MCQ
The function $f:[0,3] \rightarrow[1,29]$, defined by $f(x)=2 x^3-15 x^2+36 x+1$, is
- Aone-one and onto
- ✓onto but not one-one
- Cone-one but not onto
- Dneither one-one nor onto
✓
Answer
Correct option: B.
onto but not one-one
b
$F:[0,3] \rightarrow[1,29] $
$F:[0,3] \rightarrow[1,29] $
$f(x)=2 x^3-15 x^2+36 x+1 $
$f^{\prime}(x)=6 x^2-30 x+36 $
$=6\left(x^2-5 x+6\right) $
$=6(x-2)(x-3)$
in given domain function has local maxima, it is many-one
Now at
$x=0 $$ f(0) $$ =1 $
$x=2 $$ f(2) $$ =16-60+72+1=29 $
$x $$ =3 $$ f(3) $$ =54-135+108+1 $
$ =163-135=28$
Has range $=[1,29]$
Hence given function is onto
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Two distinct polynomials $f(x)$ and $g(x)$ are defined as follows:
→$f(x)=x^2+a x+2 ; g(x)=x^2+2 x+a$.If the equations $f(x)=0$ and $g(x)=0$ have a common root, then the sum of the roots of the equation $f(x)+g(x)=0$ is
The function $f(x) = x(x + 3){e^{ - (1/2)x}}$ satisfies all the conditions of Rolle's theorem in $ [-3, 0]$. The value of $c$ is
→If $a,\;b,\;c$ are in $A.P.$ and $a,\;b,\;d$ in $G.P.$, then $a,\;a - b,\;d - c$ will be in
→The smallest natural number $n,$ such that the coefficient of $x$ in the expansion of ${\left( {{x^2}\, + \,\frac{1}{{{x^3}}}} \right)^n}$ is $^n{C_{23}}$ is
→Let $y=y(x)$ be the solution of the differential equation $x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2} y-y-4 x^{3}\right)=0, x>1$ with $y\,(2)=-2$. Then $y\,(3)$ is equal to
→For $\alpha, \beta, \gamma \neq 0$. If $\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi$ and $(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta$, then $\gamma$ equal to
→If $f(x) = x(\sqrt x - \sqrt {x + 1} ),$ then
→The solution set of the equation ${x^{{{\log }_x}{{(1 - x)}^2}}} = 9$ is
→Equation of a line through $(7, 4)$ and touching the circle, $x^2 + y^2 - 6x + 4y - 3 = 0$ is :
→If system of equations $kx + 2y - z = 2,$$\left( {k - 1} \right)x + ky + z = 1,x + \left( {k - 1} \right)y + kz = 3$ has only one solution, then number of possible real value$(s)$ of $k$ is -
→