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CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 4 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Functions
Quiz 4
If
$$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},g(x)=\begin{cases} x\quad \quad if\quad x\leq 1 \\ 2-x\quad if\quad x>1 \end{cases}$$
and $$x\, \in\, (1, 2)$$, then $$g(f(x))$$ is equal to
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$$x^{2}\, +\, 3$$
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$$x^{2}\, -\, 3$$
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$$5\, -\, x^{2}$$
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$$1 - x$$
Explanation
$$x\, \epsilon\, (1, 2)$$
Hence for x >1, $$f(x) = 5-x^2$$and $$f(x) \epsilon (1,4) $$
$$f(x) >1$$
$$g(f(x)) = 2 - f(x)= 2-5+x^2 = x^2-3$$
Which of the functions defined below are NOT one-one function(s)
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$$f(x)\, =\, 5(x^{2}\, +\, 4),\, (x\, \in\, R)$$
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$$g(x)\, =\, 2x\, +\, \dfrac1x$$
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$$h(x)\, =\, ln(x^{2}\, +\, x\, +\, 1)\,, (x\, \in\, R)$$
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$$f(x)\, =\, e^{-x}$$
If $$g(x) = 2x + 1$$ and $$h(x) = 4x^{2} + 4x + 7$$, find a function $$f$$ such that $$f o g = h$$
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$$f(x) = x^{3} - 6$$
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$$f(x) = x^{2} + 6$$
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$$f(x) = x^{2} - 6$$
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$$f(x) = (2x+1)^2 + 6$$
Explanation
$$f(g(x)) = h = 4x^2+4x+7=(2x+1)^2+6=(g(x))^2+6$$
$$\Rightarrow f(x) =x^2+6$$
Which of the following are two distinct linear functions which map the interval $$[-1, 1]$$ onto $$[0, 2]$$
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$$f(x) = 1 + x$$ or $$1 - x$$
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$$f(x) = 1 + 2x$$ or $$1 - x$$
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$$f(x) = 1 + x$$ or $$1 - 2x$$
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$$f(x) = 1 + x$$ or $$2 - x$$
Explanation
Out of two linear functions one will be increasing and other will be decreasing.
Let $$f(x) = ax+b$$
For increasing: $$f(-1)=0=> b-a=0\Rightarrow b=a$$ and $$f(1) = 2\Rightarrow a+b=2\Rightarrow a=b=1\therefore f(x) = 1+x$$
For decreasing: $$f(-1)=2=> b-a=2$$ and $$f(1) = 0\Rightarrow a+b=2\Rightarrow b=1, a=-1\therefore f(x) = 1-x$$
Which of the following is an onto function
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$$f\, :\, [0,\, \pi]\, \rightarrow\, [-\, 1\, 1], f(x)\, =\, \sin\, x$$
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$$f\, :\, [0, \pi]\, \rightarrow\, [-1, 1],\, f(x)\, =\, \cos\, x$$
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$$f\, :\, R\, \rightarrow\, R,\, f(x)\, =\, e^{x}$$
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$$f\, :\, Q\, \rightarrow\, R, f(x)\, =\, x^{3}$$
Explanation
A function is called onto iff, Range$$=$$ co-domain.
A.
In $$[0,\pi]$$, Range of $$\sin x$$ is $$[0,1]\Rightarrow$$ Not onto.
B.
In $$[0,\pi]$$, Range of $$\cos x$$ is $$[-1,1] =$$co-domain $$\Rightarrow $$onto function.
C.
Range of $$e^x$$ is $$(0, \infty)\neq$$ co-domain.
D.
Range of
$$x^3$$
is not R, since $$x\in Q$$
If $$ f : R \rightarrow R, f(x) = (x + 1)^2$$ and $$g : R \rightarrow R, g(x) = x^2 + 1 $$ then $$(fog)(3)$$ is equal to
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$$121$$
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$$144$$
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$$112$$
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$$11$$
Explanation
Given,
$$ f : R \rightarrow R, f(x) = (x + 1)^2$$ and $$g : R \rightarrow R,g(x) = x^2 + 1 $$
$$g(3)=3^2+1=10$$
$$f(10)=(10+1)^2=121$$
$$\therefore fog(3)=f\left(g(3)\right)=f(10)=121$$
Hence, option A.
If $$f(x) = \log x$$, $$g(x) = x^3$$, then $$f[g(a)] + f[g(b)]$$ equals
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$$f[g(a) + g(b)]$$
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$$3f(ab)$$
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$$g[f(ab)]$$
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$$g[f(a) + f(b)]$$
Explanation
We have $$f(x) = \log x$$
$$g(x) = x^3$$
$$g(a) = a^3 $$
$$f(g(a)) = \log (a^3) $$
$$f(g(a)) =3 \log a $$
Similarly,
$$f(g(b)) = 3 \log b $$
$$\Rightarrow f(g(a)) + f(g(b)) = 3 \log a + 3 \log b $$
$$\Rightarrow f(g(a)) + f(g(b)) = 3 \log ab $$
$$\Rightarrow f(g(a)) + f(g(b)) = 3 f(ab)$$
If $$f(x) = x^3 $$ and $$g(x) = sin2x$$, then
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$$g[f(1)] = 1$$
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$$f(g(\pi/12) = 1/8$$
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$$g{f(2)} = \sin 2$$
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none of these
Explanation
Given,
$$ f(x)=x^3 $$ and $$g(x)=\sin 2x$$
Clearly,
$$ f(1)=1 \Rightarrow g[f(1)]=\sin 2 $$
$$ g(\cfrac{\pi}{12}) =\dfrac{1} {2} \Rightarrow f[g(\dfrac{\pi}{12})]=\dfrac{1}{8}$$
and,$$ f(2)=8 \Rightarrow g[f(2)]= \sin 16 $$
So, the correct option is (B).
If $$f(x) = (a x^n)^{1/n},$$ where $$\ n \in N$$, then $$f\{f(x)\}$$ equals
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$$0$$
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$$x$$
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$$x^n$$
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none of these
Explanation
We have, $$f(x) = (a x^n)^{1/n}=a^{1/n}x=kx$$ where $$k = a^{1/n}$$
$$\therefore f\{f(x)\} = kf(x) = k^2x =a^{2/n} x$$
The domain of the function $$\ln (x-1)$$ is.
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$$[0, 1)$$
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$$R$$
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$$R -Z$$
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$$(1, \infty )$$
Explanation
$$\log(x-1)$$
where $$ (x-1)>0$$
$$x>1$$
Domain of $$\log(x-1)$$ $$=(1,\infty)$$
If $$f(x) =\ln {\displaystyle \frac { 1+x }{ 1-x } } $$ and $$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$$, then $$f[g(x)]$$ equals.
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$$f(x)$$
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$$[f(x)]^3$$
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$$3f(x)$$
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$${f(x)}^2$$
Explanation
Given $$f(x)=\ln {\displaystyle \frac { 1+x }{ 1-x } } $$ and $$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$$
then $$f[g(x)]=\ln {\displaystyle \frac { 1+\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } }{ 1-\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } } } $$
$$=\ln {\displaystyle \frac { 1+3x^{ 2 }+3x+x^{ 3 } }{ 1+3x^{ 2 }-3x-x^{ 3 } } } $$
$$=\ln\left[ {\displaystyle \frac { 1+x }{ 1-x } }\right]^3 $$
$$=3\ln {\displaystyle \frac { 1+x }{ 1-x } } $$
$$\Rightarrow f[g(x)]=3f(x)$$
Hence, option C.
If $$f(x) = \left\{\begin{matrix} 1&x \in Q \\ 0 &x \notin Q\end{matrix}\right.$$ then $$fof(\sqrt 3 )$$ is equal to
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$$0$$
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$$1$$
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$$\sqrt 3$$
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none of these
Explanation
We know $$\sqrt{3}$$ is an irrational number
Thus $$f(\sqrt{3}) =0$$ and $$0$$ is a rational number
Hence $$fof(\sqrt 3 )= f(0) = 1$$
Let $$f(x) = e^{3x}, g(x) = \log_ex, x > 0$$, then $$fog (x)$$ is
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$$3x$$
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$$x^3$$
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$$\log_{10}3x$$
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$$\log3x$$
Explanation
Given $$f(x) = e^{3x}, g(x) = \log_ex, x > 0$$
$$fog(x)=f(\log_ex)=e^{3\log_ex}=x^3$$
Hence, option B.
$$f(x)\, >\, x;\, \forall\, x\, \epsilon\, R.$$ The equation $$f (f(x)) -x = 0$$ has
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Atleast one real root
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More than one real root
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No real root if f(x) is a polynomial & one real root if f(x) is not a polynomial
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No real root at all
Explanation
Given $$f(x) > x $$ $$\forall x \epsilon R$$
Hence, $$ f(x)-x$$ has no real roots, because $$f(x)-x >0$$
Since, $$x \epsilon R \implies f(x) \epsilon R$$
$$f(f(x))>x$$
$$\implies f(f(x))-x>0$$
Hence it will have no real root.
If functions $$f\left ( x \right )$$ and $$g\left ( x \right )$$ are defined on $$R\rightarrow R$$ such that
$$f(x)=x+3, x$$ $$\in $$ rational
$$ =4x, x$$ $$\in $$ irrational
$$g(x)=x+\sqrt{5}$$, x$$\in $$ irrational
$$ =-x, x$$ $$\in $$ rational
then $$\left ( f-g \right )\left ( x \right )$$ is
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one-one & onto
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neither one-one nor onto
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one-one but not onto
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onto but not one-one
Explanation
$$f\left( x \right) =x+3, x\epsilon$$ rational
$$=4x, x\epsilon$$ irrational
$$g\left( x \right) =x+\sqrt { 5 } , x \epsilon$$ irrational
$$= -x,x\epsilon$$ rational
$$(f-g)(x)=2x+3, x\ \epsilon$$ rational
$$+3x-\sqrt { 5\quad } x\quad \epsilon$$ irrational
For one-one;
We know that one one function is a $${ f }^{ \underline { n } }$$ for which every element of the range of the function corresponds to exactly one element of the domain.
But in this case this is not true as for all rational nos. the$$ { f }^{ \underline { n } }$$ is one one but for irrational $${ f}^{ \underline { n } }$$, every elements of the range does not corresponds to exactly one element of the domain.
$$(f-g)(x)\neq (f-g)({ x }^{ \prime })$$ when $$x \epsilon$$ rational and
$$x \epsilon$$ irrational
for onto:
The $${ f }^{ \underline { n } }(f-g)(x)$$ does not cover the whole range of the function.
The domain of the function, $$\displaystyle y=f(x)=\sqrt {\log_{10} \left ( \frac {5x-x^2}{4} \right )}$$ is
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$$[1, 4]$$
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$$(1, 4)$$
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$$[1, 4)$$
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$$(1, 4]$$
Explanation
Given,
$$\displaystyle y=f(x)=\sqrt {\log_{10} \left ( \frac {5x-x^2}{4} \right )}$$
For given function to be defined,
$$\displaystyle \log_{10} \left ( \frac {5x-x^2}{4} \right )\geq 0 \Rightarrow \left ( \frac {5x-x^2}{4} \right ) \geq 10^0=1$$
$$\Rightarrow 5x-x^2\geq 4\Rightarrow x^2-5x+4\leq 0$$
$$\Rightarrow (x-1)(x-4) \leq 0\Rightarrow x \in [1,4]$$
Let $$f(x) =\frac {ax+b} {cx+d}$$. Then fof(x) = x provided that.
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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
$$f(f(x))=\dfrac{af(x) + b}{cf(x)+d}$$
$$f(f(x))=\dfrac{a\dfrac{ax+b}{cx+d} + b}{c\dfrac{ax+b}{cx+d}+d}$$
$$f(f(x))=\dfrac{a^2x + ab+bcx+bd}{acx+bc+d^2+dcx}=x$$
$$\Rightarrow a^2x+ab+bcx+bd=acx^2+bcx+d^2x+dcx^2$$
Given $$a=-d$$, the above equation can also be verified
If $$f(x) =\dfrac {1}{1-x}, x \neq 0, 1$$ then the graph of the function $$y = f[f\{f(x)\}]$$ for $$x > 1 $$ is
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a straight line
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a circle
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an ellipse
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a pair of straight lines
Explanation
Given $$f(x) = \cfrac{1}{1-x}$$
$$\displaystyle \Rightarrow f\{f(x)\} = \frac{1}{1-f(x)} = \frac{1}{1-\frac{1}{1-x}}=\frac{1-x}{1-x-1}=1-\frac{1}{x}$$
$$\displaystyle \therefore f[f\{f(x)\}] = \frac{1}{1- f\{f(x)\}}=\frac{1}{1-1+\frac{1}{x}}=x$$, which is a straight line.
Let $$\displaystyle f\left ( x \right )=\frac{3}{2}+\sqrt{x-\frac{3}{4}}$$ be a function and $$g\left ( x \right )$$ be another function such that $$g\left ( f\left ( x \right ) \right )=x,$$ then the value of $$g\left ( 20 \right )$$ will be
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$$333$$
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$$335$$
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$$338$$
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$$343$$
Explanation
$$g\left ( x \right )$$ is inverse of $$f\left ( x \right )$$
$$\Rightarrow $$ $$\displaystyle x=\frac{3}{2}+\sqrt{y-\frac{3}{4}}$$
$$\Rightarrow $$ $$\displaystyle y=\left ( x-\frac{3}{2} \right )^2+\frac{3}{4}=f^{-1}(x)=g(x)$$
$$\therefore $$ $$\displaystyle g\left ( 20 \right )=\left ( 20-\frac{3}{2} \right )^{2}+\frac{3}{4}=343$$
Let $$f : R \rightarrow R, g : R \rightarrow R$$ be two function such that
$$f(x) = 2x- 3, g(x) = x^3 + 5$$
The function $$(fog)^{1}(x)$$ is equal to.
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$$\left ( \dfrac {x+7}{2} \right )^{1/3}$$
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$$\left (x- \dfrac {7}{2} \right )^{1/3}$$
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$$\left ( \dfrac {x-2}{7} \right )^{1/3}$$
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$$\left ( \dfrac {x-7}{2} \right )^{1/3}$$
Explanation
Given $$f(x) = 2x-3$$ and $$g(x) = x^3+ 5$$
$$(fog)(x) = fg(x) = f(x^3+5) = 2(x^3+5) -3 = 2x^3 +7$$
Let $$(fog)(x) = y = 2x^3+7$$
$$y-7 = 2x^3$$
$$x = (\dfrac{y-7}{2})^{\dfrac{1}{3}}$$
$$\therefore (fog)^{-1}(x) = (\dfrac{x-7}{2})^{\dfrac{1}{3}}$$
The domain of the function, $$f(x) = \displaystyle \dfrac {1}{\sqrt{[x]^2-[x]-6}}$$ is
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$$(-\infty , -3] \cup [4, \infty )$$
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$$(\infty , 2) \cup [4, \infty )$$
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$$(\infty , 2) \cup (4, \infty )$$
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none of these
Explanation
Given : $$f(x) = \displaystyle \frac {1}{\sqrt{[x]^2-[x]-6}}$$
$$f(x)$$ is real if
$${\sqrt{[x]^2-[x]-6}}\in R$$
i.e.
$$[x]^{ 2 }-[x]-6>0$$
$$\implies [x]^{ 2 }-3[x]+2[x]-6>0$$
$$\implies [x]([x]-3)+2([x]-3)>0$$
$$\implies ([x]-3)([x]+2)>0$$
$$ \implies \left[ x \right] <-2 $$ and $$\left[ x \right] >3$$
Therefore, $$x \in (-\infty , -3] \cup [4, \infty )$$
The domain of the function f(x)=$$\displaystyle \frac{\sqrt{-\log_{0.3}(x-1)}} {\sqrt{-x^2+2x+8}}$$ is
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$$(1, 4)$$
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$$(2, 4)$$
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$$[2, 4)$$
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none of these
Explanation
Given,
f(x)=$$\displaystyle \frac{\sqrt{-\log_{0.3}(x-1)}} {\sqrt{-x^2+2x+8}}$$
For given function to be defined,
$$-\log_{0.3}(x-1) \geq 0\Rightarrow \log_{10/3}(x-1) \geq 0\Rightarrow x-1\geq 1\Rightarrow x\geq 2$$
and,
$$-x^2+2x+8> 0\Rightarrow x^2-2x-8< 0\Rightarrow (x+2)(x-4) < 0\Rightarrow -2 < x< 4$$
Combining above two we get required domain $$[2,4)$$
If $$f : [0, \Pi ] \rightarrow [-1, 1]$$, f(x) = cosx, then f is.
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one-one
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onto
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one-one onto
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none of these
Explanation
f takes all values from $$[-1,1]$$ while travelling from $$[0,\pi]$$. None of the values are repeated either.
Hence it is one-one and onto.
The domain of the function $$f(x)=\sqrt {1-{\sqrt{1-\sqrt{1-x^2}}}}$$ is .
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$$(\infty , 1) $$
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$$(1, \infty) $$
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$$[0, 1]$$
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$$[-1, 1]$$
Explanation
For given function to be defined,
$$(1-x^2) > 0\Rightarrow x^2-1\leq 0 \Rightarrow x \in [-1,1]$$
If X = {2,3,5,7,11} and Y = {4,6,8,9,10} then find the number of one-one functions from X to Y
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720
0%
120
0%
24
0%
12
Explanation
Both sets X and Y have the same number of elements.
When two sets have same number of $$ n $$ elements, then the number of one to one functions from one set to the other is $$ n! $$
So, number of one to one functions from X to Y $$ = 5 ! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $$
If $$\displaystyle f\left ( x \right )=\left ( 1-x^{3} \right )^{\frac{1}{3}}$$, then find $$fof(x)$$
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$$\dfrac1x$$
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$$x$$
0%
$$x^2$$
0%
$$x^3$$
Explanation
Given, $$\displaystyle f\left ( x \right )=\left ( 1-x^{3} \right )^{\frac{1}{3}}$$
Therefore, $$\displaystyle fof\left ( x \right )=f\left [ f\left ( x \right ) \right ]$$
$$\displaystyle =f\left ( \left ( 1-x^{3} \right )^{\frac{1}{3}} \right )=\left [ 1-\left \{ \left ( 1-x^{3} \right )^{\frac{1}{3}} \right \}^{3} \right ]^{\frac{1}{3}}$$
$$\displaystyle =\left [ 1-\left ( 1-x \right )^{3} \right ]^{\frac{1}{3}}=\left ( x^{3} \right )^{\frac{1}{3}}=x$$
$$\displaystyle f:A\rightarrow B$$ defined by f(x) = 2x+3 and if A = {-2,-1,0,1,2} B = {-1,1,3,5,7} then which type of function is f?
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One-one
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Onto
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Bijection
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Constant
Explanation
As the function is from set A to set B, let us find the range of the elements of A.
$$ f(x) = 2x + 3 $$
$$ => f(-2) = 2(-2) + 3 = -1 $$
$$ => f(-1) = 2(-1) + 3 = 1 $$
$$ => f(0) = 2(0) + 3 = 3 $$
$$ => f(1) = 2(1) + 3 = 5 $$
$$ => f(2) = 2(2) + 3 = 7 $$
We can see that each element of A is mapped to only one and different element of B
Hence, the function is a Bijection function.
The domain of the function $$\displaystyle f\left ( x \right )=\frac{1}{\sqrt{x-3}}$$ is
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$$\displaystyle x< 3$$
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$$\displaystyle x> 3$$
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$$\displaystyle x\geq -3$$
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$$\displaystyle x\leq 3$$
Explanation
The function will have a value or range only when the denominator $$ \neq 0 $$
$$ => x \neq 3 $$ --- (1)
And since the denominator $$ \sqrt {x=3} $$ is under a square root, $$ x - 3 $$ should always be $$ > 0 $$
This means, $$ x > 3 $$ --- (2)
From 1, and 2, we understand that the domain of the function is $$ x > 3 $$
If f(x) + f(1-x) = 10 then the value of $$\displaystyle f\left ( \frac{1}{10} \right )+f\left ( \frac{2}{10} \right )+.........+f\left ( \frac{9}{10} \right )$$
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is 45
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is 50
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is 90
0%
Cannot be determined
Explanation
$$ f(\dfrac {1}{10}) + f(\dfrac {2}{10}) + f(\dfrac {3}{10}) + f(\dfrac {4}{10}) + f(\dfrac {5}{10}) + f(\dfrac {6}{10}) + f(\dfrac {7}{10}) + f(\dfrac {8}{10}) + f(\dfrac {9}{10}) $$
$$ = [f(\dfrac {1}{10}) + f(\dfrac {9}{10}) ] + [ f(\dfrac {2}{10}) + + f(\dfrac {8}{10}) ] + [ f(\dfrac {3}{10}) + + f(\dfrac {7}{10}) ] + [ f(\dfrac {4}{10}) + f(\dfrac {6}{10}) ] + f(\dfrac {5}{10}) ] $$
$$ = 10 + 10 + 10 + 10 + 5 = 45 $$
If $$f (x) = 2x - 1$$ and $$g (x) = 3x + 2$$, then find $$(fog) (x)$$ :
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$$2 (3x + 1)$$
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$$2 ( 3x + 2)$$
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$$3 (2x + 1 )$$
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$$3 ( 3x + 1 )$$
Explanation
$$fog(x)=2(3x+2)-1=6x+4-1=6x+3=3(2x+1)$$
The domain of the function f(x) = log [x-1] is_______
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R - {1,-1}
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R - {1}
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R - {-1}
0%
R - {0}
Explanation
The function will have a value or range only when $$ log [x-1] $$ is defined.
We know that logarithm of all numbers is defined except for $$ log 0 $$ is not defined. This means, $$ x - 1 \neq 0 $$
$$ => x \neq 1 $$
Thus the domain of the given function is $$R - $$ { $$ 1 $$} $$
If f = {(1,3) (2,1) (3,4) (4,2)} and g = {(1,2) (2,3) (3,4) (4,1)} then find n(fog)
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0%
12
0%
16
0%
4
0%
5
Explanation
we can find $$ n(fog) $$ by finding the range of the domain of $$f $$
In $$ (1,3) $$ as the range $$ 3 $$ is mapped to $$ 4$$ in $$g $$, the ordered pair here for $$fog $$ is $$ (1,4) $$
In $$ (2,1) $$ as the range $$ 1 $$ is mapped to $$2 $$ in $$g $$, the ordered pair here for $$fog $$ is $$ (2,2) $$
In $$ (3,4) $$ as the range $$ 4 $$ is mapped to $$ 1 $$ in $$g $$, the ordered pair here for $$fog $$ is $$ (3,1) $$
In $$ (4,2) $$ as the range $$ 2 $$ is mapped to $$ 3 $$ in $$g $$, the ordered pair here for $$fog $$ is $$ (4,3) $$
As we have $$ 4 $$ ordered pairs for $$ fog $$ , we have $$ n(fog) = 4 $$
If $$f(x) = -x^2+1, g(x) = -\sqrt[3]{x}$$ then (gofogofogogog) (x) is.
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an odd function
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an even function
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a polynomial function
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an identity function
Explanation
If $$f(x) = -x^2+1$$......
$$f(x) = -x^2+1, g(x) = -\sqrt[3]{x}$$........
$$\Rightarrow$$ $$f$$ is even and $$g$$ is odd
$$\therefore$$ $$gogog$$ is even
$$\Rightarrow$$ $$fogogog$$ is even
$$\Rightarrow$$ $$gofogofogogog$$ is even.
The domain of the function, $$\displaystyle f(x) = \frac{\left | x \right |\,-2}{\left | x \right |\,-3}$$ is ..........
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$$R$$
0%
$$R - \{2, 3\}$$
0%
$$R - \{2, -2\}$$
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$$R - \{-3, 3\}$$
Explanation
In $$f(x)$$ the only cause for function to be undefined will be if the denominator becomes 0.
Hence $$x\in R-\{-3,3\}$$
If f(x)=2x-1 and g(x)=3x+2 then find (fog) (x)
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2(3x+1)
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2(3x+2)
0%
3(2x+1)
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3(3x+1)
Explanation
$$(fog)(x) = f(g(x)) = 2(3x+2) -1 = 6x+4-1 = 6x+3 = 3(2x+1) $$
If f(x) = 2x+1 and g(x) = 3x-5 then find $$\left ( fog \right )^{-1}\left ( 0 \right )$$
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5/3
0%
3/2
0%
2/3
0%
3/5
Explanation
$$ (fog)(x) = f(g(x)) = 2(3x-5) + 1 = 6x-10 + 1 = 6x-9 $$
Let $$ (fog)(x)= y $$
Then, to find $$ (fog)^{-1} (x)$$, we find $$ x $$ in terms of $$y $$
So, $$ 6x-9 = y $$
$$ => x = \dfrac {y+9}{6} $$
$$ => (fog)^{-1} (x)= \dfrac {y+9}{6} $$
$$ => (fog)^{-1} (0)= \dfrac {0+9}{6} = \dfrac {3}{2} $$
Find $$\left( f\circ g \right) \left( 3 \right) $$ when $$f\left( x \right) =7x-6$$ and $$g\left( x \right) =5{ x }^{ 2 }-7x-6$$.
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0%
$$-36$$
0%
$$1014$$
0%
$$-90$$
0%
$$120$$
Explanation
We have, $$f(x)=7x-6$$ and $$g(x)=5x^2-7x-6$$
Then,$$(f\circ g)(x)$$
$$=f(g(x))$$
$$=f(5x^2-7x-6)$$
$$=7(5x^2-7x-6)-6$$
$$=35x^2-49x-42-6$$
$$=35x^2-49x-48$$
$$\therefore (f\circ g)(3)$$
$$=35(3)^2-49(3)-48$$
$$=315-147-48$$
$$=120$$
So, $$\text{D}$$ is the correct option.
If f is a constant function and f(100)=100 then f(2007)=_____
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0%
2007
0%
100
0%
0
0%
None of these
Explanation
A constant function $$ f(x) $$ will have the result as a constant for any value of $$ x $$
So, if $$ f(100) = 100 $$
then $$ f(2007 ) $$ is also $$ 100 $$
If $$f(x)\, =\, (p\, -\, x^n)^{1/n},\, p\, >\, 0$$ and $$n$$ is a positive integer, then $$f(f(x)) =$$
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0%
$$x$$
0%
$$x^n$$
0%
$$p^{1/n}$$
0%
$$p\, -\, x^n$$
Explanation
Given, $$f(x)=(p-x^{n})^{\dfrac{1}{n}}$$
$$\therefore f(f(x))=(p-((p-x^{n})^{\dfrac{1}{n}})^{n})^{\dfrac{1}{n}}$$
$$=(p-(p-x^{n}))^{\dfrac{1}{n}}$$
$$=(x^{n})^{\dfrac{1}{n}}$$
$$=x$$
If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ are defined by $$f\left( x \right) =\left| x \right| $$ and $$g\left( x \right) =\left[ x-3 \right] $$ for $$x\in R$$, then
$$g\left( f\left( x \right) \right) :\left\{ -\dfrac { 8 }{ 5 } < x < \dfrac { 8 }{ 5 } \right\} $$ is equal to
[.] is Greatest integer function
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0%
$$\left\{ 0,1 \right\} $$
0%
$$\left\{ 1,2 \right\} $$
0%
$$\left\{ -3,-2 \right\} $$
0%
$$\left\{ 2,3 \right\} $$
Explanation
Given that, $$f\left( x \right) =\left| x \right| $$ and $$g\left( x \right) =\left[ x-3 \right] $$
For $$-\dfrac { 8 }{ 5 } < x < \dfrac { 8 }{ 5 } , 0\le f\left( x \right) < \dfrac { 8 }{ 5 } $$
Now, for $$0 < f \left( x \right) < 1$$,
$$g\left( f\left( x \right) \right) = \left[ f\left( x \right) -3 \right] $$
$$=-3\quad \because -3\le f\left( x \right) -3 < -2$$
Again, for $$1 < f\left( x \right) < 1.6$$
$$g\left( f\left( x \right) \right) =-2$$
$$\because -2\le f\left( x \right) -3 < -1.4$$
Hence, required set is $$\left\{ -3,-2 \right\} $$
Let $$R$$ be the set of real numbers and the functions $$f: R \rightarrow R$$ and $$g: R\rightarrow R$$ be defined by $$f(x) = x^{2} + 2x - 3$$ and $$g(x) = x + 1$$. Then the value of $$x$$ for which $$f(g(x)) = g(f(x))$$ is
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0%
$$-1$$
0%
$$0$$
0%
$$1$$
0%
$$2$$
Explanation
According to the question,
$$f(g(x)) = g(f(x))$$
$$\Rightarrow f(x + 1) = g(x^{2} + 2x - 3)$$
$$\Rightarrow (x + 1)^{2} + 2 (x + 1) - 3 = x^{2} + 2x - 3 + 1$$
$$\Rightarrow x^{2} + 1 + 2x + 2x + 2 - 3 = x^{2} + 2x - 2$$
$$\Rightarrow x^{2} + 4x = x^{2} + 2x - 2$$
$$\Rightarrow x^{2} + 4x - x^{2} - 2x + 2 = 0$$
$$\Rightarrow 2x + 2 = 0$$
$$\Rightarrow 2x = -2$$
$$\Rightarrow x = -1$$
Let $$f:R\rightarrow R$$ be such that $$f$$ is injective and $$f(x)f(y)=f(x+y)$$ for all $$x,y\in R$$, if $$f(x), f(y)$$ and $$f(z)$$ are in GP, then $$x,y$$ and $$z$$ are in
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0%
AP always
0%
GP always
0%
AP depending on the values of $$x,y$$ and $$z$$
0%
GP depending on the values of $$x,y$$ and $$z$$
Explanation
Let the funtion $$f(x)={a}^{kx}$$
which define in $$f:R\rightarrow R$$ and injective also.
Now, we have
$$f(x)f(y)=f(x+y)$$
$$\Rightarrow$$ $${a}^{kx}.{a}^{ky}={a}^{k(x+y)}$$
$$\Rightarrow$$ $${a}^{k(x+y)}={a}^{k(x+y)}$$
$$\because$$ $$f(x), f(y)$$ and $$f(z)$$ are in GP
$$\therefore$$ $$f({y}^{2})=f(x).f(z)$$
$$\Rightarrow$$ $${a}^{2ky}={a}^{kx}.{a}^{kz}$$
$$\Rightarrow$$ $${e}^{2ky}={e}^{k(x+z)}$$
On comparing, we get
$$2ky=k(x+z)$$ $$\Rightarrow$$ $$2y=x+z$$
$$\Rightarrow$$ $$x,y$$ and $$z$$ are in AP
Let Q be the set of all rational numbers in [0, 1] and $$f : [0, 1]\rightarrow [0, 1]$$ be defined by $$f(x)=\begin{cases}x&for&x\in Q\\ 1-x&for&x\notin Q\end{cases}$$
Then the set $$S=\{x\in [0, 1]: (f\, o \, f)(x)=x\}$$ is equal to
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0%
[0, 1]
0%
Q
0%
[0, 1] - Q
0%
(0, 1)
Explanation
Let $$x\in Q$$, then,
$$f(x)=x$$ where $$x\in Q$$
So, $$fof(x)=f(f(x))=f(x)=x$$ as $$x\in Q$$
$$\therefore fof(x)=x$$ when $$x \in Q$$
Now,
Let $$x\notin Q$$ then
$$f(x) =1-x$$
$$\therefore fof(x)=1-(1-x) = x$$
as $$1-x\notin Q$$ as $$x\notin Q$$
where $$x\notin Q$$
$$fof(x)=\begin{cases}x \,where\, x\in Q \,\&\, x\in [0, 1] \\ x\, where x\notin Q\, \&\, x\in [0, 1] \end{cases}$$
$$\therefore$$ the set $$S = [0, 1]$$
If $$f(x)={2}^{100}x+1, g(x)={3}^{100}x+1$$, then the set of real numbers $$x$$ such that $$f\left\{ g(x) \right\} =x$$ is
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0%
empty
0%
a singleton
0%
a finite set with more than one element
0%
infinite
Explanation
Given $$f(x)={2}^{100}x+1, g(x)={3}^{100}x+1$$
Now $$fo{g(x)}=x$$
$$\Rightarrow$$ $$f({3}^{100}.x+1)=x\\$$
$$\Rightarrow$$ $${2}^{100}({3}^{100}.x+1)+1=x\\$$
$$\Rightarrow$$ $${6}^{100}.x+{2}^{100}+1=x\\$$
$$\Rightarrow$$ $$x(1-{6}^{100})=(1+{2}^{100})$$
$$\Rightarrow$$ $$x=\cfrac{1+{2}^{100}}{1-{6}^{100}}$$
Hence $$fog(x)=x$$ represent a singleton set.
If $$f: R\rightarrow R^{+}$$ and $$g: R^{+} \rightarrow R$$ are such that $$g(f(x)) = |\sin x|$$ and $$f(g(x)) = (\sin \sqrt {x})^{2}$$, then a possible choice for f and g is
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0%
$$f(x) = x^{2} , g(x) = \sin \sqrt {x}$$
0%
$$f(x) = \sin x, g(x) = |x|$$
0%
$$f(x) = \sin^{2}x, g(x) = \sqrt {x}$$
0%
$$f(x) = x^{2}, g(x) = \sqrt {x}$$
Explanation
As f returns only positive values and takes any real number it can be mod or square hence option 2 is eliminated
For g only positive values has to be given and it will return any real number
so on trying the option only option $$A$$ and $$C$$ results in the given equation ie $$g(f(x)) = |sin x|$$
Now function g should take positive values and return real values which is satisfied by only option $$A$$ as in option $$C$$ square root of positive number will always result in positive number while $$sin\sqrt{x}$$ will give -ve real numbers as well.
The domain of the function $$f(x) = \log (1 - x) + \sqrt {x^{2} - 1}$$
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0%
$$(-\infty, -1)$$
0%
$$(-\infty, -1]$$
0%
$$(-\infty, 2]$$
0%
$$(-\infty, 0)$$
Explanation
$$\log(1 - x)$$ is defined if $$1 - x > 0\Rightarrow x < 1 \Rightarrow x\epsilon (-\infty, 1) ..... (1)$$
$$\sqrt {x^{2} - 1}$$ is defined if $$x^{2} - 1 \geq 0\Rightarrow x^{2} \geq 1$$
$$\Rightarrow x \geq 1$$ or $$ x \leq -1 ..... (2)$$
$$\therefore$$ Required domain is the instruction of $$(1)$$ and $$(2)$$
i.e. $$(-\infty, -1]$$
If $$h(x)={x}^{3}+x$$ and $$g(x)=2x+3$$, then calculate $$g(h(2))$$.
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0%
$$7$$
0%
$$10$$
0%
$$17$$
0%
$$19$$
0%
$$23$$
Explanation
Given, $$h(x)=x^3+x$$
$$\therefore h(2)=2^3+2$$
$$\Rightarrow h(2)=8+2=10$$
Also given, $$g(x)=2x+3$$
$$\Rightarrow g(h(2))=g(10)$$
$$\because h(2)=10$$
Then the value of $$g(10)=2\times 10+3$$
$$\Rightarrow g(10)=20+3=23$$
$$f(x) = x^{2} + d$$ and $$g(x) = 2x^{2}$$, where d is a constant. If $$\dfrac {f(g(2))}{f(2)} = 4$$, find the value of $$d$$.
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0%
$$16$$
0%
5
0%
22
0%
18
Explanation
Given, $$f(x)=x^2+d$$ and $$g(x)=2x^2$$
$$g(2)=2(2^{2})=8$$
Hence, $$f(g(2))=f(8)=64+d$$
Therefore $$\dfrac{f(g(2))}{f(2)}=\dfrac{64+d}{4+d}=4$$
Hence, $$64+d=16+4d$$
$$\Rightarrow 48=3d$$
$$\Rightarrow d=16$$
Find the correct
expression for $$\displaystyle f\left( g\left( x \right) \right) $$
if $$\displaystyle f(x)=4x+1$$ and $$\displaystyle g\left( x \right) ={ x }^{ 2 }-2$$
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0%
$$\displaystyle -{ x }^{ 2 }+4x+1$$
0%
$$\displaystyle { x }^{ 2 }+4x-1$$
0%
$$\displaystyle 4{ x }^{ 2 }-7$$
0%
$$\displaystyle 4{ x }^{ 2 }-1$$
0%
$$\displaystyle 16{ x }^{ 2 }+8x-1$$
Explanation
$$f(x)=4x+1$$ and $$g(x)=x^2-2$$
$$\Rightarrow f(g(x))=f(x^2-2)$$
$$\Rightarrow (4(x^2-2)+1$$
$$\Rightarrow 4x^2-8+1$$
$$\Rightarrow 4x^2-7$$
If $$f(g(a)) = 0$$ where $$ g(x) = \dfrac {x}{4} + 2$$ and $$f(x) = |x^{2} - 3|$$, find the possible value of $$a.$$
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0%
$$-8+4\sqrt{3}$$
0%
$$-(8+4\sqrt{3})$$
0%
$$6$$
0%
$$18$$
Explanation
Given $$g(a)=\dfrac{a}{4}+2,f(x)=|x^2|-3$$ and $$f(g(a))=0$$
Now, $$f(g(a))=|g(a)^{2}-3|=|(\dfrac{a}{4}+2)^{2}-3|$$
$$=|\dfrac{a^{2}}{16}+a+4-3|=|\dfrac{a^{2}}{16}+a+1|=\dfrac{|a^{2}+16a+16|}{16}$$
but
$$f(g(a))=0$$
$$\Rightarrow\dfrac{|a^2+16a+16|}{16}=0$$
$$\Rightarrow a^{2}+16a+16=0$$ $$[\because |x|=0\Rightarrow x=0]$$
$$\Rightarrow a= \dfrac{-16\pm\sqrt{16^2-4(1)(64)}}{2}$$
$$\Rightarrow a=\dfrac{-16\pm\sqrt{256-64}}{2}=\dfrac{-16\pm8\sqrt{3}}{2}$$
Hence,
$$a=-(8+4\sqrt{3})$$ and $$a=-8+4\sqrt{3}$$.
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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