Relations and Functions Test

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Relations and Functions Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)={ x }^{ 3 }+{ x }^{ 2 }{ f }^{ ' }(1)+x{ f }^{ '' }(2)+{ f }^{ ''' }(3)$$ for all $$x\in R$$, then f(x) =

A
$${ x }^{ 3 }+{ 2x }^{ 2 }-5x+6$$

B
$${ x }^{ 3 }+{ 2x }^{ 2 }+6x-5$$

C
$${ x }^{ 3 }-{ 5x }^{ 2 }+6x+2$$

D
$${ x }^{ 3 }-{ 5x }^{ 2 }+2x+6$$

Solution
Question 2
1 / -0

If $$\frac { { x }^{ 4 } }{ \left( x-a \right) \left( x-b \right) \left( x-c \right) } $$
$$=p(x)+\frac { A }{ x-a } +\frac { B }{ x-b } +\frac { C }{ x-c } then\quad p(x)=$$

A
$$x-a$$

B
$$x-a-b$$

C
$$x-a-b-c$$

D
$$x+a+b+c$$

Solution

Carefully observing LHS and RHS we can

say that $$P(x)$$ has to be a linear

equation, so let $$P(x)=m x+n$$

$$\frac{x^{4}}{(x-a)(x-b)(x-c)}=(m x+n)+\frac{A}{(x-a)}+\frac{B}{(x-b)}+\frac{c}{(x-c)}$$

$$\frac{x^{4}}{(x-a)(x-b)(x-c)}=(m x+n)(x-a)(x-b)(x-c)+A(x-b)(x-c)$$

$$\quad \frac{+B(x-a)(x-c)+C(x-a)(x-b)}{(x-a)(x-b)(x-c)}$$

Comparing cofficients of $$x^{4}, x^{3}, x^{2}, x$$ in both sides

$$1=m=$$ (1)

$$0=-a m-b m-c m+n$$

$$\Rightarrow 0=-a-b-c+n$$

$$\Rightarrow a+b+c=n-$$ (2)

$$\therefore \quad P(x)=m x+n$$

$$=x+(a+b+c)$$
$$\therefore$$ Answer option (D)
Question 3
1 / -0

Let $$f_k(x)=\dfrac{1}{k}(\sin^kx+\cos^kx)$$ where $$x\in R$$ and $$k\geq 1$$. Then $$f_4(x)-f_6(x)=?$$

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{12}$$

C
$$\dfrac{1}{6}$$

D
$$\dfrac{1}{3}$$

Solution

$$\begin{aligned} f_{4}(x) &=\frac{1}{4}\left(\sin ^{4} x+\cos ^{4} x\right) \\ &=\frac{1}{4}\left\{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right\} \\ &=\frac{1}{4}\left(1-2 \sin ^{2} x \cos ^{2} x\right) \end{aligned}$$

$$\begin{aligned} f_{6}(x) &=\frac{1}{6}\left(\sin ^{6} x+\cos ^{6} x\right) \\ &=\frac{1}{6}\left\{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cos ^{2} x\right)\right\} \\ &=\frac{1}{6}\left\{(1)\left(1-3 \sin ^{2} x \cos ^{2} x\right)\right\} \\ &=\frac{1}{6}\left(1-3 \sin ^{2} x \cos ^{2} x\right) \end{aligned}$$

$$\begin{aligned} f_{4}(x)-f_{6}(x) &=\frac{1}{4}-\frac{1}{2} \sin ^{2} x \cos ^{2} x-\frac{1}{6}+\frac{1}{2} \sin ^{2} x \cos ^{2} x \\ &=\frac{1}{4}-\frac{1}{6}=\frac{1}{12} \end{aligned}$$
Answer: option (B)
Question 4
1 / -0

If $$f(x)=x^{ { 2 } }+2x+2;x\geq -1=-x^{ { 3 } }+3x+1;x<-1,$$ then the value of $$f ( f ( - 2 ) )$$ is :

A
$$15$$

B
$$17$$

C
$$22$$

D
None of these

Solution

$$f(x)=\left\{\begin{array}{l}x^{2}+2 x+2 ; x \geqslant-1 \\ -x^{3}+3 x+1 ; x<-1\end{array}\right.$$
$$f(-2)=-(-2)^{3}+3(-2)+1=8-6+1=3$$
$$f(f(-2))=f(3)=(3)^{2}+2(3)+2=17$$
Answer option: (B).
Question 5
1 / -0

Let $$f(z)\equiv z^{3}+iz^{2}+iz-1$$. Among the roots of the equation $$f(z)=0$$

A
All are real

B
All are imaginary

C
Sum of two of the roots is zero

D
One is real and two are imaginary

Solution
Question 6
1 / -0

Let $$F\left( x \right) \begin{vmatrix} 1 & 1+\sin { x } & 1+\sin { x } +\cos { x } \\ 2 & 3+2\sin { x } & 4+3\sin { x } +2\cos { x } \\ 3 & 6+3\sin { x } & 10+6\sin { x } +3\cos { x } \end{vmatrix}, x\in R$$; then $$F'\left(\dfrac{\pi}{2}\right)$$ is equla to:

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution
Question 7
1 / -0

$$f (x) = x^4 - 10x^3 + 35x^2 - 50x + c$$ is a constant. the number of real roots of . f (x) = 0 and
f'' (x) = 0 are respectively

A
1 , 0

B
3, 2

C
1 , 2

D
3 , 0

Solution

Given, $$f(x)=x^{4}-10 x^{3}+35 x^{2}-50 x+c$$
Then $$, f^{\prime}(x)=\dfrac{d}{d x}\left(-10 x^{3}+x^{4}+35 x^{2}-50 x+c\right)$$
$$\Rightarrow f^{\prime}(x)=4 x^{3}-30 x^{2}+70 x-50=0$$
$$\therefore 2 x^{3}-15 x^{2}+35 x-25=0$$
First, let us factorise $$f(x)=0$$
$$\Rightarrow x^{4}-10 x^{3}+35 x^{2}-50 x+c=0$$
$$\Rightarrow(x-1)\left(x^{3}-9 x^{2}+26 x-c\right)=0$$
$$\Rightarrow(x-1)\left((x-2)\left(x^{2}-7 x+\dfrac{c}{2}\right)\right)=0$$
$$\Rightarrow f(x)=(x-1)(x-2)(x-4)(x-3)$$
Roots of $$f^{\prime}(x)=0$$
$$\Rightarrow 2 x^{3}-15 x^{2}+35 x-25=0$$
$$\Rightarrow f^{\prime}(x)$$ has 3 roots which are
$$x=\dfrac{5}{2}, \dfrac{5+\sqrt{5}}{2}, \dfrac{-\sqrt{5}+5}{2}$$
Hence $$(A)$$ is the correct option.
Question 8
1 / -0

The function $$f$$ satisfies the functional equation $$3f(x) + 2 f\left (\dfrac {x + 59}{x - 1}\right ) = 10x + 30$$ for all real $$x\neq 1$$. The value of $$f(7)$$ is

A
$$8$$

B
$$4$$

C
$$-8$$

D
$$11$$

Solution

$$\textbf{Step 1: Form equations putting different values of x.}$$

$$\text{It is given that,}$$

$$3f(x) + 2 f\left (\dfrac {x + 59}{x - 1}\right ) = 10x + 30.....(1)$$

$$\text{Putting x=7 in eq.(1)}$$

$$\Rightarrow 3f(7) + 2 f\left (\dfrac {7 + 59}{7 - 1}\right ) = 10\times 7 + 30$$

$$\Rightarrow 3f(7) + 2 f\left (\dfrac {66}{6}\right ) = 70 + 30$$

$$\Rightarrow 3f(7) + 2 f\left (11\right ) = 100.......(2)$$

$$\text{Again putting x=11 in eq.(1)}$$

$$\Rightarrow 3f(11) + 2 f\left (\dfrac {11 + 59}{11 - 1}\right ) = 10\times 11 + 30$$

$$\Rightarrow 3f(11) + 2 f\left (\dfrac {70}{10}\right ) = 110 + 30$$

$$\Rightarrow 3f(11) + 2 f\left (7\right ) = 140.......(3)$$

$$\textbf{Step 2: Simplify to obtain the required result.}$$

$$\text{Multiplying eq(2) by 3 we get,}$$

$$\Rightarrow 9f(7) + 6 f\left (11\right ) = 300.......(4)$$

$$\text{Multiplying eq(3) by 2 we get,}$$

$$\Rightarrow 6f(11) + 4 f\left (7\right ) = 280.......(5)$$

$$\text{Subtracting eq.(5) from eq.(4) we get,}$$

$$\Rightarrow [9f(7) + 6 f\left (11\right )]-[6f(11) + 4 f\left (7\right )] = 300-280$$

$$\Rightarrow 9f(7) + 6 f\left (11\right )-6f(11) - 4 f\left (7\right ) = 20$$

$$\Rightarrow 5f\left (7\right ) = 20$$

$$\Rightarrow f\left (7\right ) = 4$$

$$\text{So, the required value is 4.}$$

$$\textbf{Hence, the correct option is B.}$$
Question 9
1 / -0

If $$f(x)=\dfrac{x^2}{2}+3x$$ where $$x=2, \delta x=0.05$$ then $$df=?$$

A
$$2.5$$

B
$$0.25$$

C
$$2.635$$

D
$$2.652$$

Solution

$$f(x)=\dfrac{x^{2}}{2}+3 x$$
$$d f=\left(\dfrac{2 x}{2}+3\right) \delta x$$
$$d f=(x+3) \delta x$$
$$d f=(2+3)(0.05)$$
$$d f=0.25$$
$$\therefore$$ Answer : option $$(B)$$
Question 10
1 / -0

If a relation R is defined on the set Z of integers as follows : (a,b) $$\epsilon \quad R\Leftrightarrow { a }^{ 2 }+{ b }^{ 2 }=25,$$ , Then domain (R)=

A
{3,4,5}

B
{0,3,4,5}

C
{0,$$\pm $$3, $$\pm $$4, $$\pm $$5}

D
{3,4}

Solution

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Relations and Functions Test - 60

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Question 1

$${ x }^{ 3 }+{ 2x }^{ 2 }-5x+6$$

$${ x }^{ 3 }+{ 2x }^{ 2 }+6x-5$$

$${ x }^{ 3 }-{ 5x }^{ 2 }+6x+2$$

$${ x }^{ 3 }-{ 5x }^{ 2 }+2x+6$$

Solution

Question 2

$$x-a$$

$$x-a-b$$

$$x-a-b-c$$

$$x+a+b+c$$

Solution

Question 3

$$\dfrac{1}{4}$$

$$\dfrac{1}{12}$$

$$\dfrac{1}{6}$$

$$\dfrac{1}{3}$$

Solution

Question 4

$$15$$

$$17$$

$$22$$

None of these

Solution

Question 5

All are real

All are imaginary

Sum of two of the roots is zero

One is real and two are imaginary

Solution

Question 6

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 7

1 , 0

3, 2

1 , 2

3 , 0

Solution

Question 8

$$8$$

$$4$$

$$-8$$

$$11$$

Solution

Question 9

$$2.5$$

$$0.25$$

$$2.635$$

$$2.652$$

Solution

Question 10

{3,4,5}

{0,3,4,5}

{0,$$\pm $$3, $$\pm $$4, $$\pm $$5}

{3,4}

Solution

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