Limits Continuity And Differentiability(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

LIMITS CONTINUITY AND DIFFERENTIABILITY MCQs

Total Questions : 45 | Page 1 of 5 pages

Question 1.

$f (x) = {\begin{matrix} 4 x - 3, & x < 1 x^{2} & x \geq 1 \end{matrix},$ then
$lim x \to 1 f (x) =$

$1$
$- 1$
$0$
$0$

Discuss Question

Answer: Option A. ->

1

:
A

lim x \to 1^{-} f (x) = lim x \to 1^{-} (4 x - 3) = 1

lim x \to 1^{+} f (x) = lim x \to 1^{+} x^{2} = 1

Since LHL = RHL

\Rightarrow lim x \to 1 f (x) = 1

Question 2.

$l i m n \to \infty \frac{{2.3}^{n} - {3.5}^{n}}{{3.3}^{n} + {4.5}^{n}} =$

$\frac{2}{3}$
$- \frac{3}{4}$
$1$
$0$

Discuss Question

Answer: Option B. ->

- \frac{3}{4}

:
B

l i m n \to \infty \frac{{2.3}^{n} - {3.5}^{n}}{{3.3}^{n} + {4.5}^{n}}

= l i m n \to \infty \frac{5^{n} (2 {(\frac{3}{5})}^{n} - 3)}{5^{n} (3 {(\frac{3}{5})}^{n} + 4)}

A s n \to \infty, {(\frac{3}{5})}^{n} \to 0

= - \frac{3}{4}

Question 3.

$lim α \to β \frac{{sin}^{2} α - {sin}^{2} β}{α^{2} - β^{2}}$ is equal to

$\frac{sin 3 β}{(4 β)}$
$\frac{sin 2 β}{(2 β)}$
$\frac{sin 8 β}{(7 β)}$
$\frac{sin 6 β}{(4 β)}$

Discuss Question

Answer: Option B. ->

\frac{sin 2 β}{(2 β)}

:
B

lim α \to β \frac{{sin}^{2} α - {sin}^{2} β}{α^{2} - β^{2}} = lim α \to β \frac{sin (α - β) sin (α + β)}{(α - β) (α + β)} = \frac{sin 2 β}{(2 β)}

Question 4.

$lim x \to \infty \frac{(2 + x)^{40} (4 + x)^{5}}{(2 - x)^{45}}$

$- 1$
$1$
$16$
$32$

Discuss Question

Answer: Option A. ->

- 1

:
A

lim x \to \infty \frac{(2 + x)^{40} (4 + x)^{5}}{(2 - x)^{45}}

= lim x \to \infty \frac{x^{45} {(\frac{2}{x} + 1)}^{40} {(\frac{4}{x} + 1)}^{5}}{x^{45} {(\frac{2}{x} - 1)}^{45}}

= \frac{(1)^{40} (1)^{5}}{(- 1)^{45}}

= \frac{(1)^{45}}{(- 1)^{45}}

= - 1

Question 5.

$lim x \to \infty \sqrt{\frac{x + sin x}{x - cos x}} =$

$0$
$1$
$- 1$
does not exist

Discuss Question

Answer: Option B. ->

1

:
B

lim x \to \infty \sqrt{\frac{x + s i n x}{x - c o s x}}

= lim x \to \infty \frac{x^{\frac{1}{2}} \sqrt{1 + \frac{s i n x}{x}}}{x^{\frac{1}{2}} \sqrt{1 - \frac{c o s x}{x}}}

We know, that for any value of x, sinx and cosx will be [-1,1]
So,

= lim x \to \infty \frac{S i n x}{x} = 0

And

= lim x \to \infty \frac{C o s x}{x} = 0

= x^{\frac{1}{2}} \sqrt{1 + \frac{s i n x}{x}} \frac{\sqrt{1 + \frac{s i n x}{x}}}{\sqrt{1 + \frac{c o s x}{x}}}

= lim x \to \infty \frac{1}{1}

= 1

Question 6.

$lim x \to 2 \frac{\sqrt{x - 2} + \sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}}$ is equal to

$\frac{1}{2}$
$1$
$2$
$0$

Discuss Question

Answer: Option A. ->

\frac{1}{2}

:
A

lim x \to 2 \frac{\sqrt{x - 2} + \sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}}

= lim x \to 2 (\frac{\sqrt{x - 2}}{\sqrt{x + 2} \sqrt{x - 2}} + \frac{\sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}})

On rationalisation -

= lim x \to 2 (\frac{1}{\sqrt{x + 2}} + \frac{x - 2}{\sqrt{x^{2} - 4} (\sqrt{x} + \sqrt{2})})

= lim x \to 2 \frac{1}{\sqrt{x + 2}} + lim x \to 2 \sqrt{\frac{x - 2}{x + 2}} \times \frac{1}{\sqrt{x} + \sqrt{2}}

= \frac{1}{2}

Question 7.

The value of $lim x \to 0 \frac{x}{a} [\frac{b}{x}]$ is, $([.] \to G . I . F)$

$0$
$\infty$
$\frac{b}{a}$
does not exist

Discuss Question

Answer: Option C. ->

\frac{b}{a}

:
C

lim x \to 0 \frac{x}{a} [\frac{b}{x}]

= lim x \to 0 \frac{x}{a} (\frac{b}{x} - {\frac{b}{x}})

Since

{\frac{b}{x}} ϵ [0, 1)

= lim x \to 0 \frac{x}{a} . {\frac{b}{x}} = 0

lim x \to 0 \frac{x}{a} [\frac{b}{x}] = lim x \to 0 (\frac{x}{a}) (\frac{b}{x})

= lim x \to 0 . \frac{b}{a}

= \frac{b}{a}

Question 8.

$lim x \to 0 \frac{2 sin x - sin 2 x}{x^{3}}$ is equal to

$1$
$- 1$
$0$
does not exist

Discuss Question

Answer: Option A. ->

1

:
A

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

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

= lim x \to 0 2 \frac{{sin}^{3} x}{x^{3}} \times \frac{1}{1 + cos x}

= 2 \times (1)^{3} \times \frac{1}{1 + 1} = 1

Question 9.

If G(x) = - $\sqrt{25 - x^{2}}$ then $lim x \to 1 \frac{G (x) - G (1)}{x - 1}$ has the value

$\frac{1}{24}$
$\frac{1}{5}$
$- \sqrt{24}$
$\frac{1}{2 \sqrt{6}}$

Discuss Question

Answer: Option D. ->

\frac{1}{2 \sqrt{6}}

:
D

lim x \to 1 \frac{- \sqrt{25 - x^{2}} - (- \sqrt{24})}{x - 1}

= lim x \to 1 \frac{\sqrt{24} - \sqrt{25 - x^{2}}}{x - 1} \times \frac{\sqrt{24} + \sqrt{25 - x^{2}}}{\sqrt{24} + \sqrt{25 - x^{2}}}

lim x \to 1 \frac{x^{2} - 1}{(x - 1) [\sqrt{24} + \sqrt{25 - x^{2}}]} = \frac{2}{2 \sqrt{24}} = \frac{1}{2 \sqrt{6}}

Question 10.

If $lim x \to 0 \frac{((a - n) n x - tan x) sin n x}{x^{2}} = 0$ where n is nonzero real number, then a is equal to

$0$
$\frac{n + 1}{n}$
$n$
$n + \frac{1}{n}$

Discuss Question

Answer: Option D. ->

n + \frac{1}{n}

:
D

lim x \to 0 \frac{((a - n) n x - tan x) sin n x}{x^{2}} = 0

lim x \to 0 ((a - n) n - (\frac{tan x}{x})) . \frac{sin n x}{n x} . n = 0

\Rightarrow ((a - n) n - 1) .1 . n = 0

\Rightarrow a = \frac{1}{n} + n

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