1, then’which of the following can be a discontinuous function
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f(x) + g(x)
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f(x) – g(x)
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f(x).g(x)
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\(\frac{g(x)}{f(x)}\)
Explanation
\(\frac{g(x)}{f(x)}\)
The function f(x) = \(\frac{4-x^2}{4x-x^3}\) is
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discontinuous at only one point at x = 0
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discontinuous at exactly two points
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discontinuous at exactly three points
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None of these
Explanation
discontinuous at only one point at x = 0
The set of points where the function f given by f (x) =| 2x – 1| sin x is differentiable is
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R
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R = {\(\frac{1}{2}\)}
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(0, ∞)
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None of these
Explanation
R = {\(\frac{1}{2}\)}
The function f(x) = cot x is discontinuous on the set
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{x = nπ, n ∈ Z}
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{x = 2nπ, n ∈ Z}
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{x = (2n + 1) \(\frac{π}{2}\) n ∈ Z}
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{x – \(\frac{nπ}{2}\) n ∈ Z}
Explanation
{x = nπ, n ∈ Z}
The function f(x) = e is
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continuous everywhere but not differentiable at x = 0
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continuous and differentiable everywhere
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not continuous at x = 0
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None of these
Explanation
continuous everywhere but not differentiable at x = 0
If f(x) = x² sin\(\frac{1}{x}\), where x ≠ 0, then the value of the function f(x) at x = 0, so that the function is continuous at x = 0 is
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0
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-1
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1
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None of these
Explanation
0
If f(x) = is continuous at x = \(\frac{π}{2}\), then
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m = 1, n = 0
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m = \(\frac{nπ}{2}\) + 1
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n = \(\frac{mπ}{2}\)
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m = n = \(\frac{π}{2}\)
Explanation
n = \(\frac{mπ}{2}\)
If y = log(\(\frac{1-x^2}{1+x^2}\)), then \(\frac{dy}{dx}\) is equal to
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\(\frac{4x^3}{1-x^4}\)
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\(\frac{-4x}{1-x^4}\)
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\(\frac{1}{4-x^4}\)
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\(\frac{-4x^3}{1-x^4}\)
Explanation
\(\frac{-4x}{1-x^4}\)
Let f(x) = |sin x| Then
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f is everywhere differentiable
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f is everywhere continuous but not differentiable at x = nπ, n ∈ Z
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f is everywhere continuous but no differentiable at x = (2n + 1) \(\frac{π}{2}\) n ∈ Z
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None of these
Explanation
f is everywhere continuous but not differentiable at x = nπ, n ∈ Z
If y = \(\sqrt{sin x+y}\) then \(\frac{dy}{dx}\) is equal to
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\(\frac{cosx}{2y-1}\)
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\(\frac{cosx}{1-2y}\)
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\(\frac{sinx}{1-xy}\)
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\(\frac{sinx}{2y-1}\)
Explanation
\(\frac{cosx}{2y-1}\)
The derivative of cos(2x² – 1) w.r.t cos x is
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2
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\(\frac{-1}{2\sqrt{1-x^2}}\)
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\(\frac{2}{x}\)
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1 – x²
Explanation
2
The value of c in Rolle’s theorem for the function f(x) = x³ – 3x in the interval [o, √3] is
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1
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-1
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\(\frac{3}{2}\)
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\(\frac{1}{3}\)
Explanation
1
For the function f(x) = x + \(\frac{1}{x}\), x ∈ [1, 3] the value of c for mean value theorem is
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1
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√3
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2
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None of these
Explanation
√3
Let f be defined on [-5, 5] as
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continuous at every x except x = 0
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discontinuous at everyx except x = 0
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continuous everywhere
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discontinuous everywhere
Explanation
discontinuous at everyx except x = 0
Let function f (x) =
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continuous at x = 1
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differentiable at x = 1
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continuous at x = -3
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All of these
Explanation
All of these
If f(x) = \(\frac{\sqrt{4+x}-2}{x}\) x ≠ 0 be continuous at x = 0, then f(o) =
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\(\frac{1}{2}\)
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\(\frac{1}{4}\)
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2
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\(\frac{3}{2}\)
Explanation
\(\frac{1}{4}\)
let f(2) = 4 then f”(2) = 4 then \(_{x→2}^{lim}\) \(\frac{xf(2)-2f(x)}{x-2}\) is given by
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2
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-2
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-4
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3
Explanation
-4
It is given that f'(a) exists, then \(_{x→2}^{lim}\) [/latex] \(\frac{xf(a)-af(x)}{(x-a)}\) is equal to
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f – af'
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f'(a)
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-f’(a)
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f (a) + af'(a)
Explanation
f(a) – af'(a)
If f(x) = \(\sqrt{25-x^2}\), then \(_{x→2}^{lim}\)\(\frac{f(x)-f(1)}{x-1}\) is equal to
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\(\frac{1}{24}\)
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\(\frac{1}{5}\)
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–\(\sqrt{24}\)
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\(\frac{1}{\sqrt{24}}\)
Explanation
\(\frac{1}{\sqrt{24}}\)
If y = ax² + b, then \(\frac{dy}{dx}\) at x = 2 is equal to ax
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4a
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3a
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2a
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None of these
Explanation
4a
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