Relations and Functions Test

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Relations and Functions Test - 40

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Question 1
1 / -0

For $$a,\ b\ \epsilon \ R-\left\{ 0 \right\}$$, let $$f(x)=ax^{2}+bx+a$$ satisfies $$f\left(x+\dfrac{7}{4}\right)=f\left(\dfrac{7}{4}-x\right) \forall \ x\ \epsilon\ R$$.
Also the equation $$f(x)=7x+a$$ has only one real distinct solution.
The value of $$(a+b)$$ is equal to

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

$$\because f\left(x+\dfrac{7}{4}\right)=f\left(\dfrac{7}{4}-x\right)$$
$$\therefore f(x)$$ is about $$x=\dfrac{7}{4}$$
$$\because f(x)=ax^{2}+bx+a$$
$$f'(x)=2ax+b\Rightarrow x=\dfrac{-b}{2a}=\dfrac{7}{4}$$
$$\Rightarrow \dfrac{b}{a}=-\dfrac{7}{2}$$
$$\Rightarrow b=-\dfrac{7}{2}a$$
Also
Question 2
1 / -0

$$f(x)$$ is a cubic polynomial with it's leading coefficient 'a'. $$x=1$$ is a point of extremum of $$f(x)$$ and $$x=2$$ is a point of extremum of $$f(x)$$. Then?

A
The other point of extremum is at $$x=3$$

B
The other point of extremum is at $$x=0$$

C
If at $$x=1$$, $$f(x)$$ has a local maximum, $$a > 0$$

D
If at $$x=1$$, $$f(x)$$ has a local minimum, $$a < 0$$

Solution

$$\begin{array}{l} f\left( x \right) =9{ x^{ 3 } }+b{ x^{ 2 } }+cx+d \\ f'\left( x \right) =3a{ x^{ 2 } }+2bx+c \\ x=1\, \, and\, \, x=2\, \, are\, \, roots\, \, of\, \, earation \\ \Rightarrow 3a+2b+c=0 \\ \Rightarrow 12a+4b+c=0 \\ \Rightarrow b=-9a/2,\, \, c=69 \\ f'\left( x \right) =3{ x^{ 2 } }-9x+6 \\ f''\left( x \right) =6x-9 \\ for\, \, x=1\, \, po{ { int } }\, \, of\, \, \max ima\, \, f''\left( x \right) \, \, '-ve' \end{array}$$
Option (c)
Question 3
1 / -0

If x is real, then $$\dfrac{x^2+2x+c}{x^2+4x+3c}$$ can take all real values if?

A
$$0 < c < 2$$

B
$$0 < c < 1$$

C
$$-1 < c < 1$$

D
None of the above

Solution

Let us assume that $$\dfrac{x^2+2x+c}{x^2+4x+3c}=y$$
$$\Rightarrow yx^2+4xy+3yc=x^2+2x+c$$
$$\Rightarrow x^2(y-1)+2x(2y-1)+3yc-c=0$$
For x to be real, Discrimination of this equation should be $$\geq 0$$
$$\therefore 4(2y-1)^2-4\times (y-1)\times (3yc-c)\geq 0$$
$$4(4y^2-4y+1)-4(3cy^2-4cy+c)\geq 0$$
Dividing both sides by $$4$$, we get
$$(4-3c)y^2+4y(c-1)+1-c\geq 0$$
For this equation to hold true, coefficient of $$y^2$$ should be greater than $$0$$ and D $$< 0$$
$$\Rightarrow 4-3c > 0$$ i.e. $$c < \dfrac{4}{3}$$ ..$$(1)$$
Also,
$$16(c^2-2c+1)-4\times (1-c)\times (4-3c) < 0$$
$$\Rightarrow 16c^2-32c+16-4(4-7a+3a^2) < 0$$
$$\Rightarrow 16c^2-32c+16-12c^2+28c-16 < 0$$
$$\Rightarrow 4c(c-1) < 0$$
$$\therefore 0 < c < 1$$.. $$(2)$$
From $$(1)$$ and $$(2)$$, we can say that $$0 < c < 1$$ is the required condition.
Question 4
1 / -0

If $$\dfrac{{{x^2} + {y^2}}}{{x + y}} = 4$$, then all possible values of $$(x-y)$$ is given by

A
$$\left[ { - 2\sqrt 2 ,2\sqrt {2} } \right]$$

B
$$ [ - 4,1] $$

C
$$\left[ { - 4,4} \right]$$

D
$$\left[ { - 2,2} \right]$$

Solution

$$\dfrac{x^2+y^2}{x+y}=4$$
$$x^2+y^2=4(x+y)$$
$$x^2-4x+y^2-4y=0$$
$$x^2-4x+4+y^2-4y+4-8=0$$
$$(x-2)^2 (y-2)^2=8$$
$$(x-2)^2=4$$ and $$(y-2)^2=4$$
$$(x-2)=-2$$ and $$ (y-2)=-2$$
$$\therefore x=0 \text{or} 4$$ and $$y=0 \text{or}4$$
$$\therefore $$ if $$x=0$$ and $$y=4$$
$$x-y=-4$$
if $$x=4$$ and $$y=0$$
$$x-y=4$$
$$(x-y)\in [-4,4]$$
Question 5
1 / -0

The area of the region $$R = \{(x, y) : |x| \le |y| \, \text{and} \, x^2 + y^2 \le 1 \}$$ is

A
$$\dfrac{3 \pi}{8}$$ sq. units

B
$$\dfrac{5 \pi}{8} $$ sq. units

C
$$\dfrac{\pi}{2}$$ sq. units

D
$$\dfrac{\pi}{8} $$ sq. units

Solution
Question 6
1 / -0

Let $$R = \left \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 2)\right \}$$ be a relation on the set $$A = \left \{1, 2, 3, 4\right \}$$. The relation $$R$$ is

A
Reflexive

B
Transitive

C
Not symmetric

D
none of these

Solution
Question 7
1 / -0

If $$f:R\rightarrow R,g:R\rightarrow R$$ are defined by $$f(x)=5x-3,g(x)={ x }^{ 2 }+3,$$ then $$(go{ f }^{ -1 })(3)=\\ $$

A
$$\frac { 25 }{ 3 } $$

B
$$\frac { 111}{ 25 } $$

C
$$\frac { 9 }{ 25 } $$

D
$$\frac { 25 }{ 111 } $$
Question 8
1 / -0

$$f:c \to c$$ is defined as $$f(x) = \dfrac{{ax + b}}{{cx + d}},bd \ne 0$$ then $$f$$ is a constant function when,

A
a=c

B
b=d

C
ad=bc

D
ab=cd

Solution

f($$x$$)=$$\frac{ax+b}{cx+d}$$ is a constant function,
then lets say it equal to same constant m.
$$m(cx+d)=ax+b$$
$$a=mc $$
$$b=md $$
$$\frac{a}{c}$$ =$$\frac{b}{d}=m$$
$$\frac{a}{b}$$ =$$\frac{c}{d}$$
$$ad=bc$$
C is correct.
Question 9
1 / -0

If $$f$$ is a real valued function such the $$\left| {f\left( x \right) - f\left( y \right)} \right| \le {\left| {x - y} \right|^3}$$ then $$f'\left( x \right)$$

A
$$\left| {\cos x\pi } \right|$$

B
$$\left| {\sin \left[ x \right]\pi } \right|$$

C
$$2\left| x \right|$$

D
not possible to find

Solution

$$ \displaystyle\left | f(x)-f(y) \right | \leq \left | x-y \right |^{3}$$
$$ \displaystyle \Rightarrow \dfrac{\left | f(x)-f(y) \right |}{\left | x-y \right |} \leq \left | x-y \right |^{2}$$
$$ \displaystyle \Rightarrow \lim_{x\rightarrow y} \left | \dfrac{f(x)-f(y)}{x-y} \right | \leq \lim_{x\rightarrow y} \left | x-y \right |^{2}$$
$$ \displaystyle \Rightarrow \lim_{x\rightarrow y} \left | \dfrac{f(x)-f(y)}{x-y} \right |\leq \lim_{x\rightarrow y} \left | x-y \right |^2 $$
$$ \Rightarrow \left | {f}' (x) \right | \leq 0.$$
which is not possible.
$$ \therefore {f}'(x) $$ is not possible to find.
Question 10
1 / -0

Let for $$a\ne a_1 \ne 0$$, $$f(x)=ax^2+bx+c$$, $$\,\,\ g(x)=a_1x^2+b_1x+c_1$$ 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
$$3$$

B
$$9$$

C
$$6$$

D
$$18$$

Solution

$$ p(x) = (a-a_{1})x^{2}+(b-b_{1})x+(c-c_{1})$$
$$ p(-1)= 0$$
$$ p(-2)=2$$
$$ p(0) =2 $$ (one to symmetry around x = -1 )
$$ p(2) = ?$$
$$ p(0) = 2 \Rightarrow c =c_{1}+2$$
$$ p(-1) = 0 \Rightarrow (a-a_{1})-(b-b_{1})+2 = 0$$ __ (1)
$$ p(-2) = 2 \Rightarrow 4(a-a_{1})-2(b-b^{1})+2=2$$
$$ eq^{n}(1)\times 2 $$
$$\Rightarrow 2(a-a_{1})-2(b-b_{1})+4 = 0$$
$$ 2(a-a_{1}) = 2+2 $$
$$ \Rightarrow a-a_{1} = 2$$
$$ \Rightarrow b-b^{1} = 4 $$
$$p(2)= 4(a-a_{1})+2(b-b_{1})+(c-c_{1})$$
$$ = 4\times 2+2\times 4+2$$
$$ = 8+8+2$$
$$ p(2) = 18 $$