NIMCET 2026 Previous Year Questions (PYQs) – Page 6 of 12

Given,

$f(x)=|x+1|e^{-x^2}$

For $x<-1$,

$f(x)=-(x+1)e^{-x^2}$

Differentiate:

$f'(x)=e^{-x^2}(2x^2+2x-1)$

For critical points,

$2x^2+2x-1=0$

Using quadratic formula,

$x=\frac{-2\pm\sqrt{4+8}}{4}$

$x=\frac{-2\pm 2\sqrt{3}}{4}$

$x=\frac{-1\pm\sqrt{3}}{2}$

For $x<-1$, the valid critical point is

$x=\frac{-1-\sqrt{3}}{2}$

This lies in the interval $(-2,-1)$.

At this point, $f$ has a point of maxima.

Using expansion near $x=0$,

$(1+x)^{\frac{1}{3}}=1+\frac{x}{3}-\frac{x^2}{9}+O(x^3)$

Also,

$(1+2x)^{\frac{1}{4}}=1+\frac{x}{2}-\frac{3x^2}{8}+O(x^3)$

Now,

$(1+x)^{\frac{1}{3}}-(1+2x)^{\frac{1}{4}}$

$=\left(1+\frac{x}{3}-\frac{x^2}{9}\right)-\left(1+\frac{x}{2}-\frac{3x^2}{8}\right)+O(x^3)$

$=-\frac{x}{6}+\left(-\frac{1}{9}+\frac{3}{8}\right)x^2+O(x^3)$

$=-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)$

Therefore,

$f(x)=\frac{-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)}{x}$

$f(x)=-\frac{1}{6}+\frac{19x}{72}+O(x^2)$

For differentiability at $x=0$, function must be continuous at $x=0$.

So,

$a=\lim_{x\to 0}f(x)=-\frac{1}{6}$

Also,

$f'(0)=\frac{19}{72}$

Hence,

$a+f'(0)=-\frac{1}{6}+\frac{19}{72}$

$=-\frac{12}{72}+\frac{19}{72}$

$=\frac{7}{72}$

We know that

$(f^{-1})'(y)=\frac{1}{f'(x)}$, where $f(x)=y$

Here, we need $(f^{-1})'(2)$.

So first find $x$ such that

$f(x)=2$

$\frac{3x^2+4x+1}{x^2+3x+2}=2$

$3x^2+4x+1=2x^2+6x+4$

$x^2-2x-3=0$

$(x-3)(x+1)=0$

Since domain is $[0,\infty)$,

$x=3$

Now,

$f(x)=\frac{3x^2+4x+1}{x^2+3x+2}$

Let

$N=3x^2+4x+1$

and

$D=x^2+3x+2$

Then,

$f'(x)=\frac{N'D-ND'}{D^2}$

Now at $x=3$,

$N=3(3)^2+4(3)+1=40$

$D=(3)^2+3(3)+2=20$

$N'=6x+4$

So,

$N'=22$

$D'=2x+3$

So,

$D'=9$

Therefore,

$f'(3)=\frac{22\cdot 20-40\cdot 9}{20^2}$

$f'(3)=\frac{440-360}{400}$

$f'(3)=\frac{80}{400}$

$f'(3)=\frac{1}{5}$

Hence,

$(f^{-1})'(2)=\frac{1}{f'(3)}$

$=5$

Given,

$x-y\tan 35^\circ=\tan 25^\circ(y+x\tan 35^\circ)$

Let

$\tan 35^\circ=A$

and

$\tan 25^\circ=B$

Then,

$x-yA=B(y+xA)$

$x-yA=By+ABx$

$x-ABx=yA+By$

$x(1-AB)=y(A+B)$

So,

$\frac{x}{y}=\frac{A+B}{1-AB}$

Now,

$\frac{\tan 35^\circ+\tan 25^\circ}{1-\tan 35^\circ\tan 25^\circ}=\tan(35^\circ+25^\circ)$

$=\tan 60^\circ$

$=\sqrt{3}$

Therefore,

$\frac{x}{y}=\sqrt{3}$

So,

$x=\sqrt{3}y$

Since $\sqrt{3}>1$, according to the given option pattern, we get

$x>y$

By sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

So,

$\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A+\sin B}$

Now,

$\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

and

$\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

Therefore,

$\frac{\sin A-\sin B}{\sin A+\sin B}=\cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right)$

Since,

$A+B=\pi-C$

So,

$\cot\left(\frac{A+B}{2}\right)=\cot\left(\frac{\pi-C}{2}\right)$

$=\cot\left(\frac{\pi}{2}-\frac{C}{2}\right)$

$=\tan\left(\frac{C}{2}\right)$

Thus,

$\frac{a-b}{a+b}=\tan\left(\frac{C}{2}\right)\tan\left(\frac{A-B}{2}\right)$

But option $4$ gives

$\frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{C}{2}\right)}$

Given,

$PA=60$

$PB=80$

$\angle APB=60^\circ$

Using cosine rule in triangle $APB$,

$AB^2=PA^2+PB^2-2(PA)(PB)\cos 60^\circ$

$AB^2=60^2+80^2-2(60)(80)\cdot \frac{1}{2}$

$AB^2=3600+6400-4800$

$AB^2=5200$

$AB=\sqrt{5200}$

$AB=\sqrt{400\cdot 13}$

$AB=20\sqrt{13}$

We know that the principal range of $\cos^{-1}x$ is $[0,\pi]$.

Now,

$\cos\left(-\frac{\pi}{6}\right)=\cos\frac{\pi}{6}$

So,

$\cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right)=\frac{\pi}{6}$

Also,

$\sin\frac{5\pi}{6}=\frac{1}{2}$

The principal range of $\sin^{-1}x$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

So,

$\sin^{-1}\left(\sin\frac{5\pi}{6}\right)=\sin^{-1}\left(\frac{1}{2}\right)$

$=\frac{\pi}{6}$

Therefore,

$\cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right)+\sin^{-1}\left(\sin\frac{5\pi}{6}\right)$

$=\frac{\pi}{6}+\frac{\pi}{6}$

$=\frac{\pi}{3}$

Given,

$\tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4}$

Taking tangent on both sides,

$\tan\left(\tan^{-1}(3x)+\tan^{-1}(2x)\right)=\tan\frac{\pi}{4}$

Using,

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

we get,

$\frac{3x+2x}{1-(3x)(2x)}=1$

$\frac{5x}{1-6x^2}=1$

So,

$5x=1-6x^2$

$6x^2+5x-1=0$

Factorizing,

$6x^2+6x-x-1=0$

$6x(x+1)-1(x+1)=0$

$(x+1)(6x-1)=0$

So,

$x=-1$ or $x=\frac{1}{6}$

Now check both values.

For $x=\frac{1}{6}$,

$\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{4}$

So, $x=\frac{1}{6}$ is valid.

For $x=-1$,

$\tan^{-1}(-3)+\tan^{-1}(-2)$ is negative, so it cannot be equal to $\frac{\pi}{4}$.

Therefore, only one solution exists.

For intersection of the two graphs,

$f(x)=g(x)$

So,

$2\cos\left(\frac{x}{2}\right)+3=4$

$2\cos\left(\frac{x}{2}\right)=1$

$\cos\left(\frac{x}{2}\right)=\frac{1}{2}$

Now,

$x\in [0,4\pi]$

So,

$\frac{x}{2}\in [0,2\pi]$

In the interval $[0,2\pi]$,

$\cos\theta=\frac{1}{2}$ has two solutions:

$\theta=\frac{\pi}{3},\frac{5\pi}{3}$

Here,

$\theta=\frac{x}{2}$

So,

$\frac{x}{2}=\frac{\pi}{3}$ or $\frac{x}{2}=\frac{5\pi}{3}$

Therefore,

$x=\frac{2\pi}{3}$ or $x=\frac{10\pi}{3}$

Hence, the number of points of intersection is $2$.

First expression is:

$\left(x^2+\frac{1}{x}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x^2)^{12-k}\left(\frac{1}{x}\right)^k$

$T_{k+1}={}^{12}C_k x^{24-2k-k}$

$T_{k+1}={}^{12}C_k x^{24-3k}$

For coefficient of $x^{10}$,

$24-3k=10$

$3k=14$

$k=\frac{14}{3}$

This is not an integer, so $x^{10}$ term is not present in the first expression.

Now second expression is:

$\left(x+\frac{1}{x^2}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x)^{12-k}\left(\frac{1}{x^2}\right)^k$

$T_{k+1}={}^{12}C_k x^{12-k-2k}$

$T_{k+1}={}^{12}C_k x^{12-3k}$

For coefficient of $x^{10}$,

$12-3k=10$

$3k=2$

$k=\frac{2}{3}$

This is also not an integer, so $x^{10}$ term is not present in the second expression.

Therefore, the coefficient of $x^{10}$ is $0$.