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

Question 1

If $f (x) = l n (\frac{x^{2} + e}{x^{2} + 1}),$ then range of f(x) is

(0,1)

(0,1]

[0,1]

{0,1}

SOLUTION

Solution : B

$f (x) = l n (\frac{x^{2} + e}{x^{2} + 1}) = l n (\frac{x^{2} + 1 - 1 + e}{x^{2} + 1}) = l n (1 + \frac{e - 1}{x^{2} + 1})$
$0 < \frac{e - 1}{x^{2} + 1} \leq (e - 1) \Rightarrow 1 < (1 + \frac{e - 1}{x^{2} + 1}) \leq e \Rightarrow 0 < l n (\frac{x^{2} + e}{x^{2} + 1}) \leq 1$
Hence range is $(0, 1]$
Hence (B) is correct answer.

Question 2

$Range of the function$ $f (x) = x^{2} + \frac{1}{x^{2} + 1}, i s$

$[1, \infty)$

$[2, \infty)$

$[\frac{3}{2}, \infty)$

$(- \infty, \infty)$

SOLUTION

Solution : A

$f (x) = x^{2} + 1 + \frac{1}{x^{2} + 1} - 1 x^{2} + 1 + \frac{1}{x^{2} + 1} \geq 2 [∵ A M \geq G M] x^{2} + \frac{1}{x^{2} + 1} \geq 1 ∴ f (x) ϵ [1, \infty)$

Question 3

Range of $f (x) = \frac{t a n (π [x^{2} - x])}{1 + s i n (c o s x)}$ is where $[x]$ denotes the greatest integer function

$(- \infty, \infty) \sim [0, t a n 1]$

$(- \infty, \infty) \sim [t a n 2, 0)$

$[t a n 2, t a n 1]$

${0}$

SOLUTION

Solution : D

$f (x) = \frac{t a n (π [x^{2} - x])}{1 + s i n (c o s x)} = {0} because of [x^{2} - x] i s integer .$

Question 4

$Let f (x) = [x] c o s (\frac{π}{[x + 2]}) where denotes the greatest integer function. Then, the domain of f is$

$x ϵ R, x n o t a n i n t e g e r$

$(- \infty, - 2) \cup [- 1, \infty)$

$x ϵ R, x \neq - 2$

$(- \infty, - 1]$

SOLUTION

Solution : B

$[x + 2] \neq 0 [x] + 2 \neq 0 [x] \neq - 2 x should not belong to [- 2, - 1) Domain of f is (\infty, - 2) \cup [- 1, \infty) .$

Question 5

$Let f (x) = \frac{x - [x]}{1 + x - [x]}, x ϵ R, [] dentoes the greatest integer function . Then, the range of f is$

(0,1)

$[0, \frac{1}{2})$

[0,1]

$[0, \frac{1}{2}]$

SOLUTION

Solution : B

The graph of y = x – [x] is as shown below

When x is an integer, x – [x] = 0

Hence, f(x) = 0 when x is an integer

$x \Rightarrow [x] as x tends to an integer. A s \to 1, \frac{x}{1 + x} \to \frac{1}{2} Hence , the range of f (x) i s [0, \frac{1}{2})$

Question 6

$The domain of the function f (x) = l o g_{e} (x^{2} + x + 1) + s i n \sqrt{x - 1} i s$

(-2,1)

$(- 2, \infty)$

$(1, \infty)$

None of these

SOLUTION

Solution : C

$We must have x - 1 \geq 0. Note that (x^{2} + x + 1) is always positive combining , the domain is [1, \infty) .$

Question 7

$The range of the function f (x) = c o s^{2} \frac{x}{4} + s i n \frac{x}{4}, x ϵ R i s$

$[0, \frac{5}{4}]$

$[1, \frac{5}{4}]$

$(- 1, \frac{5}{4})$

$[- 1, \frac{5}{4}]$

SOLUTION

Solution : D

$f (x) = 1 - s i n^{2} \frac{x}{4} + s i n \frac{x}{4} = - {s i n^{2} \frac{x}{4} - s i n \frac{x}{4}} + 1 = - {{(s i n \frac{x}{4} - \frac{1}{2})}^{2} - \frac{1}{4}} + 1 = \frac{5}{4} - {(s i n \frac{x}{4} - \frac{1}{2})}^{2} Maximum f (x) = \frac{5}{4} Minimum f (x) = \frac{5}{4} = {(- 1 - \frac{1}{2})}^{2} = \frac{5}{4} - \frac{9}{4} = - 1 Range of f (x) = [- 1, \frac{5}{4}]$

Question 8

$The range of the function f (x) = \frac{x + 3}{| x + 3 |}, x \neq - 3 i s$

{3,-3}

R-{-3}

all positive integers

{-1,1}

SOLUTION

Solution : D

$f (x) = 1 w h e n x + 3 > 0 f (x) = - 1 w h e n x + 3 < 0 Range = {- 1, 1}$

Question 9

The function $f (x) = l o g_{a} x ((a > 0 a n d a \neq 1))$

A. Odd

B. Even

C. Neither even nor odd

D. Both even and odd

SOLUTION

Solution : C

$∵$ domain is $(0, \infty)$ and is not symmetric about the origin

Question 10

$The function$ $f : R + \to (1, e) defined by f (x) = \frac{X^{2} + e}{X^{2} + 1}$ is

One-one but not onto

Onto but not one-one

Both one-one and onto

Neither one-one nor onto

SOLUTION

Solution : C

$f (x) = \frac{x^{2} + e}{x^{2} + 1} f^{^{'}} (x) = \frac{2 x (x^{2} + 1) - 2 x (x^{2} + e)}{(x^{2} + 1)^{2}} = \frac{2 x^{3} + 2 x - 2 x^{3} - 2 e x}{(x^{2} + 1)^{2}} = \frac{2 x - 2 x e}{(x^{2} + 1)^{2}} = \frac{2 x (1 - e)}{(x^{2} + 1)^{2}} < 0 f^{^{'}} (x) < 0, f (x) is decreasing$ $Hence f is one-one function .$ $x \to 0, f (x) \to e$ $x \to \infty, f (x) \to e$ $Hence range = (1, e) = co-domain$

Question 11

The entire graphs of the equation $y = x^{2} + k x - x + 9$ is strictly above the x-axis if and only if

k<7

-5<k<7

k>-5

-7<k<5

SOLUTION

Solution : B

$y = x 2 + (k - 1) x + 9 = {(x + \frac{k + 1}{2})}^{2} + 9 - {(\frac{k - 1}{2})}^{2}$
For entire graph to be above x-axis, we should have
$9 - {(\frac{x - 1}{2})}^{2}$ > 0
$\Rightarrow k^{2} - 2 k - 35 < 0 \Rightarrow (k - 7) (k + 5)$ < 0
$\Rightarrow$
i.e., -5<k<7

Question 12

If f(x+2y, x-2y)=xy, then f(x, y) equals

$\frac{x^{2} - y^{2}}{8}$

$\frac{x^{2} - y^{2}}{4}$

$\frac{x^{2} + y^{2}}{4}$

$\frac{x^{2} - y^{2}}{2}$

SOLUTION

Solution : A

Let x + 2y = p
x – 2y = q
Solving we got $x = \frac{p + q}{2}$
$y = \frac{p - q}{2} ∴ f (p, q) = \frac{p^{2} - q^{2}}{8} F (x, y) = \frac{x^{2} - y^{2}}{8}$

Question 13

If the function $f : R \to$ A given by $f (x) = \frac{x^{2}}{x^{2} + 1}$ is a surjection, then A is

A. R

B. [0,1]

C. (0,1]

D. [0,1)

SOLUTION

Solution : D

$f (x) = \frac{x^{2}}{x^{2} + 1} = 1 - \frac{1}{1 + x^{2}}$
$x \to 0, f (x) \to 0$
$x \to \pm \infty, f (x) \to 1$
$∴ A ϵ [0, 1)$

Question 14

If $f : [1, \infty) \to [0, \infty) a n d f (x) = \frac{x}{1 + x}$ then f is

one-one and into

onto but not one-one

one-one and onto

neither one-one nor onto

SOLUTION

Solution : A

Given that $f : [0, \infty) \to [0, \infty)$
$s . t . f (x) = \frac{x}{x + 1}$
then $f^{'} (x)$ = $\frac{1 + x - x}{(1 + x^{2})} = \frac{1}{(1 + x)^{2}}$ >0, $\forall x$
$∴$ f is an increasing function $\Rightarrow$ is one-one.
Also $D_{f} = [0, \infty)$
And for range let $\frac{x}{x + 1} = y \Rightarrow x = \frac{y}{y + 1}$

Question 15

If $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x}) \forall x ϵ R - 0$ where f(x) be a polynomial function and f(5) =126 then f(3)=

A. 28

B. 26

C. 27

D. 25

SOLUTION

Solution : A

$f (x) = 1 \pm x^{n} o r, f (5) = 1 \pm 5^{n}$
or, $126 = \pm 5^{n}$
or, $\pm 5^{n} = 125 \Rightarrow \pm 5^{n} = 5^{3}$
n=3
$f (3) = 1 + 3^{3} = 28$

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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