JEE Main 2025 (Jan 22nd, Shift 1) Practice Questions & Answers

Let $a_1, a_2, a_3, \dots$a1​,a2​,a3​,… be a G.P. of increasing positive terms. If $a_1 a_5 = 28$a1​a5​=28 and $a_2 + a_4 = 29$a2​+a4​=29, then $a_6$a6​ is equal to:

Let $x = x(y)$x=x(y) be the solution of the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$y2dx+(x−y1​)dy=0. If $x(1) = 1$x(1)=1, then $x (\frac{1}{e})$x(e1​) is :

Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac{m}{n}$nm​, where gcd(m, n) = 1, then m + n is equal to :

The product of all solutions of the equation $e^{5(\log_e x)^2+3} = x^8, x > 0$e5(loge​x)2+3=x8,x>0, is :

Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line $x + 2y = 2$x+2y=2. If the centroid of $\triangle PQR$△PQR is the point $(\alpha, \beta)$(α,β), then $15(\alpha - \beta)$15(α−β) is equal to :

Let for $f(x) = 7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$f(x)=7tan8x+7tan6x−3tan4x−3tan2x, $I_1 = \int_0^{\pi/4} f(x)dx$I1​=∫0π/4​f(x)dx and $I_2 = \int_0^{\pi/4} x f(x)dx$I2​=∫0π/4​xf(x)dx. Then $7I_1 + 12I_2$7I1​+12I2​ is equal to :

Let the parabola $y = x^2 + px - 3$y=x2+px−3, meet the coordinate axes at the points P, Q and R . If the circle C with centre at $(-1,-1)$(−1,−1) passes through the points P, Q and R, then the area of $\triangle PQR$△PQR is :

Let $L_1 : \frac{x+1}{2} = \frac{y-2}{3} = \frac{z+3}{4}$L1​:2x+1​=3y−2​=4z+3​ and $L_2 : \frac{x-2}{3} = \frac{y+4}{4} = \frac{z-5}{5}$L2​:3x−2​=4y+4​=5z−5​ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$L1​ and $L_2$L2​?

Let f(x) be a real differentiable function such that f(0) = 1 and $f(x + y) = f(x)f'(y) + f'(x)f(y)$f(x+y)=f(x)f′(y)+f′(x)f(y) for all $x, y \in R$x,y∈R. Then $\sum_{n=1}^{100} \log_e f(n)$∑n=1100​loge​f(n) is equal to :

From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ' M', is :