JEE MAIN Function Previous Year Questions (PYQs) – Page 1 of 16
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The position of a moving car at time t is given by f(t) = at2 + bt + c, t > 0, where a, b and c are realnumbers greater than 1. Then the average speed of the car over the time interval [t1, t2] isattained at the point :
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Solution
Let the domains of the functions $f(x)=\log_{4}\big(\log_{3}\big(\log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)\big)\big)$ and $g(x)=\sin^{-1}\left(\dfrac{7x+10}{x-2}\right)$ be $(\alpha,\beta)$ and $[\gamma,\delta]$, respectively. Then $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}$ is equal to
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Solution
For $f(x)=\log_{4}\big(\log_{3}\big(\log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)\big)\big)$
Condition:
$\log_{3}(\cdot)>0 \Rightarrow \log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)>1$
$\Rightarrow 8-\log_{2}(x^{2}+4x+5)>7$
$\Rightarrow \log_{2}(x^{2}+4x+5)<1$
$\Rightarrow x^{2}+4x+5<2$
$\Rightarrow (x+2)^2<1$
$\Rightarrow -3$
So $(\alpha,\beta)=(-3,-1)$
For $g(x)=\sin^{-1}\left(\frac{7x+10}{x-2}\right)$
Condition: $-1\le \frac{7x+10}{x-2}\le 1$
Solve:
$\frac{7x+10}{x-2}\le 1 \Rightarrow \frac{6x+12}{x-2}\le 0 \Rightarrow \frac{x+2}{x-2}\le 0$
$\Rightarrow -2\le x<2$
$\frac{7x+10}{x-2}\ge -1 \Rightarrow \frac{8x+8}{x-2}\ge 0 \Rightarrow \frac{x+1}{x-2}\ge 0$
$\Rightarrow x\le -1 \text{ or } x>2$
Intersection: $[-2,-1]$
So $(\gamma,\delta)=(-2,-1)$
Now:
$\alpha^2+\beta^2+\gamma^2+\delta^2=9+1+4+1=15$
$\boxed{15}$
Condition:
$\log_{3}(\cdot)>0 \Rightarrow \log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)>1$
$\Rightarrow 8-\log_{2}(x^{2}+4x+5)>7$
$\Rightarrow \log_{2}(x^{2}+4x+5)<1$
$\Rightarrow x^{2}+4x+5<2$
$\Rightarrow (x+2)^2<1$
$\Rightarrow -3$
So $(\alpha,\beta)=(-3,-1)$
For $g(x)=\sin^{-1}\left(\frac{7x+10}{x-2}\right)$
Condition: $-1\le \frac{7x+10}{x-2}\le 1$
Solve:
$\frac{7x+10}{x-2}\le 1 \Rightarrow \frac{6x+12}{x-2}\le 0 \Rightarrow \frac{x+2}{x-2}\le 0$
$\Rightarrow -2\le x<2$
$\frac{7x+10}{x-2}\ge -1 \Rightarrow \frac{8x+8}{x-2}\ge 0 \Rightarrow \frac{x+1}{x-2}\ge 0$
$\Rightarrow x\le -1 \text{ or } x>2$
Intersection: $[-2,-1]$
So $(\gamma,\delta)=(-2,-1)$
Now:
$\alpha^2+\beta^2+\gamma^2+\delta^2=9+1+4+1=15$
$\boxed{15}$
If the maximum value of $a$, for which the function $f_a(x)=\tan^{-1}(2x)-3ax+7$ is non-decreasing in $\left(-\tfrac{\pi}{6},\,\tfrac{\pi}{6}\right)$, is $\bar a$, then $f_{\bar a}\!\left(\tfrac{\pi}{8}\right)$ is equal to :
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Solution
If f(x + y) = f(x)f(y) and $\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$ , x, y $ \in $ N, where N is the set of all natural number, then thevalue of${{f\left( 4 \right)} \over {f\left( 2 \right)}}$ is :
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Solution
Let $f(x)=3\sqrt{x-2}+\sqrt{4-x}$ be a real-valued function.
If $\alpha$ and $\beta$ are respectively the minimum and maximum values of $f$,
then $\alpha^2+2\beta^2$ is equal to:
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Solution
Let $[ \cdot ]$ denote the greatest integer function and
$f(x) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right]$. Then $12\sum_{j=1}^{\infty} f(j)$ is equal to:
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Solution
$\sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] \le \sum_{k=1}^{n} \frac{k^2}{3^x}$
$= \frac{n(n+1)(2n+1)}{6\cdot 3^x}$
$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] = \frac{1}{3^{x+1}}$
$\Rightarrow f(x) = \frac{1}{3^{x+1}}$
$12\sum_{j=1}^{\infty} f(j) = 12\sum_{j=1}^{\infty} \frac{1}{3^{j+1}}$
$= 12\left(\frac{1}{9} + \frac{1}{27} + \cdots\right)$
$= 12\left(\frac{1/9}{1 - 1/3}\right) = 2$
Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $ \in $ R. If f(x) attains maximum value at $\alpha $ and g(x) attains
minimum value at $\beta $, then
$\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$ is equal to :
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Solution
For a suitably chosen real constant a, let afunction, $f:R - \left\{ { - a} \right\} \to R$ be defined by$f(x) = {{a - x} \over {a + x}}$. Further suppose that for any realnumber $x \ne - a$ and $f(x) \ne - a$, (fof)(x) = x. Then $f\left( { - {1 \over 2}} \right)$ is equal to :
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Solution
Let the range of the function
$f(x)=6+16\cos x\cdot \cos\!\left(\frac{\pi}{3}-x\right)\cdot \cos\!\left(\frac{\pi}{3}+x\right)\cdot \sin 3x\cdot \cos 6x,\ x\in\mathbb{R}$
be $[\alpha,\beta]$. Then the distance of the point $(\alpha,\beta)$ from the line $3x+4y+12=0$ is:
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Solution
If $a\in\mathbb{R}$ and the equation $-3(x-[x])^{2}+2(x-[x])+a^{2}=0$
(where $[x]$ denotes the greatest integer $\le x$) has no integral solution,
then all possible values of $a$ lie in the interval :
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Solution
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