NIMCET 2026 Previous Year Questions (PYQs) – Page 5 of 12

Let

$\sqrt{x+y}=a$

and

$\sqrt{y+z}=b$

Given,

$2\sqrt{x+y}-3\sqrt{y+z}=2$

So,

$2a-3b=2$

Also,

$x=a^2-y$

and

$z=b^2-y$

Now use the second condition:

$4x-5y-9z=8$

$4(a^2-y)-5y-9(b^2-y)=8$

$4a^2-4y-5y-9b^2+9y=8$

$4a^2-9b^2=8$

So,

$(2a-3b)(2a+3b)=8$

Since $2a-3b=2$,

$2(2a+3b)=8$

$2a+3b=4$

Now solve:

$2a-3b=2$

$2a+3b=4$

Adding both equations,

$4a=6$

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

Now,

$2a+3b=4$

$3+3b=4$

$3b=1$

$b=\frac{1}{3}$

Now,

$20x+38y+18z+1$

$=20(a^2-y)+38y+18(b^2-y)+1$

$=20a^2+18b^2+1$

Also,

$9y+9z+2=9(y+z)+2=9b^2+2$

Now substitute $a=\frac{3}{2}$ and $b=\frac{1}{3}$.

Numerator:

$20a^2+18b^2+1=20\left(\frac{3}{2}\right)^2+18\left(\frac{1}{3}\right)^2+1$

$=20\cdot \frac{9}{4}+18\cdot \frac{1}{9}+1$

$=45+2+1$

$=48$

Denominator:

$9b^2+2=9\left(\frac{1}{3}\right)^2+2$

$=1+2$

$=3$

Therefore,

$\sqrt{\frac{20x+38y+18z+1}{9y+9z+2}}=\sqrt{\frac{48}{3}}$

$=\sqrt{16}$

$=4$

Given equation is

$\left(\sum_{k=1}^{n}(-1)^{k-1}k\right)^2-\sum_{k=1}^{n}(-1)^{k-1}k^2+2450=0$

Check for odd $n$.

Let

$n=2m-1$

Then,

$\sum_{k=1}^{n}(-1)^{k-1}k=1-2+3-4+\cdots+(2m-1)$

Pairing terms,

$(1-2)+(3-4)+\cdots+[(2m-3)-(2m-2)]+(2m-1)$

$=-(m-1)+(2m-1)$

$=m$

So,

$\left(\sum_{k=1}^{n}(-1)^{k-1}k\right)^2=m^2$

Now,

$\sum_{k=1}^{n}(-1)^{k-1}k^2=1^2-2^2+3^2-4^2+\cdots+(2m-1)^2$

For odd $n=2m-1$,

$\sum_{k=1}^{n}(-1)^{k-1}k^2=m(2m-1)$

Now put in the equation:

$m^2-m(2m-1)+2450=0$

$m^2-2m^2+m+2450=0$

$-m^2+m+2450=0$

$m^2-m-2450=0$

Now factorize:

$m^2-m-2450=0$

$m^2-50m+49m-2450=0$

$m(m-50)+49(m-50)=0$

$(m-50)(m+49)=0$

Since $m$ is positive,

$m=50$

Therefore,

$n=2m-1$

$n=2(50)-1$

$n=99$

Given equations are:

$x+y+z=1$ .....$(1)$

$4x+9y+16z=25$ .....$(2)$

$16x+81y+256z=625$ .....$(3)$

Now subtract $4\times(1)$ from $(2)$:

$4x+9y+16z-4x-4y-4z=25-4$

$5y+12z=21$ .....$(4)$

Now subtract $16\times(1)$ from $(3)$:

$16x+81y+256z-16x-16y-16z=625-16$

$65y+240z=609$ .....$(5)$

Multiply equation $(4)$ by $13$:

$65y+156z=273$ .....$(6)$

Now subtract $(6)$ from $(5)$:

$65y+240z-(65y+156z)=609-273$

$84z=336$

$z=4$

Put $z=4$ in equation $(4)$:

$5y+12(4)=21$

$5y+48=21$

$5y=-27$

$y=-\frac{27}{5}$

Now use equation $(1)$:

$x+y+z=1$

$x-\frac{27}{5}+4=1$

$x-\frac{27}{5}=-3$

$x=-3+\frac{27}{5}$

$x=\frac{-15+27}{5}$

$x=\frac{12}{5}$

So,

$x=\frac{36}{15}$

A point equidistant from the coordinate axes satisfies:

$|x|=|y|$

So, either

$y=x$

or

$y=-x$

Given line is:

$8x-3y=11$

Case 1:

$y=x$

Put $y=x$ in the line:

$8x-3x=11$

$5x=11$

$x=\frac{11}{5}$

This is not an integer, so this case is rejected.

Case 2:

$y=-x$

Put $y=-x$ in the line:

$8x-3(-x)=11$

$8x+3x=11$

$11x=11$

$x=1$

Then,

$y=-1$

So, the point is:

$(1,-1)$

Here $x>0$ and $y<0$, so the point lies in the IV quadrant.

Since the given lines are diameters of the circle, both lines pass through the centre of the circle.

So, the centre is the intersection point of

$2x+3y=1$ .....$(1)$

$4x-3y=11$ .....$(2)$

Adding $(1)$ and $(2)$,

$6x=12$

$x=2$

Put $x=2$ in $(1)$:

$2(2)+3y=1$

$4+3y=1$

$3y=-3$

$y=-1$

So, centre of the circle is

$(2,-1)$

Now, area of circle is

$\pi r^2=153.94$

Since $153.94\approx 49\pi$,

$r^2=49$

$r=7$

Equation of circle is

$(x-2)^2+(y+1)^2=7^2$

$(x-2)^2+(y+1)^2=49$

Expanding,

$x^2-4x+4+y^2+2y+1=49$

$x^2+y^2-4x+2y-44=0$

Let the common tangent be

$y=mx+c$

For the parabola

$y=-x^2$

we have

$-x^2=mx+c$

$x^2+mx+c=0$

For tangency, discriminant must be zero.

$m^2-4c=0$

So,

$c=\frac{m^2}{4}$ .....$(1)$

Now, for the parabola

$y=(x-2)^2$

we have

$(x-2)^2=mx+c$

$x^2-4x+4=mx+c$

$x^2-(m+4)x+(4-c)=0$

For tangency,

$(m+4)^2-4(4-c)=0$

Substitute $c=\frac{m^2}{4}$:

$(m+4)^2-16+4\cdot \frac{m^2}{4}=0$

$(m+4)^2-16+m^2=0$

$m^2+8m+16-16+m^2=0$

$2m^2+8m=0$

$2m(m+4)=0$

So,

$m=0$ or $m=-4$

From the given options, $m=-4$ is possible.

Using $(1)$,

$c=\frac{(-4)^2}{4}$

$c=4$

So, the common tangent is

$y=-4x+4$

Given,

$h(x)=f[g(x)]$

Now, $f(t)=1$ when

$|t|\le 1$

So, $h(x)=1$ when

$|g(x)|\le 1$

For $|x|\le 2$,

$g(x)=2-x^2$

So,

$|2-x^2|\le 1$

This gives

$-1\le 2-x^2\le 1$

Subtract $2$ from all sides:

$-3\le -x^2\le -1$

Multiplying by $-1$ reverses the inequalities:

$1\le x^2\le 3$

So,

$1\le |x|\le \sqrt{3}$

For $|x|>2$,

$g(x)=2$

So,

$|g(x)|=2>1$

Hence, this case is not valid.

Therefore, the required interval is

$1\le |x|\le \sqrt{3}$

Given limit is

$\lim_{x\to\infty}\left(\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}\right)$

Divide numerator and denominator inside the radical by $x$.

$\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}$

$=\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

Now,

$\frac{\sqrt{x+\sqrt{x}}}{x}\to 0$ as $x\to\infty$

Therefore,

$\lim_{x\to\infty}\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

$=\sqrt{\frac{1}{1+0}}$

$=1$

Given curves are:

$y=(x-2)^2$

and

$y=-4+6x-x^2$

At the point of intersection,

$(x-2)^2=-4+6x-x^2$

$x^2-4x+4=-4+6x-x^2$

$2x^2-10x+8=0$

$x^2-5x+4=0$

$(x-1)(x-4)=0$

So,

$x=1$ or $x=4$

Now, slopes of the curves are:

For $y=(x-2)^2$,

$m_1=\frac{dy}{dx}=2(x-2)$

For $y=-4+6x-x^2$,

$m_2=\frac{dy}{dx}=6-2x$

At $x=1$,

$m_1=2(1-2)=-2$

$m_2=6-2(1)=4$

Angle between two curves is given by:

$\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$

So,

$\tan\theta=\left|\frac{4-(-2)}{1+(-2)(4)}\right|$

$=\left|\frac{6}{1-8}\right|$

$=\left|\frac{6}{-7}\right|$

$=\frac{6}{7}$

Therefore,

$\theta=\tan^{-1}\left(\frac{6}{7}\right)$

We have,

$\lim_{x\to 0}\frac{|x|\log_e(1+|\sin 2x|)}{x^2(|x|+3)}$

As $x\to 0$,

$|\sin 2x|\sim 2|x|$

Also,

$\log_e(1+|\sin 2x|)\sim |\sin 2x|$

So,

$\log_e(1+|\sin 2x|)\sim 2|x|$

Now the numerator becomes approximately:

$|x|\cdot 2|x|=2x^2$

The denominator becomes approximately:

$x^2(|x|+3)\to 3x^2$

Therefore,

$\lim_{x\to 0}\frac{|x|\log_e(1+|\sin 2x|)}{x^2(|x|+3)}=\frac{2x^2}{3x^2}$

$=\frac{2}{3}$

Hence, the limit exists and is equal to $\frac{2}{3}$.