∣1241bb21cc2∣→C2→C2−C1,C3→C3−C1∣1240b−2b2−40c−2c2−4∣=(b−2)(c−2)(c−b)
Let common difference of AP=d
∴b=2+d,c=2+2d
∣A∣=d×2d×d=2d3∈[2,16]
⇒d3∈[1,8]⇒d∈[1,2]⇒2d∈[2,4]
⇒2+2d∈[4,6]
∴c∈[4,6]
Let the numbers 2,b,c be in an A.P. and A=[1241bb21cc2] . If det(A)∈[2,16], then c lies in the interval:
Held on 8 Apr 2019 · Verified 6 Jul 2026.
[2,3)
[4,6]
[3,2+243]
(2+243,4)
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If the roots of x² - 5x + k = 0 are in the ratio 2:3, then k equals:
Let $\alpha = 3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2+x-2=0$, $\beta \in \left(0, \dfrac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to:
If the set of all solutions of $|x^2 + x - 9| = |x| + |x^2 - 9|$ is $[\alpha, \beta] \cup [\gamma, \infty)$, then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to:
The sum of squares of all the real solutions of the equation $\log_{(x+1)}(2x^2+5x+3) = 4 - \log_{(2x+3)}(x^2+2x+1)$ is equal to ________.
Let $f:(1,\infty)\to\mathbb{R}$ be a function defined as $f(x) = \dfrac{x-1}{x+1}$. Let $f^{i+1}(x) = f(f^i(x))$, $i=1, 2, \ldots, 25$, where $f^1(x)=f(x)$. If $g(x) + f^{26}(x) = 0$, $x \in (1, \infty)$, then the area of the region bounded by the curves $y=g(x)$, $2y=2x-3$, $y=0$ and $x=4$ is:
Work through every JEE Main Algebra PYQ, year by year.