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x = t² – t + 1        … (1)

y = t² + t + 1    … (2)

y – x = 2t & x + y = 2 (t² + 1)

__________on eliminating ‘t’ we get

$\Rightarrow \left(x+y-2\right)=2{\left(\frac{y-x}{2}\right)}^2$

${\left(x-y\right)}^2=2\left(x+y-2\right)$

Axis : x – y = 0

Tangent at vertex : x + y – 2 = 0

Vertex : (1, 1) = (x, y)

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$x^2\left(y+z\right)y^2\left(z+x\right)z^2\left(x+y\right)=a^3{b}^3{c}^3=x^3{y}^3{z}^3$

$\Rightarrow \left(x+y\right)\left(y+z\right)\left(z+x\right)=xyz$

$\begin{array}{l}\Rightarrow x^2\left(y+z\right)+y^2\left(z+y\right)+z^2\left(x+y\right)+xyz=0\\ \Rightarrow a^3+b^3+c^3+abc=0\end{array}$

f (x) = f (6 – x) Þ f' (x) = -f' (6 – x) …. (1)

put x = 0, 2, 5

f' (0) = f' (6) = f' (2) = f' (4) = f' (5) = f' (1) = 0

and from equation (1) we get f' (3) = -f' (3)

$?f\text{'}(3)=0$

So f' (x) = 0 has minimum 7 roots in $x?\left[0,6\right]?f\text{'}\text{'}\left(x\right)$  has min 6 roots in   $x?\left[0,6\right]$

h (x) = f' (x) . f' (x)

h' (x) = (f' (x)² + f' (x) f' (x)

h (x) = 0 has 13 roots in x $?$   [0, 6]

h' (x) = 0 has 12 roots in x $?$ [0, 6]

Angular retardation,

OP² = x² = y²

y = e^x, y’ = e^x,

slope of normal =  $-\frac{1}{e^x}$

$\frac{y}{x}=-\frac{1}{e^x}$    

$-\frac{1}{e^x}=\frac{e^x}{x}\Rightarrow -x=e^{2x}$      

By hit and trial we get  $x=-\frac{2}{5}$

$P\equiv \left(-\frac{2}{5},e^{-2/5}\right)$

$OP=\sqrt[]{\frac{4}{25}+e^{-4/5}}\Rightarrow O{P}^2=\frac{14}{25}=\frac{m}{n}$

$x_1=\underset{x\to \infty }{lim}\frac{2^{\frac{x^n}{e^x}}-3^{\frac{x^n}{e^x}}}{\frac{x^n}{e^x}},put\frac{x^n}{e^x}=t$

$x_2=\underset{x\to \infty }{lim}\frac{co{t}^{-1}\left(\sqrt[]{x+1}-\sqrt[]{x}\right)}{se{c}^{-1}\left({\left(\frac{2x+1}{x-1}\right)}^x\right)}$

$x_2=\frac{2}{\pi }\underset{x\to \infty }{lim}ta{n}^{-1}\left(\sqrt[]{x+1}+\sqrt[]{x}\right)$

$x_2=\frac{2}{\pi }.\frac{\pi }{2}=1$

$g\left(f\left(x\right)\right)=x\Rightarrow g\text{'}\left(f\left(x\right)\right).f\text{'}\left(x\right)=1$

$f\left(x\right)=1\Rightarrow x=0$

$g\text{'}\left(1\right).f\text{'}\left(0\right)=1$

$f\text{'}\left(x\right)=2x+e^{-x}$

$\begin{array}{l}f\text{'}\left(0\right)=1\\ g\text{'}\left(1\right)=1\end{array}$

$\left|\begin{array}{ccc}\hfill 1-\alpha \hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 4\hfill & \hfill -1-\alpha \hfill & \hfill 4\hfill \\ \hfill 2\alpha \hfill & \hfill 2\alpha \hfill & \hfill \alpha -5\hfill \end{array}\right|=0$

$\alpha =-5$

In disproportionation reaction, one element of a compound will simultaneously get reduced and oxidised. In ClO₄ -, oxidation number of Cl is +7 and it can not increase it further. So, ClO₄- will not get oxidised and so will not undergo disporportionation reaction.

Let S be the distance between Two ends a be the constant acceleration

As we know $v^2-u^2=2aS$

or, $aS=\frac{v^2-u^2}{2}$

Let $v_c$  be velocity at mid point.

Therefore, $v_c^2-u^2=2a\frac{S}{2}$

$v_c^2=u^2+aS$

$v_c^2=u^2+\frac{v^2-u^2}{2}$

$v_c=\sqrt[]{\frac{u^2+v^2}{2}}$

$\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{4},\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{1}{12}$

divide both $\frac{P\left(B\right)}{P\left(A\right)}=\frac{1}{3}$

$P\left(A\right)=\frac{1}{11},P\left(B\right)=\frac{1}{33}$

$\frac{dy}{dx}=3a{x}^2-2bx$

${\frac{dy}{dx}|}_{x=1}=3a-2b=3$

$a=\frac{2b+3}{3}\in \left[1,2\right]$

$2b+3\in \left[3,6\right]$

$2b\in \left[0,3\right]$

$b\in \left[0,\frac{3}{2}\right]$

P? V indicator diagram for isobaric

P? V indicator diagram for isochoric process

P? V indicator diagram for isothermal process

$\frac{3z_1+i}{4}=\frac{2z_2+2z_3}{4}$

P = point of intersection of AD & BC

$AD=1\Rightarrow DP=\frac{3}{4}$

$BP=\sqrt[]{1-\frac{9}{16}}=\frac{\sqrt[]{7}}{4}\Rightarrow BC=\frac{\sqrt[]{7}}{2}$

area of Quad. ABCD = $\frac{1}{2}.AD\times BC$

= $\frac{1}{2}\times 1\times \frac{\sqrt[]{7}}{2}=\frac{\sqrt[]{7}}{4}$

${\displaystyle \int e^{sinx}}\left(\frac{5+co{s}^{2}x}{\left(2-sinx\right)\sqrt[]{3+co{s}^{2}x}}\right)cosxdx$

put sin x = t

${\displaystyle {}^{}}\frac{{}^{}}{\text{exception 25:}}$  ${\displaystyle \int e^t}\left(\frac{6-t^2}{\left(2-t\right)\sqrt[]{4-t^2}}\right)$ dt

${\displaystyle \int e^t}\left(\sqrt[]{\frac{2+t}{2-t}}+\frac{2}{{\left(2-t\right)}^{3/2}{\left(2+t\right)}^{1/2}}\right)dt$

If $g\left(t\right)=\sqrt[]{\frac{2+t}{2-t}},g\text{'}\left(t\right)=\frac{2}{{\left(2-t\right)}^{3/2}{\left(2+t^2\right)}^{1/2}}$ $\text{exception 25:}\frac{}{{}^{}{{}^{}}^{}}$

$e^t\sqrt[]{\frac{2+t}{2-t}}+c$

$f\left(\frac{\pi }{2}\right)=\sqrt[]{3}e$

Correct order is Be

Line L is   $\frac{x-1}{1}=\frac{y-2}{2\sqrt[]{2}}=\frac{z}{0}$

this line L makes an angle of 45° with the plane  $\sqrt[]{2}x+y-z=1$

$\therefore$ Required distance PQ is perpendicular distance of plane from P is i.e.,

$PQ=\frac{\left|\sqrt[]{2}+7-\sqrt[]{2}-1\right|}{\sqrt[]{2+1+1}}=3$

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