Free Objective Test 03 Practice Test - 11th and 12th

Question 1

The value of $lim x \to 0 {(\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x}}{n})}^{a / x}$ is

(2 n!)^{\frac{a}{n}}

(n!)^{\frac{2 a}{n}}

(n!)^{\frac{a}{n}}

0

SOLUTION

Solution : C

$lim x \to 0 {(\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x}}{n})}^{a / x} (1^{\infty} f o r m)$
$= e^{lim x \to 0} (\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x}}{n} - 1) . \frac{a}{n}$
$= e^{lim x \to 0} (\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x} - n}{n}) . \frac{a}{n}$
$= e^{lim x \to 0} (\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x} - n}{x}) . \frac{a}{n}$
$= e^{lim x \to 0} {\frac{1^{x} - 1}{x} + \frac{2^{x} - 1}{x} + . . . + \frac{n^{x} - 1}{x}} \frac{a}{n}$
We know $lim x \to 0 {\frac{a^{x} - 1}{x}} = a$
So, $= e^{(log 1 + log 2 + log 3 + . . . log n) \frac{a}{n}}$
$e^{a / n} (log (1.2.3... n))$ = $e^{({log}_{e} n!)^{a / n}} = (n!)^{a / n}$

Question 2

If $α$ is a repeated root of $a x^{2} + b x + c = 0$ then $lim x \to α \frac{sin (a x^{2} + b x + c)}{(x - α)^{2}}$ is

0

a

b

c

SOLUTION

Solution : B

$lim x \to α \frac{sin (a x^{2} + b x + c)}{(x - α)^{2}} = lim x \to α \frac{sin a (x - α) (x - α)}{(x - α)^{2}}$
$= lim x \to α \frac{sin a (x - α)^{2}}{a (x - α)^{2}} \times a = a$

Question 3

$lim x \to 0 (cos x)^{1 / x^{4}}$

0

1

e

D. limit does not exist

SOLUTION

Solution : A

$lim x \to 0 (cos x)^{1 / x^{4}} = e^{lim x \to 0} \frac{cos x - 1}{x^{4}}$
$e^{lim x \to 0} \frac{- 2 sin \frac{x}{2}}{\frac{x}{2}} . \frac{sin \frac{x}{2}}{\frac{x}{2}} \times \frac{1}{4 x^{2}} = e^{- \infty} = 0$

Question 4

Let f be a function such that f(x+y)=f(x)+f(y) for all x and y and $f (x) = (2 x^{2} + 3 x) g (x)$ for all x where g(x) is continuous and g(0)=9 then f'(0) is equals to

A. 9

B. 3

C. 27

D. 6

SOLUTION

Solution : C

$f^{'} (x) = l i m h \to 0 \frac{f (x + h) - f (x)}{h}$
$= l i m h \to 0 \frac{f (x) + f (h) - f (x)}{h}$
$= l i m h \to 0 \frac{(2 h^{2} + 3 h) g (h)}{h}$
$= l i m h \to 0 (2 h + 3) g (h)$
$= 3 g (0)$
$= 27$

Question 5

ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC.The triangle ABC has perimeter $P = 2 [\sqrt{(2 h r - h^{2})} + \sqrt{2 h r}]$ and A be the area of the triangle .Find $lim h \to 0 \frac{A}{P^{3}}$

\frac{1}{r}

\frac{1}{64 r}

\frac{1}{128} r

\frac{1}{2 r}

SOLUTION

Solution : C

In $△ A B C, A B = A C$
$A D ⊥ B C (D i s m i d p o i n t o f B C)$
Let r = radius of circumcircle
$\Rightarrow$ OA = OB = OC = r
Now BD = $\sqrt{B O^{2} - O D^{2}}$
$= \sqrt{r^{2} - (h - r)^{2}} = \sqrt{2 r h - h^{2}}$
$\Rightarrow B C = 2 \sqrt{2 r h - h^{2}}$
$∴$ Area of $△ A B C =$ $\frac{1}{2} \times B C \times A D$
$= h \sqrt{2 r h - h^{2}}$
Also $lim h \to 0 \frac{A}{p^{3}} = \frac{h \sqrt{2 r h - h^{2}}}{8 (\sqrt{2 r h - h^{2}} + \sqrt{2 h r})^{3}}$
$= lim h \to 0 \frac{h^{3 / 2} \sqrt{2 r - h}}{8 h^{3 / 2} (\sqrt{2 r - h} + \sqrt{2 r})^{3}}$
$= lim h \to 0 \frac{\sqrt{2 r - h}}{8 [\sqrt{2 r - h} + \sqrt{2 r}]^{3}}$
$= \frac{\sqrt{2 r}}{8 (\sqrt{2 r} + \sqrt{2 r})^{3}} = \frac{\sqrt{2 r}}{8.8.2 r \sqrt{2 r}} = \frac{1}{128 r}$

Question 6

$lim n \to \infty$ $20 \sum x = 1$ $c o s^{2 n} (x - 10)$ is equal to

SOLUTION

Solution : B

$lim n \to \infty$ $c o s^{2 n}$ $x$ = ${\begin{matrix} 1, x = r π, r \in I 0, x \neq r π, r \in I \end{matrix}$

Here, for $x$ = 10, $lim n \to \infty c o s^{2 n} (x - 10)$ = 1

and in all other cases it is zero

$∴ lim n \to \infty 20 \sum x = 1 c o s^{2 n} (x - 10)$ = 1

Question 7

$The value of l i m x \to \infty \frac{x + c o s x}{x + s i n x} i s$

-1

SOLUTION

Solution : C

$l i m x \to \infty \frac{x + c o s x}{x + s i n x} [Putting x = \frac{1}{h}; a s x \to \infty, h \to 0] = l i m h \to 0 \frac{\frac{1}{h} + c o s (\frac{1}{h})}{\frac{1}{h} + s i n (\frac{1}{h})} = l i m h \to 0 \frac{1 + h c o s \frac{1}{h}}{1 + h c o s \frac{1}{h}} = \frac{1 + 0}{1 + 0} ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} ∵ - 1 \leq s i n \frac{1}{h} \leq 1 a n d - 1 \leq c o s \frac{1}{h} \leq 1, ∴ w h e r e h \to 0, h c o s \frac{1}{h} \to 0 a n d h s i n \frac{1}{h} \to 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ = 1$

Question 8

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

2/3

1/3

SOLUTION

Solution : A

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

Question 9

If $l i m_{x \to 0}$ kx cosec x = $l i m_{x \to 0}$ x cosec kx , then k =

-1

SOLUTION

Solution : C

$l i m_{x \to 0}$ kx cosec x = $l i m_{x \to 0}$ x cosec kx

= k $l i m_{x \to 0}$ $\frac{x}{s i n x}$ = $\frac{1}{k}$ $l i m_{x \to 0}$ $\frac{k x}{s i n k x}$ =k= $\frac{1}{k}$ =k=± 1

Question 10

${lim}_{x \to 0} \frac{t a n^{- 1} x - s i n^{- 1} x}{x^{3}} is equal to$

1/2

-1/2

-1

SOLUTION

Solution : B

$l i m x \to 0 \frac{t a n^{- 1} x - s i n^{- 1} x}{x^{3}} (\frac{0}{0} f o r m) l i m x \to 0 \frac{\frac{1}{1 + x^{2}} - \frac{1}{\sqrt{1 - x^{2}}}}{3 x^{2}} l i m x \to 0 \frac{\sqrt{1 - x^{2}} - (1 + x^{2})}{3 x^{2} (1 + x^{2}) \sqrt{1 - x^{2}}} l i m x \to 0 \frac{(1 - x^{2}) - (1 + x^{2})^{2}}{3 x^{2} (1 + x^{2}) \sqrt{1 - x^{2}} [\sqrt{1 - x^{2}} + (1 + x^{2})]} = - \frac{3}{6} = - \frac{1}{2}$

Question 11

$The value of {lim}_{x \to 0} \frac{\sqrt[3]{1 + s i n x} - \sqrt[3]{1 - s i n x}}{x} i s$

2/3

-2/3

3/2

-3/2

SOLUTION

Solution : A

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

Question 12

$If l i m x \to - \infty (\sqrt{x^{2} - x + 1} - a x - b) = 0$ then the values of a and b are given by -

$a = - 1, b = \frac{1}{2}$

$a = 1, b = \frac{1}{2}$

$a = 1, b = - \frac{1}{2}$

a=-1, b=-1/2

SOLUTION

Solution : A

$We have, l i m x \to - \infty (\sqrt{x^{2} - x + 1} - a x - b) = 0 [putting x=-y; x \to - \infty, y \to \infty] \Rightarrow l i m y \to \infty (\sqrt{y^{2} + y + 1} + a y - b) = 0 \Rightarrow l i m y \to \infty [y {(1 + \frac{1}{y} + \frac{1}{y^{2}})}^{1 / 2} + a y - b] = 0 \Rightarrow l i m y \to \infty [y {1 + \frac{1}{2} (\frac{1}{y} + \frac{1}{y^{2}} + . . . . . . . . .)} + a y - b] = 0 \Rightarrow l i m y \to \infty [y (1 + a) + (\frac{1}{2} - b) + \frac{1}{2 y} + . . . . . . . .] = 0 \Rightarrow 1 + a = 0 a n d (1 / 2) - b = 0 \Rightarrow a = - 1 a n d b = 1 / 2$

Question 13

$The values of constants a and b so that l i m x \to \infty (\frac{x^{2} + 1}{x + 1} - a x - b) = \frac{1}{2}, a r e$

$a = 1, b = - \frac{3}{2}$

$a = - 1, b = \frac{3}{2}$

a=0, b=0

a =2, b= -1

SOLUTION

Solution : A

$l i m x \to \infty (\frac{x^{2} + 1}{x + 1} - a x - b) = \frac{1}{2} \Rightarrow l i m x \to \infty \frac{(x^{2} + 1) - (a x + b) (x + 1)}{x + 1} = \frac{1}{2} \Rightarrow l i m x \to \infty \frac{x^{2} (1 - a) - (a + b) x - b + 1}{x + 1} = \frac{1}{2} \Rightarrow 1 - a = 0 a n d a + b = - \frac{1}{2} ∴ a = 1 b = - \frac{3}{2}$

Question 14

The function $f (x) = \frac{l o g (1 + a x) - l o g (1 - b x)}{x}$ is not defined at x-0 . The value which should be assigned to f at x = 0 so that it is continuos at x=0 , is

A. a-b

B. a+b

C. log a+log b

D. log a-log b

SOLUTION

Solution : B

Since limit of a function is a + b as x $\to$ 0, therefore to be continuous at a function, its value must be
a + b at x = 0 $\Rightarrow$ f(0) = a + b.

Question 15

If the derivative of the function $f (x) = b x^{2} + a x + 4; x \geq - 1 a x^{2} + b; x < - 1^{^{'}}$ , is everywhere continuous, then

A. a = 2, b = 3

B. a = 3, b = 2

C. a = - 2, b = - 3

D. a = - 3, b = - 2

SOLUTION

Solution : A

$W e h a v e, f (x) = {\begin{matrix} a x^{2} + b, & x < - 1 b x^{2} + a x + 4, & x \geq - 1 \end{matrix} ∴ f^{^{'}} (x) = {\begin{matrix} 2 a x, & < - 1 2 b x + a, & x \geq - 1 \end{matrix} S i n c e, f (x) i s d i f f e r e n t i a b l e a t x = - 1, t h e r e f o r e i t i s c o n t i n u o u s a t x = - 1 a n d h e n c e, l i m x \to - 1 - f (x) = l i m x \to - 1 + f (x) \Rightarrow a + b = b - a + 4 \Rightarrow a = 2 a n d a l s o, l i m x \to - 1 - f (x) = l i m x \to - 1 + f (x) \Rightarrow - 2 a = - 2 b + a \Rightarrow 3 a = 2 b \Rightarrow b = 3 (∵ a = 2) H e n c e, a = 2, b = 3$

Free Objective Test 03 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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