We have,
f(x)=y=(6x−α5x+3)…(i)
⇒5x+3=6xy−αy
⇒x(6y−5)=αy+3
⇒x=6y−5αy+3
Hence,
f−1(x)=(6x−5αx+3)…(ii)
Now,
(fof)(x)=x
f(x)=f−1(x)
From equation (i) and (ii), we have
(6x−α5x+3)=(6x−5αx+3)
Clearly, α=5
JEE Main 2021 — Mathematics Algebra
Let f:R−6α→R be defined by f(x)=(6x−α5x+3). Then the value of α for which (fof)(x)=x, for all x∈R−6α, is
Held on 20 Jul 2021 · Verified 6 Jul 2026.
No such α exists
5
8
6
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
The number of elements in the set $S = \left\{(r, k) : k \in \mathbb{Z} \text{ and } {}^{36}C_{r+1} = \dfrac{6\left({}^{35}C_r\right)}{(k^2 - 3)}\right\}$, is :
Suppose $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P. and $\mathrm{a}^{2}, 2 \mathrm{~b}^{2}, \mathrm{c}^{2}$ are in G.P. If $\mathrm{a}<\mathrm{b}<\mathrm{c}$ and $\mathrm{a}+\mathrm{b}+\mathrm{c}=1$, then $9\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)$ is equal to $\_\_\_\_$ .
Let $S=\left\{z \in \mathbb{C}: 4 z^{2}+\bar{z}=0\right\}$. Then $\sum_{z \in S}|z|^{2}$ is equal to:
Let $S=\{z: 3 \leqslant|2 z-3(1+i)| \leqslant 7\}$ be a set of complex numbers. Then $\min _{z \in S}\left|\left(z+\frac{1}{2}(5+3 i)\right)\right|$ is equal to :
Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}$ and $B = [b_{ij}]$, $1 \leq i, j \leq 3$. If $B = A^{99} - I$, then the value of $\dfrac{b_{31} - b_{21}}{b_{32}}$ is :
Work through every JEE Main Algebra PYQ, year by year.