JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 11 of 275
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The value of $\log_{e}2 \, \dfrac{d}{dx}\!\big(\log_{\cos x} \csc x\big)$ at $x=\tfrac{\pi}{4}$ is:
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Solution
Let $\vec a=5\hat{\imath}-\hat{\jmath}-3\hat{k}$ and $\vec b=\hat{\imath}+3\hat{\jmath}+5\hat{k}$ be two vectors. Then which one of the following statements is TRUE?
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Solution
Let $f(x)=\int_{0}^{x}\left(t+\sin(1-e^{t})\right)dt,\;x\in\mathbb{R}$.
Then, $\displaystyle\lim_{x\to0}\dfrac{f(x)}{x^{3}}$ is equal to:
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Solution
The length of the chord of the ellipse $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{2}=1$ whose midpoint is $\left(1,\dfrac{1}{2}\right)$ is:
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Solution
Given ellipse: $\frac{x^2}{4}+\frac{y^2}{2}=1$
Midpoint $(1,\frac{1}{2})$
Use midpoint (chord) formula: $T=S_1$
$\frac{xx_1}{4}+\frac{yy_1}{2}= \frac{x_1^2}{4}+\frac{y_1^2}{2}$
RHS:
$\frac{1}{4}+\frac{1/4}{2}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$
So chord equation:
$\frac{x}{4}+\frac{y}{4}=\frac{3}{8}$
⇒ $2x+2y=3$ ⇒ $x+y=\frac{3}{2}$
Now find intersection with ellipse:
$y=\frac{3}{2}-x$
Substitute:
$\frac{x^2}{4}+\frac{(3/2-x)^2}{2}=1$
Multiply by 4:
$x^2+2\left(\frac{9}{4}-3x+x^2\right)=4$
$x^2+\frac{9}{2}-6x+2x^2=4$
$3x^2-6x+\frac{9}{2}-4=0$
$3x^2-6x+\frac{1}{2}=0$
$6x^2-12x+1=0$
Roots difference:
$|x_1-x_2|=\frac{\sqrt{144-24}}{6}$
Midpoint $(1,\frac{1}{2})$
Use midpoint (chord) formula: $T=S_1$
$\frac{xx_1}{4}+\frac{yy_1}{2}= \frac{x_1^2}{4}+\frac{y_1^2}{2}$
RHS:
$\frac{1}{4}+\frac{1/4}{2}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$
So chord equation:
$\frac{x}{4}+\frac{y}{4}=\frac{3}{8}$
⇒ $2x+2y=3$ ⇒ $x+y=\frac{3}{2}$
Now find intersection with ellipse:
$y=\frac{3}{2}-x$
Substitute:
$\frac{x^2}{4}+\frac{(3/2-x)^2}{2}=1$
Multiply by 4:
$x^2+2\left(\frac{9}{4}-3x+x^2\right)=4$
$x^2+\frac{9}{2}-6x+2x^2=4$
$3x^2-6x+\frac{9}{2}-4=0$
$3x^2-6x+\frac{1}{2}=0$
$6x^2-12x+1=0$
Roots difference:
$|x_1-x_2|=\frac{\sqrt{144-24}}{6}$
$=\frac{\sqrt{120}}{6}$
$=\frac{2\sqrt{30}}{6}$
$=\frac{\sqrt{30}}{3}$
Since slope = $-1$,
Since slope = $-1$,
length of chord:
$L=\sqrt{1+1}\cdot |x_1-x_2|$
$L=\sqrt{1+1}\cdot |x_1-x_2|$
$=\sqrt{2}\cdot \frac{\sqrt{30}}{3}$
$=\frac{\sqrt{60}}{3}$
$=\frac{2\sqrt{15}}{3}$
If a curve y=y(x) passes through the point $\left(1,\dfrac{\pi}{2}\right)$ and satisfies the differential equation $(7x^{4}\cot y-e^{x}\csc y),\dfrac{dx}{dy}=x^{5},\ x\ge1$, then at $x=2$, the value of $\cos y$ is:
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Solution
Given:
$(7x^{4}\cot y - e^{x}\csc y)\frac{dx}{dy}=x^{5}$
⇒ $\frac{dy}{dx}=\frac{7x^{4}\cot y - e^{x}\csc y}{x^{5}}$
Multiply by $\sin y$:
$\sin y\frac{dy}{dx}=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$
Now note:
$\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}$
So,
$-\frac{d}{dx}(\cos y)=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$
⇒ $\frac{d}{dx}(\cos y)+\frac{7}{x}\cos y=\frac{e^{x}}{x^{5}}$
This is linear in $\cos y$
IF $=e^{\int \frac{7}{x}dx}=x^{7}$
$\frac{d}{dx}(x^{7}\cos y)=x^{2}e^{x}$
Integrate:
$x^{7}\cos y=\int x^{2}e^{x}dx$
$\int x^{2}e^{x}dx=e^{x}(x^{2}-2x+2)$
So,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)+C$
Use $(1,\frac{\pi}{2})$ ⇒ $\cos y=0$
$0=e(1-2+2)+C=e+C$ ⇒ $C=-e$
Thus,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)-e$
At $x=2$:
$2^{7}\cos y=e^{2}(4-4+2)-e=2e^{2}-e$
$\cos y=\frac{2e^{2}-e}{128}=\frac{e(2e-1)}{128}$
$\boxed{\frac{e(2e-1)}{128}}$
⇒ $\frac{dy}{dx}=\frac{7x^{4}\cot y - e^{x}\csc y}{x^{5}}$
Multiply by $\sin y$:
$\sin y\frac{dy}{dx}=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$
Now note:
$\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}$
So,
$-\frac{d}{dx}(\cos y)=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$
⇒ $\frac{d}{dx}(\cos y)+\frac{7}{x}\cos y=\frac{e^{x}}{x^{5}}$
This is linear in $\cos y$
IF $=e^{\int \frac{7}{x}dx}=x^{7}$
$\frac{d}{dx}(x^{7}\cos y)=x^{2}e^{x}$
Integrate:
$x^{7}\cos y=\int x^{2}e^{x}dx$
$\int x^{2}e^{x}dx=e^{x}(x^{2}-2x+2)$
So,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)+C$
Use $(1,\frac{\pi}{2})$ ⇒ $\cos y=0$
$0=e(1-2+2)+C=e+C$ ⇒ $C=-e$
Thus,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)-e$
At $x=2$:
$2^{7}\cos y=e^{2}(4-4+2)-e=2e^{2}-e$
$\cos y=\frac{2e^{2}-e}{128}=\frac{e(2e-1)}{128}$
$\boxed{\frac{e(2e-1)}{128}}$
The value of $\displaystyle \int_{-\pi/2}^{\pi/2} \dfrac{dx}{[x] + [\sin x] + 4}$,
where $[t]$ denotes the greatest integer less than or equal to $t$, is :
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Solution
If $\alpha, \beta$ and $\gamma$ are three consecutive terms of a non-constant G.P. such that the equations
$\alpha x^{2} + 2\beta x + \gamma = 0$ and $x^{2} + x - 1 = 0$
have a common root, then $\alpha(\beta + \gamma)$ is equal to:
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Solution
Let $A$ and $B$ be two events such that $P(\overline{A\cup B})=\dfrac{1}{6}$,
$P(A\cap B)=\dfrac{1}{4}$ and $P(\overline{A})=\dfrac{1}{4}$, where $\overline{A}$
stands for the complement of the event $A$. Then the events $A$ and $B$ are :
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Solution
The largest $n \in \mathbb{N}$, for which $7^n$ divides $101!$, is:
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Solution
Exponent of $7$ in $101!$
$= \left\lfloor \frac{101}{7} \right\rfloor + \left\lfloor \frac{101}{7^2} \right\rfloor + \left\lfloor \frac{101}{7^3} \right\rfloor + \cdots$
$= 14 + 2 = 16$
A ray of light coming from the point (2, $2\sqrt 3 $) is incident at an angle 30o on the line x = 1 at thepoint A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line ABpasses through the point :
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Solution
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