Free Objective Test 01 Practice Test - 11th and 12th

Question 1

If $lim x \to 0 (1 + a sin x)^{c o s e c x} = 3,$ then a is

A. In 2

B. In 3

C. In 4

e^{3}

SOLUTION

Solution : B

$3 = lim x \to 0 (1 + a sin x)^{c o s e c x}$
$[1^{\infty} f o r m] \Rightarrow lim x \to 0 e^{c o s e c x . a sin x} = e^{a}$
$∴ e^{a} = 3 \Rightarrow a = {log}_{e} 3 = l n 3.$

Question 2

$lim x \to 0 \frac{sin x - x + \frac{x^{3}}{6}}{x^{5}}$ is equal to

\frac{1}{120}

\frac{1}{160}

\frac{1}{2}

0

SOLUTION

Solution : A

$lim x \to 0 \frac{sin x - x + \frac{x^{3}}{6}}{x^{5}}$ ....... (1)

We know expansion of Sinx.
$s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} +$ ........ (2)

Substituting (2) in (1) we get

$l i m x \to 0 \frac{x^{5}}{5! . x^{5}}$

$= \frac{1}{5!}$

$= \frac{1}{120}$

Question 3

$lim n \to \infty \frac{a^{n} + b^{n}}{a^{n} - b^{n}}$ , where $a > b > 1$ , is equal to

- 1

1

0

\frac{a}{b}

SOLUTION

Solution : B

$lim n \to \infty \frac{a^{n} + b^{n}}{a^{n} - b^{n}}$

$a > b > 1$

$= n n \to \infty \frac{1 + {(\frac{b}{a})}^{n}}{1 - {(\frac{b}{a})}^{n}}$

$= 1$

Question 4

$lim x \to \frac{π}{2} \frac{cot x - cos x}{(π - 2 x)^{3}}$ is equal to

1

\frac{π}{2}

\frac{1}{16}

0

SOLUTION

Solution : C

$lim x \to \frac{π}{2} \frac{cot x - cos x}{(π - 2 x)^{3}}$

Let $x = \frac{π}{2} + t$

If $x \to \frac{π}{2}, t \to 0$

$lim t \to 0 \frac{sin t - tan t}{- 8 t^{3}}$

$= lim t \to 0 \frac{(t - \frac{t^{3}}{3!} + \frac{t^{5}}{5!} + \dots) - (t + \frac{t^{3}}{3} + \frac{2 t^{5}}{15} + \dots)}{- 8 t^{3}}$

$= \frac{1}{16}$

We can put $x = \frac{π}{2} - t$ and we'll get L.H.L also same.
Since L.H.L = R.H.L the limit exists and is $= \frac{1}{16}$

Question 5

$lim x \to - \infty {\frac{x^{4} sin (\frac{1}{x}) + x^{2}}{1 + | x |^{3}}}$ is equal to

2

1

- 1

D. does not exist

SOLUTION

Solution : C

$lim x \to - \infty [\frac{x^{4} sin (\frac{1}{x}) + x^{2}}{1 + | x |^{3}}]$

$lim x \to - \infty \frac{x^{3}}{1 + | x |^{3}} [x sin (\frac{1}{x}) + \frac{1}{x}]$

As $x \to - \infty$

$\frac{x^{3}}{1 + | x |^{3}} = - 1$

$x s i n (\frac{1}{x}) + \frac{1}{x} = 1$

$\Rightarrow l i m x \to - \infty \frac{x^{3}}{1 + | x |^{3}} (x s i n \frac{1}{x} + \frac{1}{x}) = - 1$

Question 6

If f(9) = 9, f'(9) = 4, then $lim x \to 9 \frac{\sqrt{f (x)} - 3}{\sqrt{x} - 3}$ equals

2

4

6

0

SOLUTION

Solution : B

Given that f(9) = 9, f’(9) = 4

$l i m x \to 9 \frac{\sqrt{f (x)} - 3}{\sqrt{x} - 3}$

On rationalisation we get -

$= l i m x \to 9 (\frac{\sqrt{f (x)} - 3}{\sqrt{x} - 3}) (\frac{\sqrt{x} + 3}{\sqrt{x} + 3}) (\frac{\sqrt{f (x)} + 3}{\sqrt{f (x)} + 3})$

$= l i m x \to 9 \frac{f (x) - f (9)}{x - 9}$ . $\frac{6}{6}$

$= l i m x \to 9 \frac{f (x) - f (9)}{x - 9}$

$= f^{^{'}} (9)$

$= 4$

Question 7

If f(x) = ${\begin{matrix} sin x & x \neq n π, n = 0, \pm 1, \pm 2... 2, & o t h e r w i s e \end{matrix}$ and g(x) = $⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{2} + 1, & x \neq 0, 2 4, & x = 0 5, & x = 2 \end{matrix}$ , then $lim x \to 0 g {f (x)}$ is

2

0

3

1

SOLUTION

Solution : D

Given that,
f(x) = ${\begin{matrix} sin x & x \neq n π, n = 0, \pm 1, \pm 2... 2, & o t h e r w i s e \end{matrix}$
and
g(x) = $⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{2} + 1, & x \neq 0, 2 4, & x = 0 5, & x = 2 \end{matrix}$
Then $lim x \to 0 g [f (x)] = lim x \to 9 g (sin x)$
$= lim x \to 0 ({sin}^{2} x + 1) = 1$

Question 8

$lim n \to \infty n . cos (\frac{π}{4 n}) . sin (\frac{π}{4 n})$ is equal to

\frac{π}{4}

\frac{π}{6}

\frac{π}{9}

\frac{π}{3}

SOLUTION

Solution : A

$lim x \to \infty n . cos (\frac{π}{4 n}) sin (\frac{π}{4 n})$

$= lim n \to \infty \frac{n}{2} sin \frac{π}{2 n}$

$= \frac{π}{4}$

Question 9

$l i m x \to 0 \frac{1 - c o s 3 x}{x (3^{x} - 1)} =$

\frac{9}{2}

\frac{9}{(2 log 3)}

\frac{9 log 3}{2}

1

SOLUTION

Solution : B

$lim x \to 0 \frac{1 - cos 3 x}{x (3^{x} - 1)}$

$c o s 3 x = 1 - \frac{9 x^{2}}{2!} + \dots$

$1 - c o s 3 x = \frac{9 x^{2}}{2!} + \dots$ ......... (1)

$l i m x \to 0 \frac{3^{x} - 1}{x} = l o g 3$ .......... (2)

$l i m x \to 0 \frac{1 - c o s 3 x}{x^{2} (\frac{3^{x} - 1}{x})}$

From (1) and (2) we get

$= \frac{9}{2 l o g 3}$

Question 10

Which of the following limit is not in the indeterminant form ?

$lim x \to a \frac{x^{3} - a^{3}}{x - a}$

$lim x \to a s i n x c o s e c x$

$lim x \to 0 x^{x}$

$lim x \to 0 \frac{0}{x}$

SOLUTION

Solution : D

$∵ lim x \to a \frac{x^{3} - a^{3}}{x - a} = (\frac{0}{0} inderminant form)$

$lim x \to a s i n x c o s e c x = (0 \times \infty inderminant form)$

$lim x \to a x^{x} = (0^{0} inderminant form)$

$lim x \to a \frac{0}{x} = 0 (not inderminant form)$

Question 11

$L e t f (x) = {\begin{matrix} x^{2} & _{\frac{k (x^{2} - 4)}{2 - x}} \end{matrix} when x is an int eger otherwise then {lim}_{x \to 2} f (x)$

exists only when k=1

exists for every real k

exists for every real k = 1

does not exist

SOLUTION

Solution : B

${lim}_{x \to 2^{+}} f (x) = {lim}_{x \to 2^{+}} \frac{k (x^{2} - 4)}{2 - x} = {lim}_{x \to 2^{+}} k (- 2 - x) = - 4 k {lim}_{x \to 2^{+}} f (x) = {lim}_{x \to 2^{+}} \frac{k (x^{2} - 4)}{2 - x} = {lim}_{x \to 2^{-}} k (- 2 - x) = - 4 k$

Hence limits exists for every real value of k.

Question 12

$l i m x \to 2 \frac{\sqrt{x - 2} + \sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}} is equal to$

1/2

-1

SOLUTION

Solution : A

$= l i m x \to 2 {\frac{1}{\sqrt{x + 2}} + \frac{\sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}}} Limit = \frac{1}{2} + l i m x \to 2 \frac{x - 2}{\sqrt{x} + \sqrt{2}} \frac{1}{\sqrt{(x + 2) (x - 2)}} = \frac{1}{2} + l i m x \to 2 \frac{1}{\sqrt{x} + \sqrt{2}} \sqrt{\frac{x - 2}{x + 2}} = \frac{1}{2}$

Question 13

The value of $lim x \to 1 [s i n s i n^{- 1} x]$ is

does not exist

$\frac{π}{2}$

SOLUTION

Solution : A

Since x→1 means x→1^-, ∵ sin^-1x is not defined for x>1
$lim x \to 1 [s i n s i n^{- 1} x] = lim x \to 1^{-} [s i n s i n^{- 1} x] = 1$

Question 14

$The value of {lim}_{x \to 0} \frac{1 - c o s^{3} x}{x s i n x c o s x}$

2/5

3/5

3/2

3/4

SOLUTION

Solution : C

${lim}_{x \to 0} \frac{1 - c o s^{3} x}{x s i n x c o s x} = {lim}_{x \to 0} \frac{(1 - c o s x) (1 + c o s x + c o s^{2} x)}{x s i n x c o s x} = {lim}_{x \to 0} \frac{2 s i n^{2} (\frac{x}{2})}{2 s i n (\frac{x}{2}) c o s (\frac{x}{2}) . x} \times \frac{(1 + c o s x + c o s^{2} x)}{c o s x} = {lim}_{x \to 0} \frac{s i n (\frac{x}{2})}{2 (\frac{x}{2})} \times \frac{1 + c o s x + c o s^{2} x}{c o s (\frac{x}{2}) c o s x} = \frac{1}{2} \times 3 = \frac{3}{2}$

Question 15

If $f (x) = {\begin{matrix} s i n x, & x \neq n π, & n ϵ I 2, & o t h e r w i s e \end{matrix}$ and $g (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{2} + 1, & x \neq 0, & 2 4, & x = 0 5, & x = 2 \end{matrix}$ , then $l i m x \to 0 g {f (x)}$

A. 5

B. 6

C. 7

D. 1

SOLUTION

Solution : D

$l i m x \to 0 g {f (x)} = l i m x \to 0 ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} (f (x))^{2} + 1, & f (x) \neq 0, 2 4, & f (x) = 0 5, & f (x) = 2 \end{matrix} = l i m x \to 0 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} s i n^{2} x + 1, & s i n x \neq 0, 2 & a n d x \neq n π 4, & s i n x = 0 & a n d x \neq n π 5, & s i n x = 2 & a n d x \neq n π 5, & 2 \neq 0, 2 & a n d x \neq n π 4, & 2 = 0 & a n d x = n π 5, & 2 = 2 & a n d x = n π \end{matrix} = l i m x \to 0 {\begin{matrix} 1 + s i n^{2} x, & x \neq n π 5, & x = n π \end{matrix} = 1 + s i n^{2} 0 = 1 + 0 = 1$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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