Free Objective Test 02 Practice Test - 11th and 12th

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

SOLUTION

Solution : 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

SOLUTION

Solution : 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)}

SOLUTION

Solution : 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

SOLUTION

Solution : 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

D. does not exist

SOLUTION

Solution : 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

SOLUTION

Solution : 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}

D. does not exist

SOLUTION

Solution : 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

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}}

SOLUTION

Solution : 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 9

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

1

- 1

0

D. does not exist

SOLUTION

Solution : 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 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}

SOLUTION

Solution : D

$lim x \to 0 \frac{((a - n) n x - tan x) sin n x}{x^{2}} = 0$
$lim x \to 0 ((a - n) n - (\frac{tan x}{x})) . \frac{sin n x}{n x} . n = 0$
$\Rightarrow ((a - n) n - 1) .1 . n = 0$
$\Rightarrow a = \frac{1}{n} + n$

Question 11

The value of $lim x \to 1 {(tan \frac{x π}{4})}^{tan \frac{π x}{2}}$ is

e^{- 2}

e^{- 1}

e

1

SOLUTION

Solution : B

$lim x \to 1 {(tan \frac{x π}{4})}^{tan \frac{π x}{2}} (1^{\infty} f r o m)$
$e^{lim x \to 1} (tan \frac{x π}{4} - 1) tan \frac{π x}{2}$
$= e^{lim x \to 1} (\frac{sin \frac{π x}{4} - cos \frac{π x}{4}}{cos \frac{π x}{4}}) \frac{2 sin \frac{π x}{4} cos \frac{π x}{4}}{cos \frac{π x}{2}}$
$= e^{lim x \to 1} \frac{2 {sin}^{2} \frac{π x}{4} - 2 sin \frac{π x}{4} cos \frac{π x}{4}}{cos \frac{π x}{2}}$
$= e^{lim x \to 1} (\frac{1 - cos \frac{π x}{2} - sin \frac{π x}{2}}{cos \frac{π x}{2}})$
$= e^{lim x \to 1} (\frac{1 - sin \frac{π x}{2}}{cos \frac{π x}{2}} - 1)$
$= e^{lim x \to 1} (\frac{cos \frac{π x}{2}}{1 + sin \frac{π x}{2}} - 1)$
= $e^{- 1}$

Question 12

$l i m x \to 0 {(\frac{1 + 2 x}{1 + 3 x})}^{\frac{1}{x^{2}}} . e^{\frac{1}{x}} =$

e^{\frac{5}{2}}

e^{2}

C. 2

D. 1

SOLUTION

Solution : A

$l i m x \to 0 e^{\frac{1}{x^{2}} (l o g (1 + 2 x) - l o g (1 + 3 x)) + \frac{1}{x}} e^{l i m x \to 0 \frac{(l o g (1 + 2 x) - l o g (1 + 3 x)) + x}{x^{2}}}$
Using the expansion we get -
$= e^{\frac{5}{2}}$

Question 13

$lim n \to \infty n \sum r = 1 c o t^{- 1} (r^{2} + \frac{3}{4})$ is

$t a n^{- 1} 1$

$t a n^{- 1} 2$

${tan}^{- 1} \frac{1}{2}$

SOLUTION

Solution : C

$lim n \to \infty n \sum r = 1 t a n^{- 1} (\frac{4}{4 r^{2} + 3})$

= $lim n \to \infty n \sum r = 1 t a n^{- 1} (\frac{1}{r^{2} - \frac{1}{4} + 1})$

= $lim n \to \infty n \sum r = 1 t a n^{- 1} (\frac{(r + \frac{1}{2}) - (r - \frac{1}{2})}{1 + (r + \frac{1}{2}) (r - \frac{1}{2})})$

= $lim n \to \infty n \sum r = 1 t a n^{- 1} {t a n^{- 1} (r + \frac{1}{2}) - t a n^{- 1} (r - \frac{1}{2})}$

= $lim n \to \infty {t a n^{- 1} (n + \frac{1}{2}) - t a n^{- 1} (\frac{1}{2})}$

= $\frac{π}{2} - t a n^{- 1} \frac{1}{2}$

= $t a n^{- 1} 2$

Question 14

The value of $lim x \to \frac{π}{2} [s i n^{- 1} s i n x]$ ,[x] is the greatest integer function of x, is

$\frac{π}{2}$

$\frac{1}{2}$

SOLUTION

Solution : A

$lim x \to \frac{π}{2} [s i n^{- 1} s i n x]$
= $lim x \to \frac{π}{2} [x]$
= 1

Question 15

$I f f (x) = {\begin{matrix} x^{2} - 3, & 2 < x < 3 2 x + 5, & 3 < x < 4 \end{matrix}$ , the equation whose roots are $lim x \to 3 - f (x)$ and $lim x \to 3 + f (x) i s$

x^{2} - 12 x + 36 = 0

x^{2} - 26 x + 66 = 0

x^{2} - 17 x + 66 = 0

x^{2} - 22 x + 121 = 0

SOLUTION

Solution : C

$f (x) = {\begin{matrix} x^{2} - 3, & 2 < x < 3 2 x + 5, & 3 < x < 4 \end{matrix}$
$∴ lim x \to 3 - f (x) = lim x \to 3 - (x^{2} - 3) = 6$
and $lim x \to 3 + f (x) = lim x \to 3 + (2 x + 5) = 11$
Hence, the required equation will be
$x^{2}$ - (sum of roots)x + (Products of roots) = 0

Free Objective Test 02 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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