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CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 10 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Functions
Quiz 10
For set $$A, B$$ and $$C$$, let $$f:A\to B, g:B\to C$$ be functions such that $$gof $$ is Injective.
Then $$f$$ is injective.
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True
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False
Explanation
Suppose that $$g\circ f$$ is injective;
we show that $$f$$ is injective.
To this end, let $$x_1, x_2\in A$$
and suppose that $$f(x_1) =f(x_2)$$.
Then $$(g \circ f)(x_1) =g(f(x_1)) =g(f(x_2)) = (g\circ f)(x_2).$$
But since $$g\circ f$$ is injective,
this implies that $$x_1=x_2.$$
Therefore $$f$$ is injective.
Which of the following functions from $$Z$$ to $$Z$$ are bijection?
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$$f(x)=x^3$$
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$$f(x)=x+2$$
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$$f(x)=2x^2+1$$
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$$f(x)=x^2+1$$
Explanation
Here, $$f(x)=x+2\Rightarrow f(x_1)=f(x_2)$$
$$x_1 +2=x_2+2\Rightarrow x_1=x_2$$
Let $$y=x+2$$
$$x=y-2\in Z, \forall \in x$$
Here, $$f(x)$$ is one-one is onto.
Let $$f(x)$$ and $$g(x)$$ be differentiable for $$0\times < 1$$ such that $$f(0)=0, g(0), f(1)=6$$. Let there exist a real number $$c$$ in $$(0,1)$$ such that $$f'(c)=2g'(c)$$, then the value of $$g(1)$$ must be
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$$1$$
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$$3$$
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$$2.5$$
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$$-1$$
If g is the inverse of function $$f$$ and $$f'(x) = \frac{1}{1 + x}$$, then the value of g'(x) is equal to:
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$$1 + x^7$$
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$$\frac{1}{1 + [g(x)]^7}$$
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$$1 + [g(x)]^7$$
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$$7x^6$$
Explanation
Since g is the inverse of $$f, f^{-1}(x) = g(x)$$
$$\therefore f[f^{-1}(x)] = f [g(x)] = x$$
$$\therefore f'[g(x)] \cdot \frac{d}{dx} [g(x)] = 1$$
$$\therefore f'[g(x)] \times g'(x) = 1$$
$$\therefore g'(x) = \frac{1}{f'[g(x)]}$$, where $$f'(x) = \frac{1}{1 + x^7}$$
$$\therefore g(x) = 1 + [g(x)]^7$$
Which one of the following is onto function define R to R .
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$$ f(x) = |x| $$
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$$ f(x) = e^{x}$$
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$$ f(x) = x^{3}$$
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$$ f(x) = \sin x $$
Which of the following in one -one function defined from R to R
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$$ f(x) = |x| $$
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$$ f(x) = \cos x $$
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$$ f(x) =e^{x}$$
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$$ f(x) = x^{2} $$
Which of the following is onto function-
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$$ f : Z \rightarrow Z , f (x) = |x| $$
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$$ f : N \rightarrow N , f (x) = |x| $$
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$$ f : R_{0} \rightarrow R^{+} , f (x) = |x| $$
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$$ f : C \rightarrow R , f (x) = |x| $$
From $$ ] \dfrac{- \pi}{2} , \dfrac{- \pi}{2}[ $$ which of the following is one - one onto function defined in R
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$$ f(x) = \tan x $$
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$$ f(x) = \sin x $$
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$$ f(x) = \cos x $$
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$$ f(x) = e^{x} + r^{-x}$$
Let $$ f: [ -1,1] \rightarrow [0,2]$$ be a linear function which is onto then f(x) is/are
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$$ 1 - x $$
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$$ 1 + x $$
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$$ x - 1$$
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$$ x - 2 $$
Explanation
If $$f(x) = x^{2} + x$$ and $$g(x) = \sqrt {x}$$, then the value of $$f(g(3))$$ is
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$$1.73$$
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$$3.46$$
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$$4.73$$
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$$7.34$$
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$$12.00$$
Explanation
Given, $$f(x)=x^2+x, g(x)=\sqrt x$$
Then we need to find the value of $$f(g(3))$$
$$\Rightarrow g(3) = \sqrt3$$ ..... given $$g(x) = \sqrt x $$
$$\Rightarrow f(g(3)) = f(\sqrt3) = 3 + \sqrt3 $$ ..... given $$f(x) = { x }^{ 2 }+x$$
If $$f\left( x \right)=\sqrt { 3\left| x \right| -x-2 } $$ and $$g(x)=\sin(x)$$, then the domain of the definition of $$f\circ g\left( x \right) $$ is
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$$\displaystyle \left\{ 2n\pi +\frac { \pi }{ 2 } \right\} $$
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$$\displaystyle \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) $$
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$$\displaystyle \left\{ 2n\pi +\frac { 7\pi }{ 6 } \right\} $$
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$$\displaystyle \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) \bigcup _{ n,m\in I }^{ }{ \left( 2n\pi +\frac { \pi }{ 2 } \right) } $$
Explanation
We have $$\displaystyle f\left( x \right) =\sqrt { 3\left| x \right| -x-2 } $$ and $$g\left( x \right) =\sin { x } $$
$$\therefore f\circ g\left( x \right) =\sqrt { 3\left| \sin { x } \right| -\sin { x } -2 } $$ which is defined,
if $$3\left| \sin { x } \right| -\sin { x } -2\ge 0$$
If $$\sin { x } >0,$$ then $$2\sin { x } -2\ge 0\quad $$
$$\Rightarrow \sin { x } \ge 1$$
$$\displaystyle \Rightarrow \sin { x } =1\Rightarrow x=\frac { \pi }{ 2 } $$
If $$\sin { x } <0,$$ then $$\displaystyle -4\sin { x } -2\ge 0\Rightarrow -1\le -\frac { 1 }{ 2 } $$
$$\displaystyle \therefore x\in \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) \bigcup _{ n,m\in I }^{ }{ \left( 2n\pi +\frac { \pi }{ 2 } \right) } n,m\in I$$
Domain of the function,
$$f(x)=\sqrt{\cos(\sin x)}+\sin^{-1}(x^2-1)$$ is
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$$[-1, 1]$$
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$$[-2, 2]$$
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$$[-\pi, -\sqrt{2}] \cup [\sqrt{2}, \pi]$$
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$$[-\sqrt{2}, \sqrt{2}]$$
Explanation
$$f(x)=\sqrt{\cos(\sin x)}+\sin^{-1}(x^2-1)$$
$$-1 \leq \sin x \leq 1$$ $$\forall x \in R$$
$$\therefore \cos(\sin x) > 0$$ $$\forall x \in R$$
$$\therefore$$ domain of $$\sqrt{\cos(\sin x)}$$ is $$R$$ .....(i)
$$-1 \leq x^2-1 \leq 1$$ i.e. $$0 \leq x^2 \leq 2$$
i.e. $$x \in [-\sqrt{2}, \sqrt{2}]$$
$$\therefore$$ domain of $$\sin^{-1}(x^2-1)$$ is $$ [-\sqrt{2}, \sqrt{2}]$$ ....(ii)
From (i) and (ii)
Domain of $$f(x) $$ is $$ [-\sqrt{2}, \sqrt{2}]$$
If $$f(x) = x^{2}$$, $$-1\leq x \leq 4$$ ,$$ g(x) = sec^{-1}x$$, $$x\geq 1$$ then
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Domain of gof(x) is $$[1, 4] \cup {-1}$$
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Domain of gof(x) is $$[1, 4] $$
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Range of gof(x) is $$[0, sec^{-1}16]$$
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Range of fog(x) is $$\left ( 0,\frac{\pi^{2}}{4} \right )$$
Explanation
gof(x) exists when $$f(x) \geq 1$$ and $$x \epsilon [-1, 4]$$
$$x^{2}\geq 1\Rightarrow X \epsilon (-\infty,-1]\cup[1,\infty)$$
$$\Rightarrow X \epsilon [1,4]\cup {-1}$$
Range of $$gof(x) = [0, sec^{-1}16]$$
If $$f (x) = x + 2, g (x) = 2 x +3,$$ then find gof
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$$2x -7 $$
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$$7x + 2$$
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$$2x + 7$$
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$$7 + 2x$$
The domain of the function $$\displaystyle f(x)=\sin^{-1}\dfrac {1}{|x^2-1|}+\dfrac {1}{\sqrt {\sin^2x+\sin x+1}}$$ is
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$$\displaystyle (-\infty, \infty)$$
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$$\displaystyle (-\infty, -\sqrt 2]\cup [\sqrt 2, \infty)$$
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$$\displaystyle (-\infty , -\sqrt 2]\cup [\sqrt 2, \infty)\cup \left \{0\right \}$$
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none of these
Explanation
As, $$\sin^2x+\sin x+1 > 0, \forall x\in R$$
$$\displaystyle \therefore \frac {1}{\sqrt {\sin^2x+\sin x+1}}$$ is always exists.
$$\displaystyle \therefore$$ For $$\sin^{-1}\left (\dfrac {1}{|x^2-1|}\right )$$ to exists,
$$\displaystyle 0 < \frac {1}{|x^2-1|} \leq 1\Rightarrow |x^2-1|\geq 1$$
$$\displaystyle \Rightarrow x^2-1 \leq -1$$ or $$x^2-1\geq 1$$
$$\displaystyle \Rightarrow x^2\leq 0$$ or $$x^2\geq 2$$
$$\displaystyle \Rightarrow x=0$$ or $$(x\leq -\sqrt 2$$ or $$x\geq \sqrt 2)$$
$$\displaystyle \therefore x\in (-\infty, -\sqrt 2]\cup [\sqrt 2, \infty)\cup \left \{0\right \}$$
Hence, (c) is the correct answer.
Let $$\displaystyle \mathrm{f}:\mathrm{R}\rightarrow \left[0,\frac{\pi}{2}\right)$$ be defined by
$$\mathrm{f}(\mathrm{x})=\mathrm{t}\mathrm{a}\mathrm{n}^{-1}(\mathrm{x}^{2}+\mathrm{x}+\mathrm{a})$$. Then the set of values of '$$\mathrm{a}$$' for which $$\mathrm{f}$$ is onto is
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$$[0,\infty)$$
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$$[2, 1]$$
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$$ \displaystyle \left\{ \frac{1}{4}\right\}$$
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$$ \displaystyle \left[ \frac{1}{4}, \infty\right)$$
Explanation
$$f(x)=\tan^{-1}(x^2+x+a)$$
For, $$f(x)$$ to be onto, codomain should be exactly equal to range.
That is, range of funnction $$ \displaystyle f(x) = \left[ 0,\frac { \pi }{ 2 } \right) $$
So, $$0\leqslant \tan^{-1}(x^2+x+a)<\dfrac{\pi}{2}$$
Now, $$ \displaystyle { x }^{ 2 }+x+a={ \left( x+\dfrac { 1 }{ 2 } \right) }^{ 2 }+a-\dfrac { 1 }{ 4 } $$
The above expression will take all real values from $$ \left[0, \infty \right)$$, only if $$ \displaystyle a = \dfrac{1}{4}$$
Hence, only for $$ a= \dfrac{1}{4}$$, the function is onto.
Let $$\displaystyle {f}({x})=\frac{{a}{x}+{b}}{{c}{x}+{d}}$$, then $$fof(x)={x}$$, provided
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$$d = -a$$
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$$d = a$$
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$$a = b = c = d = 1$$
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$$a = b = 1$$
Explanation
Given,
$$f(x)=\dfrac{ax+b}{cx+d}$$
$$f(f(x))=\dfrac{a\left ( \dfrac{ax+b}{cx+d}\right )+b}{c\left ( \dfrac{ax+b}{cx+d} \right )+d}=x$$
$$=\dfrac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}=x$$
$$a(ax+b)+b(cx+d)=x[c(ax+b)+d(cx+d)]$$
$$(a^2+bc)x+(ab+bd)=(ac+cd)x^2+(bc+d^2)x$$
matching the coefficients of $$x$$, we get,
$$a^2+bc=bc+d^2$$
$$a^2=d^2$$
$$d=\pm a$$
matching the coefficients of $$x^2$$, we get,
$$ac+cd=0$$
$$\therefore d=-a$$
matching constant terms, we get,
$$ab+bd=0$$
$$d=-a$$
Which of the following functions is/are injective map(s) ?
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$$f(x)=x^2+2, x \in (-\infty,\infty)$$
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$$f(x)=|x+2|, x \in [-2,\infty)$$
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$$f(x)=(x-4)(x-5), x \in (-\infty,\infty)$$
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$$f(x)=\dfrac{4x^2 + 3x -5}{4+3x-5x^2}, x\in(-\infty, \infty)$$
Explanation
The function $$f(x)=x^2 + 2, x \in (-\infty, \infty)$$ is not injective as
$$f(1)=f(-1) $$ but $$1 \neq -1.$$
The function $$f(x)=(x-4)(x-5), x \in (-\infty, \infty)$$ is not one-one as $$f(4)=f(5)$$ but $$4 \neq 5.$$
The functin $$f(x)=\dfrac{4x^2 + 3x -5}{4 + 3x - 5x^2}, x \in (-\infty,\infty)$$ is also not injective as $$f(1)=f(-1)$$ but $$1 \neq -1$$.
For the function , $$f(x)=-|x+2|, x\in [-2,\infty)$$
Let $$f(x)=f(y),x,y \in [-2,\infty) \Rightarrow |x+2| = |y+2| $$
$$\Rightarrow x+2 = y+2$$
$$\Rightarrow x=y$$
So , $$f$$ is an injective map.
Which of the following functions is not injective ?
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$$f:R \rightarrow R, f(x)=2x+7$$
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$$f:[0,\pi]\rightarrow[-1,1],f(x)=\cos x$$
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$$f:\left [ -\dfrac{\pi}{2},\dfrac{\pi}{2} \right ]\rightarrow R, f(x)=2 \sin x +3$$
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$$f:R\rightarrow [-1,1],f(x)=\sin x$$
Explanation
A:
$$f:R\rightarrow R, f(x)=2x+7$$
$$\dfrac{dy}{dx}=2>0 \rightarrow$$ one-one (Injective)
B:
$$f:[0,\pi]\rightarrow [-1,1],f(x)=\cos x$$
$$\dfrac{dy}{dx}=\sin x=(+ve) \rightarrow$$ one-one (Injective)
C:
$$f: \left[-\dfrac{\pi}2, \dfrac{\pi}2\right], f(x) = 2\sin x+3$$
$$\dfrac{dy}{dx}= 2\cos x=+ve \rightarrow$$ one-one (Injective)
D:
$$f:R\rightarrow [-1,1],f(x)=\sin x$$
$$\dfrac{dy}{dx}=\cos x=+ve$$ & $$-ve \rightarrow$$ many one
Hence, option D.
The domain of the function $$f(x)=\log_3 \log_{1/3}(x^2+10x+25)+\dfrac {1}{[x]+5}$$ where [.] denotes the greatest integer function) is
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$$(-4, -3)$$
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$$(-6, -5)$$
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$$(-6, 4)$$
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None of these
Explanation
For given function to be defined,
(i) $$x^2+10x+25>0\Rightarrow (x+5)^2>0\Rightarrow x\neq -5=D_1$$
(ii) $$\log_{1/3}(x^2+10x+25)>0\Rightarrow x^2+10x+25<1$$
$$\Rightarrow x^2+10x+24<0\Rightarrow (x+4)(x+6)<0\Rightarrow x\in(-6,-4)=D_2$$
and
(iii) $$[x]+5\neq 0\Rightarrow x\notin [-5,4)=D_3$$
Hence domain of f is $$D_!\cap D_2\cap D_3 =(-6,-5)$$
Suppose $$f(x)=ax+b$$ and $$g(x)=bx+a$$, where $$a$$ and $$b$$ are positive integers. If $$f\left ( g(50) \right )-g\left ( f(50) \right )=28$$ then the product $$(ab)$$ can have the value equal to
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$$12$$
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$$48$$
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$$180$$
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$$210$$
Explanation
$$f(g(x))=a(bx+a)+b$$
$$=abx+a^2+b$$
$$g(f(x))=b(ax+b)+a$$
$$=abx+b^2+a.$$
Therefore $$f(g(x))-g(f(x))$$
$$=abx+a^2+b-(abx+b^2+a)$$
$$=a^2-b^2+b-a$$
$$=(a-b)(a+b-1)$$
Since it is a constant function.
$$(a-b)(a+b-1)=28$$
$$(a-b)(a+b-1)=4(7)$$
Hence $$a-b=4$$
$$a+b=8$$
$$a=6$$ and $$b=2$$
Hence $$ab=12$$
Also
$$(a-b)(a+b-1)=(1)(28)$$
$$a-b=1$$
$$a+b=29$$
$$a=15$$ and $$b=14$$
Therefore $$ab=14(15)=210$$
Which one of the following functions is not one-one?
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$$f:(-1,\infty )\rightarrow R$$ given by $$ f(x)={ x }^{ 2 }+2x\quad $$
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$$g:(2,\infty )\rightarrow R$$ given by $$g(x)={ e }^{ { x }^{ 3 }-3x+2 }\quad $$
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$$h:R\rightarrow R$$ given by $$h(x)={ 2 }^{ { x }(x-1) }\quad $$
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$$\phi :(-\infty ,0)\rightarrow R$$ given by $$\phi (x)=\cfrac { { x }^{ 2 } }{ { x }^{ 2 }+1 } $$
Explanation
$$ x(x-1)$$ is not a one -one function in whole $$R$$.
This function has a minima at $$x=\dfrac12$$, therefore repeats value after $$x=\dfrac12$$.
We can also see that $$x(x-1) = 0$$ for $$x=0,1$$ and therefore $$e^{x(x-1)} =1$$ for both $$x=0 \&\ 1$$, which rules it out as a one to one function.
$$l= \lim_{x\rightarrow \alpha}\displaystyle \frac{f(x)}{x(x-\alpha)(x-2)}$$ is
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positive
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negative
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$$0$$
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sign of $$l$$ depends upon $$\alpha\pi$$
Explanation
For the quadratic function $$f(x) = ax^2 + bx +c$$, it's given that $$a, b, c$$ are non-zero.
Hence, $$f(0) = c \neq 0$$
It's given that $$f:[0,2] \rightarrow [0,2]$$ is bijective.
Since $$f(0) \neq 0$$, $$f(0) = 2$$ for it to be bijective.
$$ \Rightarrow f(2) = 0$$
So, three cases possible as shown in the figure, which will satisfy the second condition as well.
For cases 1 and 2,
$$ \displaystyle \lim_{x\rightarrow \alpha}\displaystyle \frac{f(x)}{x(x-\alpha)(x-2)}$$
$$ \displaystyle =\lim _{ x\rightarrow \alpha } \frac { 1 }{ x\left( x-2 \right) } \lim _{ x\rightarrow \alpha } \frac { f(x) }{ x-\alpha } \\ \displaystyle =\frac { 1 }{ \alpha \left( \alpha -2 \right) } \lim _{ x\rightarrow \alpha } \frac { f'(x) }{ 1 } =\frac { f'(\alpha ) }{ \alpha \left( \alpha -2 \right) } $$
$$f'(\alpha)$$ is positive for both cases. $$ \alpha( \alpha -2)$$ is also positive for both cases. Hence, the limit value is also positive.
For the case in figure-3,
$$ \displaystyle \lim _{ x\rightarrow \alpha } \frac { f(x) }{ x(x-\alpha )(x-2) } =\lim _{ x\rightarrow \alpha } \frac { 1 }{ x } \lim _{ x\rightarrow \alpha } \frac { f(x) }{ { \left( x-\alpha \right) }^{ 2 } } \\ \displaystyle =\frac { 1 }{ \alpha } \lim _{ x\rightarrow \alpha } \frac { f'(x) }{ 2\left( x-\alpha \right) } =\frac { 1 }{ \alpha } \lim _{ x\rightarrow \alpha } \frac { f''(x) }{ 2 } =\frac { 1 }{ \alpha } \frac { f''\left( \alpha \right) }{ 2 } $$
$$f''(\alpha) > 0$$ since the graph is facing upwards. Also $$ \alpha$$ is positive.
Hence, in this case also , the value of limit is positive. Hence, option A is correct.
Domain of the function $$f(x)=\dfrac {1}{\sqrt {4x-|x^2-10x+9|}}$$, is
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$$(7-\sqrt {40}, 7+\sqrt {40})$$
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$$(0, 7+\sqrt {40})$$
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$$(7-\sqrt {40}, \infty)$$
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none of these
Explanation
Here, $$f(x)=\dfrac {1}{\sqrt {4x-|x^2-10x+9|}}$$ would exist, if
$$\displaystyle 4x-|x^2-10x+9| > 0$$
ie, $$\displaystyle |x^2-10x+9| < 4x$$,
where $$\displaystyle |x^2-10x+9|=\begin{cases}x^2-10x+9 & x\leq 1 \text{ or } x \geq 9\\ -(x^2-10x+9) &1 < x < 9 \end{cases}$$
Case I: When $$x\leq 1$$ or $$x\geq 9$$
$$\therefore x^2-10x+9 < 4x$$
$$\Rightarrow x^2-14x+9 < 0 \Rightarrow (x-7)^2 < 40$$
$$\Rightarrow x\in (7-\sqrt {40}, 7+\sqrt {40})$$ (But $$x\leq 1$$ or $$x\geq 9$$)
$$\Rightarrow x\in (7-\sqrt {40}, 1]\cup [9, 7+\sqrt {40})$$ .....(i)
Case II: When $$1 < x < 9$$
$$-x^2+10x-9 < 4x\Rightarrow x^2-6x+9 > 0$$
$$\Rightarrow (x-3)^2 > 0$$ which is always true except $$x=\left \{3\right \}$$
$$\therefore x\in (1, 9)=\left \{3\right \}$$.....(ii)
From Eqs. (i) and (ii), domain of $$f(x)\in (7-\sqrt {40}, 7+\sqrt {40})-\left \{3\right \}$$
Hence, (d) is the correct answer.
If $$f:R\rightarrow \left [\dfrac {\pi}{6}, \dfrac {\pi}{2}\right ), f(x)=\sin^{-1}\left (\dfrac {x^2-a}{x^2+1}\right )$$ is a onto function, then set of values of $$a$$ is
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$$\left \{-\dfrac {1}{2}\right \}$$
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$$\left [-\dfrac {1}{2}, -1\right )$$
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$$(-1, \infty)$$
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none of these
Explanation
Given, $$f(x)$$ is onto.
$$\therefore \displaystyle \frac {\pi}{6}\leq \sin^{-1}\left (\frac {-x^2-a}{x^2+1}\right ) < \frac {\pi}{2}$$
$$\Rightarrow \displaystyle \frac {1}{2}\leq \frac {x^2-a}{x^2+1} < 1$$
$$\Rightarrow \displaystyle \frac {1}{2}\leq 1-\frac {(a+1)}{x^2+1} < 1, \forall x\epsilon R $$
$$\Rightarrow a+1 > 0$$
$$\Rightarrow a\in (-1, \infty)$$
Hence, (c) is the correct answer.
In the following functions defined from $$[-1, 1]$$ to $$[-1, 1]$$, then functions which are not bijective are
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$$\displaystyle \sin (\sin^{-1} \: x)$$
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$$\displaystyle \frac {2}{\pi} \: \sin^{-1} (\sin \: x)$$
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$$\displaystyle (sgn \: x) \ lne^x$$
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$$\displaystyle x^3 \: sgn \: x$$
Explanation
$$sin\sin ^{ -1 }{ x } = x$$ hence it is bijective.
$$\dfrac { 2 }{ \pi } \sin ^{ -1 }{ sinx } $$= $$\dfrac { 2x }{ \pi } $$. Hence it is one one but not onto, hence not bijective
$$sgn(x).x=\dfrac { \left| x \right| }{ x }.x= |x|, x\neq$$0
Hence it is not one one or onto, hence not bijective
$${ x }^{ 3 }sgnx={ x }^{ 3 }\dfrac { |x| }{ x } ={ x }^{ 2 }|x|={ |x| }^{ 3 },x\neq 0$$. Hence function is not one one or onto, hence not bijective.
Which of the function defined below are one-one function(s)?
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$$f(x)=x+1,(x\geq-1)$$
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$$g(x)=x+\dfrac1x,(x\geq0)$$
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$$h(x)=x^2+4x-5,(x>0)$$
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$$f(x)=e^{-x},(x\geq0)$$
Explanation
A function is called one-one if it is monotonic
A. $$f(x)= x+1)\Rightarrow f'(x) =1>0\Rightarrow f$$ is monotonically increasing
B. $$g(x) =x+\dfrac{1}{x}\Rightarrow g'(x)=1-\dfrac{1}{x^2},$$ which is -ve for $$0\le x<1$$ and +ve for $$x>0\Rightarrow f$$ is not monotonic
C. $$h(x) =x^2+4x-5\Rightarrow h'(x)=2x+4>0\forall x>0\Rightarrow f$$ is monotonically increasing
D. $$f(x) =e^{-x}\Rightarrow f'(x) =-e^{-x}<0\forall x\le 0\Rightarrow f$$ is monotonically decreasing
$$f(x)=x^3+3x^2+4x+b \sin x+c \cos x, \forall x\in R$$ is a one-one function, then the value of $$b^2+c^2$$ is
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$$\geq 1$$
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$$\geq 2$$
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$$\leq 1$$
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none of these
Explanation
Here, $$f(x)=x^3+3x^2+4x+b \sin x+c \cos x$$
$$f'(x)=3x^2+6x+4+b \cos x-c \sin x$$
Now, for $$f(x)$$ to be one-one, the only possibility is
$$f'(x)\geq 0, \forall x\in R$$
ie, $$3x^2+6x+4+b \cos x-c \sin x\geq 0, \forall x\in R$$
ie, $$3x^2+6x+4 \geq c \sin x-b \cos x, \forall x\in R$$
As we are approximating lower bound for the function by a number instead of a function, hence we chose the number.
ie, $$3x^2+6x+4 \geq \sqrt {b^2+c^2}, \forall x\in R$$
ie, $$\sqrt {b^2+c^2} \leq 3(x^2+2x+1)+1,\forall x\in R$$
$$\sqrt {b^2+c^2} \leq 3(x+1)^2+1, \forall x\in R$$
$$\sqrt {b^2+c^2} \leq 1, \forall x \in R$$
$$\Rightarrow b^2+c^2 \leq 1, \forall x \in R$$
Hence, (c) is the correct answer.
Find the domain of the function $$f(x) = \dfrac {\sqrt {x - 1}}{x}$$
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All real numbers except for $$0$$
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All real numbers greater than or equal to $$1$$
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All real numbers less than or equal to $$1$$
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All real numbers greater than or equal to $$-1$$ but less than or equal to $$1$$
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All real numbers less than or equal to $$-1$$
Explanation
$$f(x) = \dfrac {\sqrt {x-1}}{x}$$, for $$f(x)$$ to exist, the expression in the square root must be non - negative and the expression in denominator should not be equal to $$0$$.
So, $$x-1 \ge 0$$ and $$x \ne 0$$
$$x \ge 1$$ and $$x \ne 0$$, which implies $$x \ge 1$$.
If $$f(x)=2x+|x|, g(x)=\dfrac {1}{3}(2x-|x|)$$ and $$h(x)=f(g(x))$$, then domain of $$\sin^{-1}\underset {\text {n times}}{\underbrace {(h(h(h(h.....h(x).....))))}}$$ is
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$$[-1, 1]$$
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$$\left [-1, -\dfrac {1}{2}\right ]\cup \left [\dfrac {1}{2}, 1\right ]$$
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$$\left [-1, -\dfrac {1}{2}\right ]$$
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$$\left [\dfrac {1}{2}, 1\right ]$$
Explanation
Since, $$f(x)=\left\{\begin{matrix}2x+x, & x\geq 0\\ 2x-x, & x < 0\end{matrix}\right.=\left\{\begin{matrix}3x, & x\geq 0\\ x, & x < 0\end{matrix}\right.$$
and $$g(x)=\dfrac {1}{3}\left\{\begin{matrix}2x-x, & x\geq 0\\ 2x+x, & x < 0\end{matrix}\right.=\left\{\begin{matrix}\dfrac {x}{3}, & x\geq 0\\ x, & x < 0\end{matrix}\right.$$
$$\therefore f(g(x))=\left\{\begin{matrix}3\left (\dfrac {x}{3} \right ) & x\geq0\\ x & x < 0\end{matrix}\right.$$
$$\Rightarrow f(g(x))=x, \forall x\in R$$
$$\therefore h(x)=x$$
$$\Rightarrow \sin^{-1} (h(h(h....h(x)....)))=\sin^{-1}x$$
$$\therefore$$ Domain of $$\sin^{-1} (h(h(h(h.....h(x)....))))$$ is $$[-1, 1]$$
Hence, A is the correct answer.
Let $$f(x)=max\left\{1+\sin x,1,1-\cos x \right\}, x\in \left [ 0,2\pi \right ]$$ and $$g(x)=max\left\{ 1,\left | x-1 \right |\right\},x\in R$$ , then
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$$g(f(0))=1$$
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$$g(f(1))=1$$
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$$f(g(1))=1$$
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$$f(g(0))=\sin 1$$
Explanation
$$f(x)=(max [1+sinx,1,1-cos(x)]$$ for $$x\epsilon[0,2\pi]$$
$$f(0)$$
$$=(max(1+0,1,0))$$
$$=1$$
$$f(1)$$
$$=(max(1+sin(1)),1,1-cos1)$$
$$=1+sin(1)$$
$$g(x)=max(1,|x-1|)$$
Hence,
$$g(f(0))$$
$$=g(1)$$
$$=max(1,0)$$
$$=1$$
And, $$g(f(1))$$
$$=max(1,|1+sin(1)-1|)$$
$$=max(1,|sin(1)|)$$
$$=1$$
$$g(0) = 1 $$
$$f(g(0)) = 1+sin 1$$
The domain of function $$\displaystyle f(x)=\sqrt{x-\sqrt{1-x^{2}}}$$ is
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$$\displaystyle \left [ -1,-\frac{1}{\sqrt{2}} \right ]\cup \left [ \frac{1}{\sqrt{2}},1 \right ]$$
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$$\displaystyle [-1,1]$$
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$$\displaystyle \left ( -\infty,-\frac{1}{2} \right ]\cup \left [ \frac{1}{\sqrt{2}},+\infty \right )$$
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$$\displaystyle \left [ \frac{1}{\sqrt{2}},1 \right ]$$
Explanation
$$\displaystyle f(x)=\sqrt{x-\sqrt{1-x^{2}}}$$
$$ x-\sqrt{1-x^{2}} \geq 0 $$ and $$ 1 -x^{2}\geq0 $$
$$\Rightarrow x\geq\sqrt{1-x^{2}}$$ and $$ -1\leq x \leq1$$ ... (i)
Now, $$ x^2\geq(\sqrt{1-x^{2}})^2$$, for positive $$x$$
$$x\leq-\dfrac{1}{\sqrt{2}} $$ or $$ x\geq\dfrac{1}{\sqrt{2}}$$ ... (ii)
From (i) and (ii), we get
$$x\in\displaystyle \left [ -1,-\frac{1}{\sqrt{2}} \right ]\cup \left [ \frac{1}{\sqrt{2}},1 \right ]$$
Hence, option 'D' is correct.
The domain of the function $$\displaystyle f(x)=\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}$$ is
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$$\displaystyle \left \{ x|x< 1 \right ) $$
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$$\displaystyle \left \{ x|x> -1 \right \}$$
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$$[0,1]$$
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$$[-1,1]$$
Explanation
For real range of $$f(x)$$
$$1-x^{2}\geq 0$$
$$x^2\leq 1$$
$$|x|\leq 1$$
$$-1 \leq x \leq 1$$
$$x\in [-1,1]$$
Hence the domain for the above function is $$[-1,1]$$
The function $$f$$ is one to one and the sum of all the intercepts of the graph is $$5$$. The sum of all the intercept of the graph $$\displaystyle y = f^{-1} \left ( x \right )$$ is
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$$5$$
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$$\dfrac15$$
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$$\dfrac25$$
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$$-5$$
Explanation
Since the function $$f$$ is one-one there exist only one $$x$$-intercept and only one $$y$$-intercept for the function.
Let $$a$$ be the $$y$$-intercept of the function $$f$$. Hence $$f\left( 0 \right) = a$$, but then we have $${f^{ - 1}}\left( a \right) = 0$$.
Therefore $$a$$ is an $$x$$-intercept of the function $${f^{ - 1}}$$.
Similarly the $$x$$-intercept of the function$$f$$ is $$y$$-intercept of the function $${f^{ - 1}}$$, say $$b$$.
Hence the sum of the intercepts of the function $$f$$ is same as the sum of the intercepts of the function $${f^{ - 1}}$$ which is equal to 5.
Let $$f\left( x \right) =\left\{ \begin{matrix} 1+|x|,\; x<-1 \\ \left[ x \right] ,\; x\ge -1 \end{matrix} \right. $$ where $$[\cdot]$$ denotes the greatest integer function. Then $$\displaystyle f\left \{f(-2.3) \right\}$$ is equal to
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$$4$$
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$$2$$
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$$-3$$
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$$3$$
Explanation
$$f\left( x \right) =\left\{ \begin{matrix} 1+|x|,\; x<-1 \\ \left[ x \right] ,\; x\ge -1 \end{matrix} \right. $$
.
$$f\left( -2.3 \right) =1+\left| -2.3 \right| =3.3\dots ( \because x < -1$$)
.
hence, $$f\left( f\left( -2.3 \right) \right) =f\left( 3.3 \right) =\left[ 3.3 \right] =3$$
$$f\left( f\left( -2.3 \right) \right)=3$$
The largest set of real values of $$x$$ for which $$\displaystyle f(x)=\sqrt{(x+2)(5-x)}-\frac{1}{\sqrt{x^{2}-4}}$$ is a real function is
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$$\displaystyle [1,2)\cup (2,5]$$
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$$\displaystyle (2,5]$$
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$$\displaystyle [3,4]$$
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$$none\:of\:these$$
Explanation
$$\sqrt { \left( x+2 \right) \left( 5-x \right) } $$ is defined for
$$\left( x+2 \right) \left( 5-x \right) \ge 0\\ \Rightarrow \left( x+2 \right) \left( x-5 \right) \le 0\\ \Rightarrow x\in \left[ -2,5 \right] .........(1)$$
.
Similarly, $$\dfrac { 1 }{ \sqrt { { x }^{ 2 }-4 } } $$ is defined for
$${ x }^{ 2 }-4>0\\ x\in \left( -\infty ,-2 \right) \bigcup \left( 2,\infty \right) ......(2)$$
.
From $$(1)$$ and $$(2)$$, the domain of $$f(x)$$ is $$x \in \left( 2, 5 \right] $$
If $$\displaystyle f \left ( x \right ) = px + q$$ and $$\displaystyle f \left ( f\left ( f\left ( x \right ) \right ) \right ) = 8x + 21$$, where $$p$$ and $$q$$ are real numbers, the $$ p + q$$ equals
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$$3$$
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$$5$$
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$$7$$
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$$11$$
Explanation
$$f(x)=px+q$$
$$f(f(x))=p(px+q)+q$$
$$=p^2x+q(p+1)$$
$$f(f(f(x)))=p(p^2x+q(p+1))+q$$
$$=p^{3}x+q(p(p+1)+1)$$
$$=8x+21$$
By comparing coefficients, we get
$$p^{3}=8$$
$$p=2$$
And $$q(2(3)+1)=21$$
$$7q=21$$
$$q=3$$
Hence
$$f(x)=2x+3$$
Therefore, $$p+q=2+3=5$$
The value of $$(a + b)$$ is equal to
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$$-2$$
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$$-1$$
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$$0$$
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$$1$$
Explanation
$$x^{2}-2x-1$$ is an upwards facing parabola. Therefore, it has a point of minimum $$=-\dfrac{b}{2a}=\dfrac12\times$$ Sum of roots
After the point of minimum the curve slopes up and the function no longer remain one-one. Hence the largest interval for it to be injecvtive is upto the point of minimum.
The roots of the quadratic are,
$$\rightarrow x=1+\sqrt{2} $$ and $$x=1-\sqrt{2}$$
So $$a=\dfrac{1+\sqrt{2}+1-\sqrt{2}}{2}$$
$$=1$$
$$f(1)=(1-1)^{2}-2$$
$$=-2$$
$$=b$$
Hence $$a+b$$
$$=1-2$$
$$=-1$$
$$K(x)$$ is a function such that $$K(f(x))=a+b+c+d$$,
Where,
$$a=\begin{cases}
0 & \text{ if f(x) is even} \\
-1 & \text{ if f(x) is odd} \\
2 & \text{ if f(x) is neither even nor odd}
\end{cases}$$
$$b=\begin{cases}
3 & \text{ if f(x) is periodic} \\
4 & \text{ if f(x) is aperiodic}
\end{cases}$$
$$c=\begin{cases}
5 & \text{ if f(x) is one one} \\
6 & \text{ if f(x) is many one}
\end{cases}$$
$$d=\begin{cases}
7 & \text{ if f(x) is onto} \\
8 & \text{ if f(x) is into}
\end{cases}$$
$$h:R\rightarrow R,h(x)=\left ( \displaystyle \frac{e^{2x}+e^{x}+1}{e^{2x}-e^{x}+1} \right )$$
On the basis of above information, answer the following questions.
$$K(\phi(x)) $$
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$$15$$
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$$16$$
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$$17$$
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$$18$$
Explanation
$$\phi(x)\rightarrow \left(\dfrac{-\pi}{2},\dfrac{\pi}{2}\right)$$
Now we know that for the given domain, $$\tan(x)$$ is one-one, periodic, odd and an into function.
Hence $$f(\phi(x))$$
$$=-1+3+5+8$$
$$=16-1$$
$$=15$$
Let $$f:{x, y, z}\rightarrow (a, b, c)$$ be a one-one function. It is known that only one of the following statements is true:
(i) $$f(x)\neq b$$
(ii)$$f(y)=b$$
(iii)$$f(z)\neq a$$
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$$f=\{(x, a), (y, b), (z, c)\}$$
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$$f=\{(x, b), (y, a), (z, c)\}$$
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$$f=\{(x, b), (y, c), (z, c)\}$$
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$$f=\{(x, b), (y, c), (z, a)\}$$
Explanation
When (i) is true, then $$f(x) \neq b, f(y) \neq b , f(z) = a $$
$$\Rightarrow$$ Two ordered pair function is possible $$ f(x) = a, f(y) = c, f(z) = a$$ or $$f(x) = c, f(y) = a, f(z) = a$$
But given $$f$$ is one-one and only one such function is possible. Hence (i) can't be true.
When (ii) is true, then $$f(y) = b, f(z) =a , f(x) = b$$. This is also not possible.
Clearly if (iii) is true then it is satisfying every condition.
Hence ordered pair of $$f$$ is $$\{(x,a), (y,b), (z,c)\}$$
Let $$\displaystyle f(x)=\begin{cases}x^{2} & \mbox{if} \quad0< x< 2\\2x-3 & \mbox{if} \quad2\leq x< 3 \\ x+2 & \mbox{if}\quad x\geq 3\end{cases}$$.
Then
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$$\displaystyle f\left \{ f\left ( f\left ( \frac{3}{2} \right ) \right ) \right \}=f\left ( \frac{3}{2} \right )$$
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$$\displaystyle 1+f\left \{ f\left ( f\left ( \frac{5}{2} \right ) \right ) \right \}=f\left ( \frac{5}{2} \right )$$
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$$\displaystyle f\left \{ f(0) \right \}=f\left ( 1 \right )=1$$
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none of these
Explanation
consider option A)
$$\displaystyle f\left( {\frac{3}{2}} \right) = \frac{9}{4}$$ as $$\displaystyle \frac{3}{2} < 2$$
and $$\displaystyle f\left( {\frac{9}{4}} \right) =2 \times \frac{9}{4} - 3 = \frac{3}{2}$$ as $$\displaystyle \frac{9}{4} > 2$$.
Therefore $$\displaystyle f\left( {f\left( {f\left( {\frac{3}{2}} \right)} \right)} \right) = f\left( {\frac{3}{2}} \right)$$.
Hence option A is correct answer.
Consider option B)
$$\displaystyle f\left( {\frac{5}{2}} \right) = 2 \times \frac{5}{2} - 3 = 2$$ as $$\displaystyle \frac{5}{2} > 2$$
and $$\displaystyle f\left( 2 \right) = 2 \times 2 - 3 = 1$$.
Also, $$\displaystyle f\left( 1 \right) = 1$$
Hence, $$\displaystyle 1+f\left \{ f\left ( f\left ( \frac{5}{2} \right ) \right ) \right \}= 2 =f\left( {\frac{5}{2}} \right)$$
Hence option B is also a correct answer .
Since the function is not defined for $$x=0$$ option C can be eliminated.
If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ are given by $$f(x)=|x|$$ and $$g(x)=[x]$$ for each $$x\in R,$$ then $$\left\{ x\in R:g\left( f\left( x \right) \right) \le f\left( g\left( x \right) \right) \right\} =$$
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$$Z\cup \left( -\infty ,0 \right) $$
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$$\left( -\infty ,0 \right) $$
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$$Z$$
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$$R$$
Explanation
$$g\left( f\left( x \right) \right) \le f\left( g\left( x \right) \right) \Rightarrow g\left( \left| x \right| \right) \le f\left( \left[ x \right] \right) \Rightarrow \left[ \left| x \right| \right] \le \left[ \left| x \right| \right] $$
This is true for each $$x\in R$$
Let $$f$$ and $$g$$ be increasing and decreasing functions respectively from $$\displaystyle \left ( 0,\infty \right )$$ to $$\left ( 0,\infty \right )$$ and let $$h\left ( x \right )=f\left [ g\left ( x \right ) \right ]$$. If $$h\left ( 0 \right )=0$$ then $$ h\left ( x \right )-h\left ( 1 \right )$$ is
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always zero
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always negative
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always positive
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strictly increasing
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None of these
Explanation
Let $$\displaystyle F\left ( x \right )=h\left ( x \right )-h\left ( 1 \right )=f\left ( g\left ( x \right ) \right )-h\left ( 1 \right )$$ $$\displaystyle F'\left ( x \right )=f'\left ( g\left ( x \right ) \right ).g'\left ( x \right )=\left ( + \right )\left ( - \right )=-ive.$$ (As f is increasing function $$\displaystyle f'\left ( g\left ( x \right ) \right )$$ is +ive and as g is decreasing function $$\displaystyle g'\left ( x \right )$$ is-ive.)
Since F'(x) is-ive therefore F(x)i.e.h(x)-h(1) is decreasing function.
Now split the interval $$\displaystyle I=\left [ 0,\infty \right ]$$ into two intervals $$\displaystyle I_{1},0\leq x< 1and I_{2}, 1\leq x< \infty ,$$
Apply the definition of decreasing function on $$\displaystyle h\left ( x \right )-h\left ( 1 \right ):$$ on $$\displaystyle I_{1},0\leq x< 1,\underset{\left ( Big \right )}{h\left ( x \right )}-\underset{\left ( Less \right )}{h\left ( 1 \right )}=+ive$$ On $$\displaystyle I_{2},1\leq x< \infty ,\underset{\left ( Less \right )}{h\left ( x \right )}-\underset{\left ( Big \right )}{h\left ( 1 \right )}=-ive$$ Hence for $$\displaystyle I,h\left ( x \right )-h\left ( 1 \right )$$ is neither always zero nor always +ive nor always-ive,nor strictly increasing throughout.
Hence (v) is the correct answer.
If $$\displaystyle f(x)=27x^{3}+\frac{1}{x^{3}}$$ and $$\alpha,\beta$$ are the roots of $$\displaystyle 3x+\frac{1}{x}=2$$ is
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$$f(\alpha)=f(\beta)$$
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$$f(\alpha)=10$$
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$$f(\beta)=-10$$
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none of these
Explanation
$$3x+\dfrac{1}{x}=2$$
Cubing the above equation, we get
$$ \left(3x+\dfrac{1}{x}\right)^3=2^3$$
$$\therefore 27x^3+\dfrac{1}{x^3}+3(3x)\left(\dfrac{1}{x}\right)\left(3x+\dfrac{1}{x}\right)=8$$
$$\therefore 27x^3+\dfrac{1}{x^3}+9\left(3x+\dfrac{1}{x}\right)=8$$
$$\therefore 27x^3+\dfrac{1}{x^3}+9(2)=8$$
$$\therefore 27x^3+\dfrac{1}{x^3}=-10$$
$$\alpha$$ and $$\beta$$ are roots of above equation.
$$\therefore 27\alpha^3+\dfrac{1}{\alpha^3}=-10$$ ...(1)
and $$27\beta^3+\dfrac{1}{\beta^3}=-10$$ ...(2)
$$f(x)=27x^3+\dfrac{1}{x^3}$$
$$f(\alpha)=27\alpha^3+\dfrac{1}{\alpha^3}$$
$$\implies f(\alpha) =-10$$ ...(from 1)
Similarly, $$f(\beta)=27\beta^3+\dfrac{1}{\beta^3}$$
$$\implies f(\beta)=-10$$ ...(from 2)
$$\therefore f(\alpha)=f(\beta)=-10$$
So, the correct options are option (A) and (C)
The value of $$x$$ satisfying the equation $$\displaystyle \left | x-1 \right |^{\log_{3}x^{2}-2\log_{9}x}= (x-1)^7$$ is
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$$3^4$$
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$$3^5$$
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$$3^6$$
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$$3^7$$
Explanation
$$\displaystyle (i) \log_{a}b$$ hold good if $$\displaystyle a> 0,a\neq 1,a> 1$$
i.e. $$a-1> 0$$
$$\therefore \left | a-1 \right |= a-1$$
$$(ii) a^{b}> 0\,\,\forall\, b\in R $$
Now from given $$\displaystyle \left | x-1 \right |^{\log_{3}x^{2}-2\log_{9}x}= (x-1)^{7}$$
$$\displaystyle \Rightarrow \left ( x-1 \right )^{\log_{3}x^{2}-2\log_{9}x}= (x-1)^{7}$$
$$\displaystyle\Rightarrow 2\log_{3}x-2\log^{x}{9}= 7$$ taking log at base $$(x -1)$$ bothsides. $$log_{x}{3}=7,x=3^7$$
The domain of $$\displaystyle f(x)=\frac{1}{\sqrt{|\cos\:x|+\cos\:x}}$$ is
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$$[-2n\pi,2n\pi]$$
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$$(2n\pi,\overline{2n+1}\pi)$$
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$$\displaystyle \left ( \frac{(4n+1)\pi}{2} ,\frac{(4n+3)\pi}{2}\right )$$
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$$\displaystyle \left ( \frac{(4n-1)\pi}{2} ,\frac{(4n+1)\pi}{2}\right )$$
If $$f(x)=2x^3$$ and $$g(x)=3x$$, calculate the value of $$g(f(-2))-f(g(2))$$.
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$$-480$$
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$$-384$$
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$$0$$
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$$384$$
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$$480$$
Explanation
$$f(-2) = 2 \times { (-2) }^{ 3 } = -16$$
$$g(2) = 3 \times 2 =6$$
$$g(f(-2)) = g(-16) = 3 \times -16 = -48$$
$$f(g(2)) = f(6) = 2 \times { (6) }^{ 3 } = 432$$
$$g(f(-2)) - f(g(2)) = -48-432 = -480$$
Find the
maximum value of $$g(f(x))$$ if:
$$f(x) = x + 4$$ and
$$g(x) = 6 - x^{2}$$
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$$-6$$
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$$-4$$
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$$2$$
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$$4$$
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$$6$$
Explanation
Given, $$f(x)=x+4, g(x)=6-x^2$$
$$\therefore g(f(x)) = g(x+4) = 6-{(x+4)}^{2}$$
The minimum value of $${(x+4)}^{2}$$ is $$0$$ which occurs at $$x=-4$$.
So, the maximum value of $$g(f(x))$$ is $$6-0 = 6$$.
Let $$\displaystyle f\left ( x \right )=\frac{ax^{2}+2x+1}{2x^{2}-2x+1}$$, the value of $$a$$ for which $$\displaystyle f:R\rightarrow \left [ -1,2 \right ]$$ is onto , is
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$$\displaystyle \left [ 2,5 \right ]$$
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$$\displaystyle \left [ -5,-2 \right ]$$
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$$\displaystyle \left [ 0,5 \right ]$$
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None of these.
Explanation
Given $$\displaystyle f\left ( x \right )=\frac{ax^{2}+2x+1}{2x^{2}-2x+1}$$
Since, $$\displaystyle f:R\rightarrow \left [ -1,2 \right ]$$ is onto
$$R(f)=Co-domain [-1,2]$$
$$-1 \le \dfrac{ax^{2}+2x+1}{2x^{2}-2x+1}\le 2$$
$$\Rightarrow -(2x^{ 2 }-2x+1)\le ax^{ 2 }+2x+1\le 2(2x^{ 2 }-2x+1)$$
$$\Rightarrow -2x^{ 2 }+2x-1\le ax^{ 2 }+2x+1\le 4x^{ 2 }-4x+2$$ ....(1)
$$\Rightarrow -2x^{ 2 }+2x-1\le ax^{ 2 }+2x+1$$
$$\Rightarrow (a+2)x^2+2\ge 0$$
So for all $$x\in R$$ , $$a+2\ge 0$$
$$\Rightarrow a\ge -2$$ .....(2)
From the inequality (1), it follows that
$$ax^{ 2 }+2x+1\le 4x^{ 2 }-4x+2$$
$$\Rightarrow (a-4)x^2+6x-1 \le 0$$
$$\Rightarrow a-4 < 0$$ and $$D\le 0$$
$$\Rightarrow a<4 $$ and $$36+4a-16 \le 0$$
$$\Rightarrow a<4 $$ and $$a\le -5$$ ....(3)
From (2) and (3), we get
$$a\in (-\infty,-5]\cup [-2,4)$$
The total number of injective mappings from a set with $$m$$ elements to a set with $$n$$ elements, $$m \leq n $$ is
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$$\displaystyle m^{n}$$
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$$\displaystyle n^{m}$$
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$$\displaystyle \frac{n!}{(n-m)!}$$
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$$n!$$
Explanation
$$a_{1} \in $$A can have $$n$$ images in $$B$$, but the element $$a_{2}$$ will have only $$(n-1)$$ images as the mappings are to be one-one (injective).
Similarly the elements $$a_{3}$$ will have $$(n-2)$$ images.
Hence the total number of mappings will be,
$$n(n-1)(n-2)...(n-\overline{m-1})=n(n-1)(n-2)...(n-m+1)$$
Multiply above and below by $$(n-m)(n-m-1)...3.2.1$$
$$\therefore$$ Required numbers is $$\displaystyle \frac{n!}{(n-m)!} $$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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