JEE MAIN 2026 Previous Year Questions (PYQs) – Page 6 of 25
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The area of the region $A=\{(x,y):4x^2+y^2\le 8 \text{ and } y^2\le 4x\}$ is :
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Solution
From $y^2\le 4x$, we have $x\ge \dfrac{y^2}{4}$. From $4x^2+y^2\le 8$, right boundary is $x=\dfrac12\sqrt{8-y^2}$. Intersection occurs when $4\left(\dfrac{y^2}{4}\right)^2+y^2=8\Rightarrow y^2=4$, so $-2\le y\le 2$. Required area $=\int_{-2}^{2}\left(\dfrac12\sqrt{8-y^2}-\dfrac{y^2}{4}\right)dy = \int_0^2\sqrt{8-y^2}\,dy-\int_0^2\dfrac{y^2}{2}\,dy = (\pi+2)-\dfrac43 = \pi+\dfrac23$.
Let $\alpha,\beta$ be the roots of the quadratic equation $12x^2-20x+3\lambda=0$, $\lambda\in \mathbb{Z}$. If $\dfrac12\le |\beta-\alpha|\le \dfrac32$, then the sum of all possible values of $\lambda$ is :
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Solution
For $12x^2-20x+3\lambda=0$, $|\beta-\alpha|=\dfrac{\sqrt{D}}{12}=\dfrac{\sqrt{400-144\lambda}}{12}=\dfrac13\sqrt{25-9\lambda}$. Given $\dfrac12\le \dfrac13\sqrt{25-9\lambda}\le \dfrac32$. So $\dfrac32\le \sqrt{25-9\lambda}\le \dfrac92$. Squaring gives $\dfrac94\le 25-9\lambda\le \dfrac{81}{4}$. Hence $\dfrac{19}{36}\le \lambda\le \dfrac{91}{36}$. Since $\lambda\in\mathbb Z$, possible values are $1,2$. Their sum is $3$.
A function f(x) is given by $f(x) = {{{5^x}} \over {{5^x} + 5}}$, then the sum of the series $f\left( {{1 \over {20}}} \right) + f\left( {{2 \over {20}}} \right) + f\left( {{3 \over {20}}} \right) + ....... + f\left( {{{39} \over {20}}} \right)$ is equal to :
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Solution
Let the domain of the function $f(x)=\log_3\log_5\left(7-\log_2(x^2-10x+85)\right)+\sin^{-1}\left(\left|\dfrac{3x-7}{17-x}\right|\right)$ be $(\alpha,\beta]$. Then $\alpha+\beta$ is equal to :
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Solution
For $\log_3\log_5(7-\log_2(\cdots))$ to exist, we need $\log_5(7-\log_2(x^2-10x+85))>0$, so $7-\log_2(x^2-10x+85)>1$. Thus $\log_2(x^2-10x+85)<6\Rightarrow x^2-10x+21<0\Rightarrow 3
Let $[\,\cdot\,]$ denote the greatest integer function, and let $f(x)=\min\{\sqrt{2}\,x,x^2\}$. Let $S=\{x\in(-2,2):\text{ the function } g(x)=|x|[x^2] \text{ is discontinuous at }x\}$. Then $\sum_{x\in S} f(x)$ equals :
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Solution
The function $[x^2]$ changes value when $x^2$ crosses an integer. In $(-2,2)$, discontinuities occur at $x=\pm1$. Hence $S=\{-1,1\}$. Now $f(-1)=\min\{-\sqrt2,1\}=-\sqrt2$ and $f(1)=\min\{\sqrt2,1\}=1$. Therefore $\sum_{x\in S}f(x)=1-\sqrt2$.
Let $S$ and $S'$ be the foci of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ and $P(\alpha,\beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2+(S'P)^2-SP\cdot S'P=37$, then $\alpha^2+\beta^2$ is equal to :
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Solution
For the ellipse, $a=5$, $b=3$, so $c=4$. Let $SP=r_1$, $S'P=r_2$. Since $P$ lies on ellipse, $r_1+r_2=2a=10$. Given $r_1^2+r_2^2-r_1r_2=37$. Using $(r_1+r_2)^2=r_1^2+r_2^2+2r_1r_2=100$, we get $3r_1r_2=63\Rightarrow r_1r_2=21$, and hence $r_1^2+r_2^2=58$. But $r_1^2+r_2^2=2(\alpha^2+\beta^2+c^2)=2(\alpha^2+\beta^2+16)$. Therefore $2(\alpha^2+\beta^2+16)=58\Rightarrow \alpha^2+\beta^2=13$.
Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^2=4x$ be the curve $S$. Let $P$ be any point on $S$. Then the locus of the point, which internally divides $OP$ in the ratio $3:1$, is :
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Solution
A chord through origin has equation $y=mx$. Its second intersection with $y^2=4x$ is found from $m^2x^2=4x$, giving $x=\dfrac{4}{m^2}, y=\dfrac{4}{m}$. Hence midpoint is $\left(\dfrac{2}{m^2},\dfrac{2}{m}\right)$, so its locus is $y^2=2x$. Let $P(x_1,y_1)$ lie on this curve. If a point $Q(x,y)$ divides $OP$ internally in ratio $3:1$, then $Q=\left(\dfrac{3x_1}{4},\dfrac{3y_1}{4}\right)$. Using $y_1^2=2x_1$, we get $\left(\dfrac{4y}{3}\right)^2=2\left(\dfrac{4x}{3}\right)$, which simplifies to $2y^2=3x$.
Let $f(x)=[x]^2-[x+3]-3$, $x\in\mathbb R$, where $[\,\cdot\,]$ is the greatest integer function. Then
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Solution
Let $n=[x]$. Since $[x+3]=[x]+3=n+3$, we get $f(x)=n^2-(n+3)-3=n^2-n-6=(n-3)(n+2)$. This is negative for $-2
Let $f$ and $g$ be functions satisfying $f(x+y)=f(x)f(y)$, $f(1)=7$ and $g(x+y)=g(xy)$, $g(1)=1$, for all $x,y\in \mathbb N$. If $\sum_{x=1}^{n}\left(\dfrac{f(x)}{g(x)}\right)=19607$, then $n$ is equal to :
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Solution
From $f(x+y)=f(x)f(y)$ and $f(1)=7$, for natural numbers we get $f(n)=7^n$. From $g(x+y)=g(xy)$ and $g(1)=1$, putting $y=1$ gives $g(x+1)=g(x)$, so $g(n)=1$ for all $n\in\mathbb N$. Hence $\sum_{x=1}^{n}\dfrac{f(x)}{g(x)}=\sum_{x=1}^{n}7^x$. Now $7+49+343+2401+16807=19607$, so $n=5$.
Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1+x)^n$, $n\in \mathbb N$, $0\le r\le n$. If $P_n=C_0-C_1+\dfrac{2^2}{3}C_2-\dfrac{2^3}{4}C_3+\cdots+\dfrac{(-2)^n}{n+1}C_n$, then the value of $\sum_{n=1}^{25}\dfrac{1}{P_{2n}}$ equals.
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Solution
We have $P_n=\sum_{r=0}^{n}\dfrac{(-2)^r}{r+1}C_r$. Using $\sum_{r=0}^{n} C_r\dfrac{a^{r+1}}{r+1}=\dfrac{(1+a)^{n+1}-1}{n+1}$ with $a=-2$, we get $P_n=-\dfrac12\cdot\dfrac{(-1)^{n+1}-1}{n+1}$. For even $n=2m$, $(-1)^{2m+1}=-1$, so $P_{2m}=\dfrac{1}{2m+1}$. Therefore $\dfrac{1}{P_{2n}}=2n+1$. Hence $\sum_{n=1}^{25}\dfrac{1}{P_{2n}}=\sum_{n=1}^{25}(2n+1)=2\cdot\dfrac{25\cdot26}{2}+25=650+25=675$.
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