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Day 189: JEE Mains Maths

5 questions. Try each one first, then open its answer.

  1. Question 1

    Sequences and Series. Asked once: 6 Apr 2023 (same type asked 3 times) See all 3 of this type

    75
    Let a1,a2,a3…ana_{1},a_{2},a_{3}\ldots a_{n} be nn positive consecutive terms of an arithmetic progression. If d>0d > 0 is its common difference, then
    lim⁡n→∞dn(1a1+a2+1a2+a3+……….+1an−1+an)\lim_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left( \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+ \ldots\ldots\ldots. +\frac{1}{\sqrt{a_{n - 1}}+\sqrt{a_{n}}} \right)

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  2. Question 2

    Differential Equations. Asked once: 1 Feb 2024 (same type asked 3 times) See all 3 of this type

    74
    Let y=y(x)y = y(x) be the solution of the differential equation dydx=2x(x+y)3−x(x+y)−1,y(0)=1\frac{dy}{dx}= 2x(x + y)^{3}- x(x + y) - 1,y(0) = 1. Then, (12+y(12))2\left( \frac{1}{\sqrt{2}}+ y\left( \frac{1}{\sqrt{2}} \right) \right)^{2} equals :

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  3. Question 3

    Limits and Continuity. Asked once: 27 Jul 2022 (same type asked 3 times) See all 3 of this type

    155
    If for p≠q≠0p \neq q \neq 0, then function f(x)=p(729+x)7−3729+qx3−9f(x) =\frac{\sqrt[7]{p(729 + x)}- 3}{\sqrt[3]{729 + qx}- 9} is continuous at x=0x = 0, then:

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  4. Question 4

    Quadratic Equations. Asked once: 28 Jul 2022 (same type asked 3 times) See all 3 of this type

    152
    Let α,β\alpha,\beta be the roots of the equation x2−2x+6=0x^{2}-\sqrt{2}x +\sqrt{6}= 0 and 1α2+1,1β2+1\frac{1}{\alpha^{2}}+ 1,\frac{1}{\beta^{2}}+ 1 be the roots of the equation x2+ax+b=0x^{2}+ ax + b = 0. Then the roots of the equation x2−(a+b−2)x+(a+b+2)x^{2}- (a + b - 2)x + (a + b + 2) =0= 0 are:

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  5. Question 5

    Relations and Functions. Asked once: 30 Jan 2023 (same type asked 3 times) See all 3 of this type

    87
    Let f1(x)=3x+22x+3,x∈R−{−32}f^{1}\left( x \right)=\frac{3x + 2}{2x + 3},x \in R -\left\{ \frac{- 3}{2} \right\}For n≥2n \geq 2, define fn(x)=f1f^{n}(x) =f^{1} of fn−1(x)f^{n - 1}\left( x \right)If f5(x)=ax+bbx+a,gcd(a,b)=1f^{5}(x) =\frac{ax + b}{bx + a},gcd(a,b) = 1, then a+ba + b is equal to