Continuity and Differentiability Online Objective Test | Continuity and Differentiability Online Objective Test

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q1.

\underset{x \to 0}{L t} \frac{1 - \cos x}{x}

is equal to

A. 1

B. none of these

C. 0

D. 1/2

Answer : Option C
Explaination / Solution:

lim_{x \to 0} \frac{1 - \cos x}{x}

= lim_{x \to 0} \frac{(1 - \cos x) (1 + \cos x)}{x (1 + \cos x)}

= lim_{x \to 0} \frac{s i n x}{x} . \frac{\sin x}{(1 + \cos x)} = 0

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q2.

lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}

is equal to

A. none of these.

B. 1

C. 1.1/2

D. 0

Answer : Option B
Explaination / Solution:

lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}

= lim_{x \to 0} \frac{2 \sin x (1 - \cos x)}{x^{3}}

= 2 lim_{x \to 0} \frac{\sin x}{x} . \frac{1 - cosx}{x^{2}}

= 2.1. \frac{1}{2} = 1

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q3.

\underset{x \to π / 4}{L t} \frac{\cos x - \sin x}{x - \frac{π}{4}}

is equal to

A. −1/√2

B. 2/√2

C. -1

D. −2/√2

Answer : Option D
Explaination / Solution:

lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{x - \frac{π}{4}}

= lim_{x \to \frac{π}{4}} \frac{- \sin x - \cos x}{1}

= - \sin \frac{π}{4} - \cos \frac{π}{4} = - \frac{2}{\sqrt{2}} = - \sqrt{2}

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q4. If [x] stands for the integral part of x, then

A. Ltx→1[x]=1

B. Ltx→1−[x]=1

C. None of these

D. Ltx→1+[x]=1

Answer : Option D
Explaination / Solution:

The

[x]

is defined as the value which takes the nearest small integer in every integer interval . For example in the interval

0 \leq x < 1

the value of the function is 0 since it is the lowest integer in that interval . Hence

lim_{x \to 1^{+}} [x] = 1 [\sin c e 1 \leq x < 2]

which gives the value as 1,which is the lowest integer of that interval.

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q5.

\underset{x \to \frac{π}{2}}{L t} [\sin x]

is equal to

A. none of these.

B. 0

C. -1

D. 1

Answer : Option B
Explaination / Solution:

limx→π2[sinx]=0[sinceπ2−h≤x≤π2+h]

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q6. limx→01−cosxx2 is equal to

A. 1/2

B. -1

C. 0

D. 1

Answer : Option A
Explaination / Solution:

lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = lim_{x \to 0} \frac{\sin x}{2 x} = \frac{1}{2}

(UsingL’hospital Rule ).

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q7.

\underset{x \to - 2}{L t} \frac{\sqrt{x^{2} + 5} - 3}{x + 2}

is equal to

A. 0

B. none of these

C. 2/3

D. -2/3

Answer : Option D
Explaination / Solution:

lim_{x \to - 2} \frac{\sqrt{x^{2} + 5} - 3}{x + 2}

= lim_{x \to - 2} \frac{x^{2} + 5 - 9}{(x + 2) (\sqrt{x^{2} + 5} + 3)} = lim_{x \to - 2} \frac{x - 2}{(\sqrt{x^{2} + 5} + 3)} = \frac{- 4}{3 + 3} = - \frac{2}{3}

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q8.

\underset{h \to 0}{L t} (\frac{1}{h \sqrt[3]{8 + h}} - \frac{1}{2 h})

is equal to

A. -1/48

B. none of these

C. 1/24

D. 1/48

Answer : Option A
Explaination / Solution:

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q9. If k be an integer, then

\underset{x \to k}{L t}

(x –[x])

A. is equal to – 1

B. does not exists.

C. 1

D. is equal to 0

Answer : Option C
Explaination / Solution:

\begin{matrix} lim_{x \to k^{-}} (x - [x]) = lim_{x \to k^{-}} x^{-} - lim_{x \to k^{-}} [x] \end{matrix}

= k - (k - 1) = 1 for all

x \in (k - 1, k), [x] = (k - 1)

# Continuity and Differentiability # CBSE 12th Mathematics Prepare / Learn

Q10.

$\underset{x \to 0}{L t} \frac{x \cos x - \log (1 + x)}{x^{2}}$ is equal to

A. none of these.

B. 0

C. 1

D. 1/2

Answer : Option D
Explaination / Solution:

\begin{matrix} lim_{x \to 0} \frac{x \cos x - \log (1 + x)}{x^{2}} \end{matrix}

= lim_{x \to 0} \frac{\cos x - x \sin x - \frac{1}{1 + x}}{2 x} = lim_{x \to 0} \frac{- s i n x - (x \cos x + \sin x) + \frac{1}{{(1 + x)}^{2}}}{2} = \frac{1}{2}

( using L’Hospital Rule)

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