−1≤2x+35−x≤ & 10−x>0, 10−x=1
2x+35−x≥1 & x<10 & x=9
(5−x)2−(2x+3)2≤0 & x<10 & 4x=9
(x+8)(3x−2)≥0 & x<10 & x=9
⇒(−∞,−8]∪[32,10)−{9}
⇒(α+β+γ+δ)=6(−8+32+10+9)
=70
If the domain of the function f(x)=sin−1(3+2x5−x)+loge(10−x)1 is (−∞,α]∪[β,γ)−{δ}, then 6(α+β+γ+δ) is equal to
Held on 22 Jan 2026 · Verified 6 Jul 2026.
67
66
70
68
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let $a_{1}=1$ and for $n \geqslant 1, a_{n+1}=\frac{1}{2} a_{n}+\frac{n^{2}-2 n-1}{n^{2}(n+1)^{2}}$. Then $\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right|$ is equal to $\_\_\_\_$.
Let $a_{1}, a_{2}, a_{3}, \ldots$ be G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4}=64$ and $a_{1}+a_{3}+a_{5}=\frac{813}{7}$. Then $a_{3}+a_{5}+a_{7}$ is equal to :
Let $\sum_{k=1}^{n} a_{k}=\alpha n^{2}+\beta n$. If $a_{10}=59$ and $a_{6}=7 a_{1}$, then $\alpha+\beta$ is equal to
Consider an A.P.: $a_{1}, a_{2}, \ldots, a_{\mathrm{n}} ; a_{1}>0$. If $a_{2}-a_{1}=\frac{-3}{4}, a_{\mathrm{n}}=\frac{1}{4} a_{1}$, and $\sum_{\mathrm{i}=1}^{\mathrm{n}} a_{\mathrm{i}}=\frac{525}{2}$, then $\sum_{\mathrm{i}=1}^{17} a_{\mathrm{i}}$ is equal to
Let $a_{1}, \frac{a_{2}}{2}, \frac{a_{3}}{2^{2}}, \ldots, \frac{a_{10}}{2^{9}}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_{1}+a_{2}+\ldots+a_{10}=62$, then $a_{1}$ is equal to :
Work through every JEE Main Algebra PYQ, year by year.