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Volume = |a b c; b c a; c a b| = |3abc – a³ – b³ – c³|
= | (a + b + c) (a² + b² + c² – ab – bc – ca)| = | (a + b + c) (a + b + c)² – 3 (ab + bc + ca)|
= |9 (81-3 × 15)| = 324

Centre of C₃ will lie on the radical axis of C₁ and C₂ which is 10x + 6y + 26 = 0.

Let center of C₃ is (h, k).

Equation of chord of contact through (h, k) to the C₁ may be given as hx + ky = 25 … (I)

Let the mid point of the chord is (x₁, y₁) the equation of the chord with the help of mid point may be given as  

$x{x}_1+y{y}_1=x_1^2+y_1^2$

Since (I) and (II) represents same straight line 10x + 6y + 26 = 0

$\Rightarrow \frac{h}{25}=\frac{x_1}{x_1^2+y_1^2},\frac{k}{25}=\frac{y_1}{x_1^2+y_1^2}$            

The locus of (x₁, y₁) is

$5x+3y+\frac{13}{25}\left(x^2+y^2\right)=0$          

$|\mathrm{A}\mathrm{d}\mathrm{j}?A|=64=|A{|}^{n-1}\Rightarrow |A|=\pm 8$ $\alpha =\pm 8$

$\beta =|\mathrm{a}\mathrm{d}\mathrm{j}?(\mathrm{a}\mathrm{d}\mathrm{j}?A\left)\right|=|A{|}^{(n-1{)}^2}=8^4\mathrm{}\text{Also}\left|\frac{\sqrt[]{\beta }}{\alpha }\right|=\frac{8^2}{8}=8$

a², b², c² . A.P.
a² + 1, b² + 1, c² + 1 . P.
a² + ab + bc + ca, b² + ab + bc + ca, c² + ab + bc + CA . P.
⇒ (a + b) (a + c), (a + b) (b + c), (b + c) (c + a) . P.
⇒ 1/ (b+c), 1/ (c+a), 1/ (a+b) . A.P.
⇒ b + c, c + a, a + b . P.

lim (x→∞) (∫? ^ (√x²+1) tan? ¹t dt) / x = lim (x→∞) (tan? ¹ (√x²+1) * (x/√ (x²+1) = lim (x→∞) (tan? ¹ x) * (x/√ (x²+1) = π/2

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  $56=\frac{50n+60m}{m+n}$

56 n + 56 m = 50 n + 60 m

6n = 4m

$\frac{n}{m}=2/3$            

$\therefore 9.\frac{n}{m}=9\times \frac{2}{3}=6$            

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Let S = z + 2z² + 3z³ + . + 18z¹?
zS = z² + 2z³ + . + 18z¹?
(1 − z)S = z + z² + z³ + . - 18z¹?
(1 − z)S = z (z¹? - 1)/ (z-1) - 18z¹?
Now z = cos 20° + isin 20° ⇒ z¹? = 1
Also |z| = 1
⇒ (1 − z)S = -18z
⇒ S = -18z / (1-z)
|S|? ¹ = | (1-z)/ (-18z)| = |1-z|/18
= (1/18) |1 – cos 20° - isin 20°| = (1/18) |2sin² 10° — 2isin 10°cos 10°|
= (1/18) |2sin 10° (sin 10° – icos 10°)| = (1/9)sin 10°

I? = ∫? ¹ (1 − x? )? · 1 dx
= (1 - x? )? x|? ¹ - ∫? ¹ 7 (1 - x? )? (-4x³)xdx = -28∫? ¹ (1 − x? )? (1 − x? − 1)dx
I? = −28∫? ¹ (1 − x? )? dx + 28∫? ¹ (1 − x? )? dx
I? = -28I? + 28I?
29I? = 28I?
I? /I? = 28/29 ⇒ (29/4) * (I? /I? ) = (29/4) * (28/29) = 7

Kindly consider the following figure

The two altitudes are

$\left(x-\sqrt[]{3}y+1\right)\left(x+\sqrt[]{3}y-5\right)=0$            

Point of int. of the 2 altitudes is  $\left(2,\sqrt[]{3}\right)$  

Let slope of 3^rd altitude be ‘m’

then   $\left|\frac{m-\frac{1}{\sqrt[]{3}}}{1+\frac{1}{\sqrt[]{3}}}\right|=\sqrt[]{3}\Rightarrow \frac{\sqrt[]{3}m-1}{\sqrt[]{3}+m}=\pm \sqrt[]{3}$

$\sqrt[]{3}m-1=\pm 3\pm \sqrt[]{3}m$            

  $m=\infty ,=\frac{-2}{2\sqrt[]{3}}=\frac{-1}{\sqrt[]{3}}$          

$\therefore$ The third altitude is x = 2

A = ∫? ² lnx dx = 2ln2 – 1
A' = 4 - 2 (2ln2 – 1) = 6 – 4ln2

${\mathrm{M}\mathrm{a}}_2{\mathrm{b}}_2{\mathrm{c}}_2=5$ Geometrical Isomers

M (AA)b2c2:3  $\mathrm{M}\left(\mathrm{A}\mathrm{A}\right){\mathrm{b}}_2{\mathrm{c}}_2:3$  Geometrical Isomers

  $\left(\lambda ,2\lambda ,3\lambda \right)$

$cos\theta =\frac{6}{\sqrt[]{42}}\Rightarrow sin\theta =\frac{\sqrt[]{6}}{\sqrt[]{42}}$ (where q is the angle between L₁ & L₂)

$Area\Delta OAB=\frac{1}{2}\left(OA\right)\left(OB\right)sin\theta$

$=\frac{1}{2}\left(\sqrt[]{3}\right)\left|\lambda \right|\left(\sqrt[]{14}\right)\frac{\sqrt[]{6}}{\sqrt[]{42}}=\sqrt[]{6}\Rightarrow \lambda =\pm 2$

Amphiprotic species are those which behave like acid and base both. They can donate and accept a proton.

dy/dx = (e? – x)? ¹
dx/dy = e? - x
dx/dy + x = e? ⇒ I.F. = e^ (∫dy) = e?
∴ xe? = ∫e? dy + c
xe? = e²? /2 + c
y (0) = 0 ⇒ c = -1/2
∴ x = e? /2 - (1/2)e? ⇒ e? – e? = 2x
e²? – 2xe? - 1 = 0
e? = x ± √ (x² + 1) ⇒ e? = x + √ (x² + 1)
y = ln (x + √ (x² + 1)