Mathematics Test

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Mathematics Test - 39

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Question 1
1 / -0

Let $f(x)=e^{(x-1)}-a x^{2}+b \& g(x)=\left[\begin{array}{ll}e^{x-1} & \text { for } x \leq 1 \\ x^{2}+1 & \text { for } x>1\end{array}\right.$ with $f^{\prime}(1)=2 \cdot$ The values of a and $b$ for which the function $h(x)=f(x) \cdot g(x),$ is differentiable at $x=1,$ are respectively

A
$-\frac{1}{2},-\frac{3}{2}$

B
$-\frac{1}{2}, \frac{3}{2}$

C
$-\frac{1}{2},-2$

D
none of these

Solution

$f(x)=e^{(x-1)}-a x^{2}+b$
$\Rightarrow f^{\prime}(x)=e^{(x-1)}-2 a x$
$\Rightarrow f^{\prime}(1)=e^{0}-2 a=1-2 a$
$\therefore f^{\prime}(1)=2 \Rightarrow 1-2 \mathrm{a}=2 \Rightarrow \mathrm{a}=-\frac{1}{2}$
$\therefore f(x)=e^{(x-1)}+\frac{x^{2}}{2}+b$
$\mathrm{h}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x}) \Rightarrow \mathrm{h}(\mathrm{x})=\left(\mathrm{e}^{(\mathrm{x}-1)}+\frac{\mathrm{x}^{2}}{2}+\mathrm{b} \mathrm{e}^{(\mathrm{x}-1)}, \mathrm{x} \leq 1\right.$
$=\left(e^{(x-1)}+\frac{x^{2}}{2}+b\right)\left(x^{2}+1\right) x>1$
For $\mathrm{h}(\mathrm{x})$ to be differentiable at $\mathrm{x}=1$ it must be continuous at $\mathrm{x}=1$
$\therefore \lim _{h \rightarrow 0} h(1-h)=\lim _{h \rightarrow 0} h(1+h)=h(1)$
$\therefore \lim _{h \rightarrow 0} h(1-h)=\lim _{h \rightarrow 0} h(1+h)=h(1)$
$\Rightarrow\left(e^{0}+\frac{1}{2}+b\right) e^{0}=\left(e^{0}+\frac{1}{2}+b\right)(2)=\left(e^{0}+\frac{1}{2}+b\right) e^{0}$
$\Rightarrow b+\frac{3}{2}=2\left(b+\frac{3}{2}\right) \Rightarrow b+\frac{3}{2}=0 \Rightarrow b=-\frac{3}{2}$
$\therefore h(x)=\left(e^{x-1}+\frac{x^{2}}{2}-\frac{3}{2}\right) e^{x-1}, x \leq 1$
$=\left(e^{x-1}+\frac{x^{2}}{2}-\frac{3}{2}\right)\left(x^{2}+1\right), x>1$
$h^{\prime}(x)=\left(e^{x-1}+\frac{x^{2}}{2}-\frac{3}{2}\right) e^{(x-1)}+e^{(x-1)}\left(e^{x-1}+x\right), x<1$
$h^{\prime}(\mathrm{x})=\varphi, \mathrm{x}=1$
$h^{\prime}(x)=\left(e^{x-1}+\frac{x^{2}}{2}-\frac{3}{2}\right) 2 x+\left(x^{2}+1\right)\left(e^{x-1}+x\right), x>1$
$\left(\because h^{\prime}(1-)=2 \& h^{\prime}(1+)=4\right.$ therefore $\left.h^{\prime}(1)=\varphi\right)$
Hence there are no value of a $\& \mathrm{b}$ for which $\mathrm{h}(\mathrm{x})$ is differentiable at $\mathrm{x}=1$
Question 2
1 / -0

The function $f(x)=\frac{\log _{e}(\pi+x)}{\log _{e}(e+x)}$ is

A
Increasing on$(0, \infty)$

B
Decreasing on$(0, \infty)$

C
Increasing on $\left(0, \frac{\pi}{e}\right),$ decreasing on $\left(\frac{\pi}{e}, \infty\right)$

D
none of these

Solution

$f(x)=\frac{\ln (\pi+x)}{\ln (e+x)}$
$f^{\prime}(x)=\left[\frac{\ln (e+x)}{(\pi+x)}-\frac{\ln (\pi+x)}{(e+x)}\right] \times \frac{1}{(\ln (e+x))^{2}}$
$=\frac{(e+x) \ln (e+x)-(\pi+x) \ln (\pi+x)}{(\pi+x)(e+x)(\ln (e+x))^{2}}$
$\Rightarrow$ But $e<\pi$
$\Rightarrow(e+x)<(\pi+x) \forall x \in R$
$\Rightarrow \ln (e+x)<\ln (\pi+x),$ provided $e+x>0 \& \pi+x>0$
$\Rightarrow(e+x) \ln (e+x)<(\pi+x) \ln (\pi+x)$ for $x \in(-e, \infty)$
$\Rightarrow(e+x) \ln (e+x)-(\pi+x) \ln (\pi+x)<0$ for $x \in(-e, \infty)$
$\Rightarrow f^{\prime}(x)<0 \forall x \in(-e, \infty)$
$\Rightarrow f^{\prime}(x)<0 \forall x \in(0, \infty)$
Question 3
1 / -0

$\int x^{\mathrm{x}^{2}}\left(2 \mathrm{x} \log _{\mathrm{e}} \mathrm{x}+\mathrm{x}\right) \mathrm{dx}$ is equal

A
$\mathrm{x}^{\mathrm{x}^{2}}+\mathrm{C}$

B
$x^{\mathrm{x}^{\log x}}+C$

C
$x^{x} \cdot \log _{4} x+c$

D
none of these

Solution

$I=\int x^{x^{2}}\left(2 x \log _{e} x+x\right) d x$
Let $x^{\mathrm{x}^{2}}=t$
Take loge on both sides of the equation $\Rightarrow \ln t=x^{2} \ln x$
$\Rightarrow \frac{1}{t} \frac{d t}{d x}=x^{2} \frac{1}{x}+2 x \ln x$
$\Rightarrow \frac{1}{t} \mathrm{dt}=(\mathrm{x}+2 \mathrm{x} \ln \mathrm{x}) \mathrm{dx}$
$\therefore I=\int t \frac{1}{t} d t=\int d t=t+C$
$x^{\mathrm{x}^{2}}+\mathrm{C}$
Question 4
1 / -0

The value of the definite integral $\int_{0}^{1} \frac{d x}{\left(x^{2}+2 x \cos \alpha+1\right)}$ for $0<\alpha<\pi,$ equal to

A
$\sin \alpha$

B
$\tan ^{-1}(\sin \alpha)$

C
$\alpha \sin \alpha$

D
$\frac{\alpha}{2 \sin \alpha}$

Solution

$I=\int_{0}^{1} \frac{d x}{x^{2}+2 x \cos \alpha+1}=\int_{0}^{1} \frac{d x}{(x+\cos \alpha)^{2}+\sin ^{2} \alpha}$
$=\frac{1}{\sin \alpha}\left[\tan ^{-}\left(\frac{x+\cos \alpha}{\sin \alpha}\right)\right]_{0}^{1}$
$=\frac{1}{\sin \alpha}\left[\tan ^{-1}\left(\frac{1+\cos \alpha}{\sin \alpha}\right)-\tan ^{-1}\left(\frac{\cos \alpha}{\sin \alpha}\right)\right]$
$=\frac{1}{\sin \alpha}\left[\tan ^{-1} \cot \alpha / 2-\tan ^{-1} \cot \alpha\right]$
$=\frac{1}{\sin \alpha}\left[\left(\frac{\pi}{2}-\frac{\alpha}{2}\right)-\left(\frac{\pi}{2}-\alpha\right)\right]=\frac{\alpha}{2 \sin \alpha}$
Question 5
1 / -0

The solution to $\frac{d y}{d x}+\frac{\tan y}{x}=\frac{1}{x^{2}} \tan y \sin y$ is :

A
$\left[\frac{1}{2 x}-\frac{1}{\sin y}\right]=C$

B
$\frac{1}{x}\left[\frac{1}{2 x}-\frac{1}{\sin y}\right]=C$

C
$\frac{1}{x}\left[\frac{1}{2 x}+\frac{1}{\sin y}\right]=C$

D
$\left[\frac{1}{2 x}+\frac{1}{\sin y}\right]=C$

Solution

$\frac{d y}{d x}+\frac{\tan y}{x}=\frac{1}{x^{2}} \tan y \cdot \sin y$
$\Rightarrow \cos \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{1}{\mathrm{x}}\right) \sin \mathrm{y}=\frac{1}{\mathrm{x}^{2}}(\sin \mathrm{y})^{2},$ put $\sin \mathrm{y}=\mathrm{z}$
$\Rightarrow \frac{d z}{d x}+\left(\frac{1}{x}\right) z=\left(\frac{1}{x^{2}}\right) z^{2}$
$\Rightarrow \frac{1}{z^{2}} \frac{d z}{d x}+\left(\frac{1}{x}\right)\left(\frac{1}{z}\right)=\frac{1}{x^{2}}$
Let $\frac{1}{z^{2}} \frac{d z}{d x}=\frac{d R}{d x} \Rightarrow \int \frac{d z}{z^{2}}=\int d R \Rightarrow R=\frac{-1}{z}$
$\Rightarrow \frac{d R}{d x}+\left(\frac{1}{x}\right)(-R)=\frac{1}{x^{2}}, \quad I=e^{\int-\frac{1}{x} d x}=\frac{1}{x}$
$\Rightarrow \frac{1}{x} \frac{d R}{d x}+\left(-\frac{1}{x^{2}}\right) R=\frac{1}{x^{3}}$
$\Rightarrow \int \mathrm{d}\left(\frac{\mathrm{R}}{\mathrm{x}}\right)=\int \frac{\mathrm{dx}}{\mathrm{x}^{3}} \Rightarrow \frac{\mathrm{R}}{\mathrm{x}}=\frac{-1}{2 \mathrm{x}^{2}}+\mathrm{C}$
$\Rightarrow \frac{-1}{(\sin y) x}=\frac{-1}{2 x^{2}}+C \Rightarrow \frac{1}{x}\left[\frac{1}{2 x}-\frac{1}{\sin y}\right]=C$
Question 6
1 / -0

In a triangle ABC, the angle B is greater than angle A . If the values of the angle A and B satisfy the equation $3 \sin x-4 \sin ^{3} x-k=0$, 0 < k < 1, then value of C is

A
$\frac{π}{3}$

B
$\frac{π}{2}$

C
$\frac{2 π}{3}$

D
$\frac{5 π}{6}$

Solution

$\sin 3 x=k$ is satisfied by $A \& B$
$\Rightarrow \sin 3 A=k \& \sin 3 B=k$
$\Rightarrow \sin 3 A=\sin 3 B \Rightarrow 3 A=3 B$ or $3 A=\pi-3 B$
But $3 \mathrm{A}=3 \mathrm{B}$ rejected $(\because \mathrm{B}>\mathrm{A})$
$\Rightarrow 3 \mathrm{A}=\pi-3 \mathrm{B} \Rightarrow \mathrm{A}+\mathrm{B}=\frac{\pi}{3} \Rightarrow \mathrm{C}=\frac{2 \pi}{3}$
Question 7
1 / -0

Coordinates of the orthocentre of the triangle whose sides are x=3, y=4 and 3x+4y=6, will be:

A
(0,0)

B
(3,0)

C
(0,4)

D
(3,4)

Solution

Sides of a triangle ABC are given byx=3, y=4 and 3x+4y=6,. It forms a right angle triangle ABC with B(3,4)as right angle.

Hence B is the orthocentre as perpendiculars drawn from A and C meet at B.
Question 8
1 / -0

The area bounded by the curves $x+2|y|=1$ and $x=0$ is:

A
1/3

B
1/2

C
2

D
3

Solution
Question 9
1 / -0

none of these equation of the hyperbola whose foci are (6,5), (-4,5)and eccentricity $\frac{5}{4}$ is

A
$\frac{(x-1)^{2}}{16}-\frac{(y-5)^{2}}{9}=1$

B
$\frac{(x+1)^{2}}{25}-\frac{(y+5)^{2}}{16}=1$

C
$\frac{(x-1)^{2}}{16}-\frac{(y-5)^{2}}{36}=1$

D
$\frac{(x-1)^{2}}{25}-\frac{(y-5)^{2}}{16}=1$

Solution

Center of the hyperbola is the mid-point of the line segment joining two foci. Therefore, coordinates of the center are (1,5). Also since the foci have same y coordinate therefore the line joining foci is parallel to x-axis. Hence the transverse axis is parallel to x-axis. Clearly the$\frac{(x-1)^{2}}{a^{2}}-\frac{(y-5)^{2}}{b^{2}}=1$

Again the distance between the foci $=10$

$\Rightarrow 2$ ae $=10 \Rightarrow$ ae $=5$

$\Rightarrow a=4$

$\therefore b^{2}=a^{2}\left(e^{2}-1\right) \Rightarrow b=3$

Hence, the equation of the hyperbola is

$\frac{(x-1)^{2}}{16}-\frac{(y-5)^{2}}{9}=1$
Question 10
1 / -0

An equilateral triangle is inscribed in the parabola y²=4axwhose vertex is at the vertex of the parabola. The length of its side is

A
$2 a \sqrt{3}$

B
$4 a \sqrt{3}$

C
6 $6 a \sqrt{3}$

D
$8 a \sqrt{3}$

Solution

$\left(\cos 30^{\circ},\left(\sin 30^{\circ}\right)\right.$

$\equiv\left(\left(\frac{\sqrt{3}}{2}, \frac{l}{2}\right)\right.$ lies on $y^{2}=4 a x$

Thus, the point will satisfy the equation of the parabola $\left(\frac{l}{2}\right)^{2}=4 a\left(\frac{l \sqrt{3}}{2}\right)$

$\therefore C=8 \mathrm{a} \sqrt{3}$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 39

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Mathematics Test - 39

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SHARING IS CARING

Question 1

\(-\frac{1}{2},-\frac{3}{2}\)

\(-\frac{1}{2}, \frac{3}{2}\)

\(-\frac{1}{2},-2\)

none of these

Solution

Question 2

Increasing on\((0, \infty)\)

Decreasing on\((0, \infty)\)

Increasing on \(\left(0, \frac{\pi}{e}\right),\) decreasing on \(\left(\frac{\pi}{e}, \infty\right)\)

none of these

Solution

Question 3

\(\mathrm{x}^{\mathrm{x}^{2}}+\mathrm{C}\)

\(x^{\mathrm{x}^{\log x}}+C\)

\(x^{x} \cdot \log _{4} x+c\)

none of these

Solution

Question 4

\(\sin \alpha\)

\(\tan ^{-1}(\sin \alpha)\)

\(\alpha \sin \alpha\)

\(\frac{\alpha}{2 \sin \alpha}\)

Solution

Question 5

\(\left[\frac{1}{2 x}-\frac{1}{\sin y}\right]=C\)

\(\frac{1}{x}\left[\frac{1}{2 x}-\frac{1}{\sin y}\right]=C\)

\(\frac{1}{x}\left[\frac{1}{2 x}+\frac{1}{\sin y}\right]=C\)

\(\left[\frac{1}{2 x}+\frac{1}{\sin y}\right]=C\)

Solution

Question 6

π3

π2

2π3

5π6

Solution

Question 7

(0,0)

(3,0)

(0,4)

(3,4)

Solution

Question 8

1/3

1/2

2

3

Solution

Question 9

\(\frac{(x-1)^{2}}{16}-\frac{(y-5)^{2}}{9}=1\)

\(\frac{(x+1)^{2}}{25}-\frac{(y+5)^{2}}{16}=1\)

\(\frac{(x-1)^{2}}{16}-\frac{(y-5)^{2}}{36}=1\)

\(\frac{(x-1)^{2}}{25}-\frac{(y-5)^{2}}{16}=1\)

Solution

Question 10

2a3

4a3

66a3

8a3

Solution

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$\frac{π}{3}$

$\frac{π}{2}$

$\frac{2 π}{3}$

$\frac{5 π}{6}$

$2 a \sqrt{3}$

$4 a \sqrt{3}$

6 $6 a \sqrt{3}$

$8 a \sqrt{3}$