Functions Test 50

Result

Functions Test 50

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$f:R\rightarrow S$$ defined by
$$f(x)=4\sin { x } -3\cos { x } +1$$ is onto, then $$S$$ is equal to

A
$$[-5,5]$$

B
$$(-5,5)$$

C
$$(-4,6)$$

D
$$[-4,6]$$

Solution

We have $$f(x)=4sin\:x-3cos\:x+1$$

Therefore
maximum of $$f=\sqrt{4^2+(-3)^2}+1=\sqrt{16+9}+1=\sqrt{25}+1=5+1=6$$ and
minimum of $$f=-\sqrt{4^2+(-3)^2}+1=-\sqrt{16+9}+1=-\sqrt{25}+1=-5+1=-4$$

$$S=$$ Range of $$f$$=$$[minimum\:f,maximum\:f]=[-4,6]$$
Question 2
1 / -0

Domain of definition of the function $$f(x) = \dfrac{3}{4 - x^2} + \log_{10} (x^3 - x),$$ is

A
$$(a , 2)$$

B
$$(-1, 0) \cup (1, 2)$$

C
$$(1, 2) \cup (2, \infty)$$

D
$$(-1, 0) \cup (1, 2) \cup (2, \infty)$$

Solution
Question 3
1 / -0

A function f has domain [-1, 2] and range [0, 1]. The domain and range respectively of the function g defined by g (x) = 1 -f (x+1) is

A
[-1, 1] ; [-1, 0]

B
[-2, 1] ; [0, 1]

C
[0, 2]; [-1, 0]

D
[1, 3] ; [-1, 0]

Solution
Question 4
1 / -0

The domain of definition of the function $$\displaystyle f(x) = \sqrt{-cos x} + \sqrt{sin x}$$ is:

A
$$\displaystyle \left[ 2 n \pi, 2 n \pi + \frac{\pi}{2} \right] , n \epsilon I$$

B
$$\displaystyle \left[ 2 n \pi, (2 n + 1) \pi \right] , n \epsilon I$$

C
$$\displaystyle \left[ \left( 2 n + \frac{1}{2} \right) \pi , (2 n + 1) \pi \right] , n \epsilon I$$

D
$$\displaystyle \left[( 2 n + 1 ) \pi , \left( 2 n + \frac{3}{2} \right) \pi \right] , n \epsilon I$$
Question 5
1 / -0

Let for $$a \neq a_{1} \neq 0,\ f(x)=ax^{2}+bx+c,\ g(x)=a_{1}x^{2}+b_{1}x+c_{1}$$ and $$p(x)=f(x)-g(x)$$. If $$p(x)=0$$ only for $$x=-1$$ and $$p(-2)=2$$, then the value of $$p(2)$$ is

A
$$6$$

B
$$18$$

C
$$3$$

D
$$9$$

Solution

Given, $$f\left(x\right)=ax^{2}+bx+c,g\left(x\right)=ax^{2}+bx+c,$$

and $$p\left(x\right)=f\left(x\right)-g\left(x\right)$$

$$p\left(x\right)=0$$ for only $$x=-1,p\left(-2\right)=2$$ and $$a\neq a_{1}\neq 0$$

$$\therefore p\left(x\right)=(a-a_{1})x^{2}+(b-b_{1})x+(c-c_{1})$$

The standard quadratic equation is $$ax^2+bx+c=0$$

For roots to be equal $$ \displaystyle \Rightarrow b^{2}-4ac= 0 $$

Then Sum of roots = $$-\dfrac {b}{a}$$

Here, Only root is $$-1 \Rightarrow D=0 \Rightarrow (b-b_{1})^{2}=4(a-a_{1})(c-c_{1})\rightarrow(1)$$

Sum of roots $$\Rightarrow (-1)+(-1)=\dfrac{-(b-b_{1})}{(a-a_{1})} \Rightarrow (b-b_{1})=2(a-a_{1}) \rightarrow(2)$$

From (1) and (2), $$c-c_{1}=(a-a_{1})\rightarrow(3)$$

$$\Rightarrow p\left(x\right)=(a-a_{1})x^{2}+2(a-a_{1})x+(a-a_{1})$$

Now, $$p\left(-2\right)=2$$

$$\Rightarrow 2=(a-a_{1})(-2)^{2}-2(a-a_{1})(-2)+(a-a_{1})$$

$$\Rightarrow (a-a_{1})=2$$

Now $$p(2)=2(2)^{2}+2(2)(2)+2=18$$
Question 6
1 / -0

Let $$f(-2, 2)\rightarrow(-2, 2)$$ be a continuous function given $$f(x)=f{(x}^{2})$$. Given $$f(0)=\dfrac{1}{2}$$ then the $$4f(\dfrac{1}{2})$$

A
$$4$$

B
$$2$$

C
$$-2$$

D
$$1$$

Solution
Question 7
1 / -0

Given that $$f'(x) > g'(x)$$ for all real x, and $$f(0) = g(0)$$. Then $$f(x) < g(x)$$ for all x belongs to

A
$$(0, \infty)$$

B
$$(- \infty, 0)$$

C
$$(- \infty, \infty)$$

D
none of these

Solution
Question 8
1 / -0

If $$ \phi (x) = 3
f(\frac{x^2}{3} ) + f(3-x^2) \forall x \in (3,4)$$ where $$f(x) >0 \forall x (-3,4)$$ then $$\phi (x)$$ is ____________.

A
(a) increasing in $$( \frac{3}{2} ,4)$$

B
(b) decreasing in $$( 3, \frac{3}{2} )$$

C
(c) increasing $$( -\frac{3}{2} , 0)$$

D
decreasing in $$( 0, \frac{3}{2})$$

Solution
Question 9
1 / -0

The domain of the function $$ f(x) = log_{1/2}\Big(-log_2(1+\frac{1}{\sqrt{4}})-1\Big)$$

A
0

B
0< x $$\leq$$ 1

C
x$$\geq$$ 1

D
Null set

Solution
Question 10
1 / -0

Let f(x) and g(x) be the differentiable functions for $$1\le x\le 3$$ such that f(1)=2=g(1) and f(3)=10. Let there exist exactly one real number $$cE (1,3)$$ such that 3f'(c)=g'(c), then the value of g(3) must be

A
12

B
13

C
16

D
26

Solution