ISC Class 12 Mathematics Question Paper - 2019 Board Examinations

Question 1

(a) If  \[(\vec{a})\] and \[(\vec{b})\] are perpendicular vectors,  \[\left( \left| \vec{a}+\vec{b} \right|=13 \right)13\text{ and }\left( \left| {\vec{a}} \right| \right)=5.\]

find the value of  \[\left( \left| {\vec{b}} \right| \right)\]

(b) Find the length of the perpendicular from origin to the plane $(\vec{r}\cdot \left( 3\hat{i}-4\hat{j}-12\hat{k} \right)+39=0)$

(c) Find the angle between the two lines \[2x=3y=-z\] and \[6x=-y=-4z.\]

Solution:

(a) Here, $(\vec { a })$ and $(\vec { b })$ are perpendicular vectors

\[\vec{a} \cdot \vec{b} =0\]

Now \[|\vec{a}+\vec{b}|^{2} =(\vec{a}+\vec{b})(\vec{a}+\vec{b})\]

\[|\vec{a}+\vec{b}|^{2} =|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \vec{b}\]

\[(13)^{2} =5^{2}+|\vec{b}|^{2}+2(0)\]

\[169=25+|\vec{b}{{|}^{2}}\]

\[|\vec{b}|=\sqrt{144}=12\]

(b) Length of the perpendicular from the origin \[O\left( 0,0,0 \right)\] to the given plane

\[=\left|\dfrac{3(0)-4(0)-12(0)+39}{\sqrt{3^{2}+(-4)^{2}+(-12)^{2}}}\right|\]

\[=\dfrac{39}{\sqrt{9+16+144}}=\dfrac{39}{\sqrt{169}}=\dfrac{39}{13}=3 \text { units }\]

(c) Given lines are:

\[2x=3 y=-z \Rightarrow \dfrac{2 x-0}{1}=\dfrac{3 y-0}{1}=\dfrac{-z}{1} \Rightarrow \dfrac{x}{1 / 2}=\dfrac{y}{1 / 3}=\dfrac{z}{-1} \quad \text { or } \quad \dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{-6}\]

\[\text { And } 6 x=-y=-4 z \Rightarrow \dfrac{6 x-0}{1}=\dfrac{-y}{1}=\dfrac{-4 z}{1} \Rightarrow \dfrac{x}{1 / 6}=\dfrac{y}{-1}=\dfrac{z}{-\dfrac{1}{4}} \text { or } \dfrac{x}{2}=\dfrac{y}{-12}=\dfrac{z}{-3}\]

\[\cos\theta=\dfrac{3(2)+2(-12)-6(-3)}{\sqrt{3^{2}+2^{2}+(-6)^{2}} \sqrt{2^{2}+(-12)^{2}+(-3)^{2}}}=\dfrac{6-24+18}{\sqrt{49} \sqrt{157}}=\dfrac{0}{\sqrt{49} \sqrt{157}} = 0\]

\[\Rightarrow \theta=\dfrac{\pi}{2}\]

Hence, the lines are perpendicular to each other.

2019 ISC Maths Paper Solved: Question 2

(a) The following results were obtained with respect to two variables \[x\] and \[y:\]

\[\Sigma x=15,\Sigma y=25,\Sigma xy=83,\Sigma {{x}^{2}}=55,\Sigma {{y}^{2}}=135\] and \[n=5\]

(i) Find the regression coefficient \[bxy.\]

(ii) Find the regression equation of \[x\] on \[y.\]

Or

(b) Find the equation of the regression line of \[y\] on \[x,\] if the observations \[\left( x,y \right)\] are as follows:

\[\left( 1,4 \right),\text{ }\left( 2,8 \right),\text{ }\left( 3,2 \right),\text{ }\left( 4,12 \right),\text{ }\left( 5,10 \right),\text{ }\left( 6,14 \right),\text{ }\left( 7,16 \right),\text{ }\left( 8,6 \right),\text{ }\left( 9,18 \right)\]

Also, find the estimated value of \[y\] when \[x=14.\]

Solution:

(a) $\Sigma x=15, \Sigma y=25, \Sigma x y=83, \Sigma x^{2}=55, \Sigma y^{2}=135$ and $n=5$

\[b_{x y}=\dfrac{\Sigma x y-\dfrac{\Sigma x \Sigma y}{n}}{\Sigma y^{2}-\dfrac{(\Sigma y)^{2}}{n}}=\dfrac{83-\dfrac{15 \times 25}{5}}{135-\dfrac{25 \times 25}{5}}=\dfrac{83-75}{135-125}=\dfrac{8}{10}\]

\[b_{x y}=\dfrac{4}{5}\]

Regression equation of $x$ and $y$ is given as:

\[x-\bar{x}\text{ }={{b}_{xy}}(y-\bar{y})\]

\[x-\dfrac{15}{5}\text{ }=\dfrac{4}{5}\left( y-\dfrac{25}{5} \right)\]

\[5(x-3)=4(y-5)\]

\[5x-4y+5=0\]

(b) Here, observations are : $(1,4),(2,8),(3,2),(4,12),(5,10),(6,14),(7,16),(8,6)\text{ and }(9,18)$

Now, $\Sigma x=1+2+3+4+5+6+7+8+9=45$

\[\Sigma y=4+8+2+12+10+14+16+6+18=90\]

\[\Sigma x^{2}=1+4+9+16+25+36+49+64+81=285\]

\[\Sigma y^{2}=16+64+4+144+100+196+256+36+324=1140\]

\[\Sigma x y=4+16+6+48+50+84+112+48+162=530\]

\[b_{y x}=\dfrac{\sum x y-\dfrac{\Sigma x \Sigma y}{n}}{\Sigma x^{2}-\dfrac{(\Sigma x)^{2}}{n}}=\dfrac{530-\dfrac{45 \times 90}{9}}{285-\dfrac{45 \times 45}{9}}=\dfrac{530-450}{285-225}=\dfrac{80}{60}=\dfrac{4}{3}\]

Regression line of $y$ on $x$ is given as

\[y-\bar{y}\text{ }={{b}_{yx}}(x-\bar{x})\]

\[y-\dfrac{90}{9}=\dfrac{4}{3}\left( x-\dfrac{45}{9} \right)\]

\[y-10=\dfrac{4}{3}(x-5)\]

\[3y-30=4x-20\]

\[4x-3y+10=0\]

\[\text { When } x=14, \text { then } 4(14)-3 y+10=0\]

\[\Rightarrow 3 y=56+10 \Rightarrow y=\dfrac{66}{3}=22\]

ISC 2019 Maths Question Paper: Question 3

$f:A\to A$ and $A=Rn\left( \dfrac{8}{5} \right),$ .show that the function $f(x)=\left( \dfrac{8x+3}{5x-8} \right)$ is one-one onto. Hence, find \[{{f}^{-1}}.\]

Solution: Given function is:

\[f:\mathbf{A}\to \mathbf{A}\text{ and }\mathbf{A}=\mathbf{R}-\left\{ \dfrac{\mathbf{8}}{5} \right\}\]

\[f(x)=\dfrac{\mathbf{8}x+3}{5x-\mathbf{8}}\]

Let $x_{1}, x_{2}$ be any two elements of set. $A,$ such that

\[f\left(x_{1}\right)=f\left(x_{2}\right)\]

\[\dfrac{8 x_{1}+3}{5 x_{1}-8}=\dfrac{8 x_{2}+3}{5 x_{2}-8}\]

\[\Rightarrow \quad 40 x_{1} x_{2}-64 x_{1}+15 x_{2}-24=40 x_{1} x_{2}+15 x_{1}-64 x_{2}-24\]

\[15 x_{2}+64 x_{2}=15 x_{1}+64 x_{2}\]

\[79 x_{2}=79 x_{1}\]

\[x_{2}=x_{1}\]

Thus, $f$ is one-one, for all $x_{1}, x_{2} \in \mathrm{A}$. Let $y$ be an arbitrary element of $\mathrm{A},$ then

\[f(x)=y\]

\[\dfrac{8 x+3}{5 x-8}=y\]

\[8 x+3=5 x y-8 y\]

\[8 x-5 x y=-8 y-3\]

\[x(5 y-8) =8 y+3\]

\[x=\dfrac{8 y+3}{5 y-8}\]

Clearly, $x=\dfrac{8 y+3}{5 y-8}$ is a real number for all $y \neq \dfrac{8}{5}$.

$\Rightarrow$ Corresponding to each $y \in \mathrm{A},$ there exists $\dfrac{8 y+3}{5 y-8} \in \mathrm{A},$ such that $f\left(\dfrac{8 y+3}{5 y-8}\right)=y$

Thus, $f$ is onto. Hence, $f$ is one-one and onto.

Now,

\[x=\dfrac{8 y+3}{5 y-8}\]

$\Rightarrow f^{-1}(y)=\dfrac{8 y+3}{5 y-8}$ for all $y \in \mathrm{R}-\left\{\dfrac{8}{5}\right\}$

ISC Class 12 Maths Paper 2019: Question 4

A \[13m\] long ladder is leaning against a wall, touching the wall at a certain height from the ground level. The bottom of the ladder is pulled away from the wall, along the ground, at the rate of\[2m/s\]. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall?

Solution:

Let at any instant of time \[t,\] the height of the top of the ladder be \[y\] and its foot be at distance \[x\] from the wall, then

Or

\[\sqrt{{{x}^{2}}+{{y}^{2}}}=13\]

\[{{x}^{2}}+{{y}^{2}}=169\]

Differentiating w.r.t. $x$, we have

\[2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\]

\[\dfrac{dy}{dt}=-\dfrac{x}{y}\cdot \dfrac{dx}{dt}\]

(Image Will be Uploaded Soon)

We are given $x=5 \mathrm{~m}$ and $\dfrac{d y}{d t}=2 \mathrm{~m} / \mathrm{sec}$

And

\[y=\sqrt{169-{{x}^{2}}}\]

\[=\sqrt{169-25}=\sqrt{144}=12~\text{m}\]

$\therefore$

\[\dfrac{dy}{dt}=-\dfrac{5}{12}.(2)=-\dfrac{5}{6}~\text{m}/\text{sec}\]

ISC Maths 2019 Question Paper: Question 5

Solve the differential equation:

$\dfrac{dy}{dx}=\dfrac{x+y+2}{2\left( x+y \right)-1}$

Solution:

Given the differential equation is

\[\dfrac{d y}{d x}=\dfrac{x+y+2}{2(x+y)-1}\]

Put $x+y=t$

$\Rightarrow$

\[1+\dfrac{d y}{d x}=\dfrac{d t}{d x} \Rightarrow \dfrac{d y}{d x}=\dfrac{d t}{d x}-1\]

\[\dfrac{d t}{d x}-1=\dfrac{t+2}{2 t-1}\]

\[\dfrac{d t}{d x}=\dfrac{t+2+2 t-1}{2 t-1}\]

\[\dfrac{d t}{d x}=\dfrac{3 t+1}{2 t-1}\]

\[\left(\dfrac{2 t-1}{3 t+1}\right) d t=d x\]

\[\left(\dfrac{2}{3}-\dfrac{5}{3} \cdot \dfrac{1}{3 t+1}\right) d t =d x\]

Integrating both sides, we have

\[\dfrac{2}{3} t-\dfrac{5}{3} \dfrac{\log |3 t+1|}{3}=x+\mathrm{C}\]

\[6 t-5 \log |3 t+1|=9 x+9 \mathrm{C}\]

\[6(x+y)-5 \log |3 x+3 y+1|=9 x+\mathrm{A}, \text { where } \mathrm{A}=9 \mathrm{C}\]

\[6 y-3 x-5 \log |3 x+3 y+1|=\mathrm{A}\]

2019 ISC Maths Question Paper: Question 6

Solution:

Evaluate: $\int\limits_{0}^{\pi }{\dfrac{x\tan x}{\sec x+\tan x}dx}$

Let $I=\int_{0}^{\pi} \dfrac{x \tan x}{\sec x+\tan x} d x$ $...(i)$

then $\mathrm{I}=\int_{0}^{\pi} \dfrac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)} d x \quad\left(\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

or $I=\int_{0}^{\pi} \dfrac{(\pi-x) \tan x}{\sec x+\tan x} d x$  $...(ii)$

$\operatorname{Add}(i)$ and $(i i)$

\[2\text{I}=\int_{0}^{\pi }{\dfrac{\pi \tan x}{\sec x+\tan x}}dx=2\pi \int_{0}^{\pi /2}{\dfrac{\tan x}{\sec x+\tan x}}dx=2\pi \int_{0}^{\pi /2}{\dfrac{\sin x}{1+\sin x}}dx\]

\[=2\pi \int_{0}^{\pi /2}{\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}}dx=2\pi \int_{0}^{\pi /2}{\left( \sec x\tan x-{{\tan }^{2}}x \right)}dx\]

\[=2\pi \int_{0}^{\pi /2}{\left( \sec x\tan x-\left( {{\sec }^{2}}x-1 \right)dx=2\pi [\sec x-\tan x+x]_{0}^{\pi /2} \right.}\]

\[=2 \pi\left\{\lim _{x \rightarrow \dfrac{-\pi}{2}}(\sec x-\tan x)+\dfrac{\pi}{2}-(1-0+0)\right\}=2 \pi\left(0+\dfrac{\pi}{2}-1\right)\]

\[\left(\because \lim _{x \rightarrow \dfrac{\pi}{2}}(\sec x-\tan x)=\right.\left.\lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{1-\sin x}{\cos x}=\lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{(1-\sin x)(1+\sin x)}{\cos x(1+\sin x)}=\lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{\cos ^{2} x}{\cos x(1+\sin x)}=\lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{\cos x}{1+\sin x}=0\right)\]

\[\therefore  \mathrm{I}=\pi\left(\dfrac{\pi}{2}-1\right)\]

2019 Maths ISC Board Paper Solved: Question 7

(a) If $\vec{a}=\hat{i}-2\hat{j}+3\hat{k},\vec{b}=2\hat{i}+3\hat{j}-5\hat{k},$ prove that $\vec{a}$ and $\vec{a}\times \vec{b}$ are perpendicular.

Or

(b) If $\vec{a}$ and $\vec{b}$ are non-collinear vector, find the value of $x$ such that the vector $\vec{\alpha }=\left( x-2 \right)\vec{a}+\vec{b}$ and $\vec{\beta }=\left( 3+2x \right)\vec{a}-2\vec{b}$ are collinear.

Solution:

(a) Here, $\vec{a}=\hat{i}-2 \hat{j}+3 k$ and $\vec{b}=2 i+3 j-5 k$

Now, $\vec{a}$ and $\vec{a} \times \vec{b}$ are perpendicular, if there scalar triple product is zero.

i.e.

\[\vec{a} \cdot(\vec{b} \times \vec{c})=0\]

\[\therefore \vec{a}\cdot (\vec{b}\times \vec{c})=\left| \begin{matrix} 1 & -2 & 3 \\ 1 & -2 & 3 \\ 2 & 3 & -5 \\ \end{matrix} \right|=0\]

\[\left[ \because {{\text{R}}_{1}}\text{ and }{{\text{R}}_{2}} \right.\text{ are idendical }\!\!]\!\!\text{ }\]

Hence, $\vec{a}$ and $\vec{a} \times \vec{b}$ are perpendicular. 

Or

(b) Here, $\vec{a}$ and $\vec{b}$ are non-collinear vectors and $\vec{\alpha}, \vec{\beta}$ are collinear vectors.

\[\therefore\] $\vec{\alpha}=\lambda \vec{\beta},$ where $\lambda$ is some scalar.

\[\Rightarrow (x-2)\vec{a}+\vec{b}\text{ }=\lambda \{(3+2x)\vec{a}-2\vec{b}\}\]

\[(x-2)\vec{a}+\vec{b}\text{ }=\lambda (3+2x)\vec{a}-2\lambda \vec{b}\]

\[x-2=\lambda (3+2x)\text{ and }1=-2\lambda \Rightarrow \lambda =-\dfrac{1}{2}\]

\[\Rightarrow x-2=-\dfrac{1}{2}(3+2 x)\]

\[\Rightarrow 2 x-4=-3-2 x\]

\[\Rightarrow 4x=1\]

\[\Rightarrow x=\dfrac{1}{4}\]

ISC Class 12 Maths Question Paper 2019: Question 8

Draw a rough sketch and find the area bounded by the curve \[{{x}^{2}}~=y\] and \[x+y=2.\]

Solution:

(Image will be uploaded soon)

The given curves are: \[{{x}^{2}}=y\]

Which is an upward parabola with vertex at origin

And line \[x+y=2\to y=2x\]

$x2=2x$

$x2+x,2=0$

$(x+2)(x1)=0$

$x=-2$ and $x=1$

Now, \[y=2-\left( -2 \right)=4\]

and \[y=21~\Rightarrow y=1\]

\[\Rightarrow y=4\] and \[\Rightarrow y=1\]

Thus, the points of intersection are \[\left( -2,\text{ }4 \right)\] and \[\left( 1,\text{ }1 \right)\]

Required area of shaded region

\[=\int_{-2}^{1}(2-x) d x-\int_{-2}^{1} x^{2} d x\]

\[=\left|2 x-\dfrac{x^{2}}{2}\right|_{-2}^{1}-\left|\dfrac{x^{3}}{3}\right|_{-2}^{1}\]

\[=2-\dfrac{1}{2}+4+\dfrac{4}{2}-\dfrac{1}{3}-\dfrac{8}{3}\]

\[=\dfrac{12-3+24+12-2-16}{6}\]

\[=\dfrac{9}{2} \text { sq. units }\]

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