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ISC Class 12 Mathematics Question Paper - 2019 Board Examinations

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Last updated date: 19th Jun 2024
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Previous Year Mathematics Question Paper for ISC Class 12 Board - 2019 - Free PDF Download

The ISC Class 12 Mathematics examination was conducted by the ICSE Board on 5th March 2019. The students were allotted three hours to finish their examination (2 pm to 5 pm). Here, we have provided the ISC 2019 Maths Paper for all the students who want to study online. Solutions of some of the most important things are provided here by our experienced teachers. Students can look at the solved sums by referring to the paper. This can help them prepare well and get good marks in their examination. So, get a glimpse of the ISC Maths Question Paper 2019 and start preparing with lock, stock, and barrel.

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ISC Maths Paper 2019 Solved

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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Mathematics is a subject in which good marks can only be attained through rigorous practice. ISC gets a little tougher than ICSE board and so, the amount of hard work and productive sources through which we can attain efficiency in this subject is huge. Maths exercises in your textbooks, class notes, formulas, all of these play a very important role contributing to the level of preparation of your subject - Maths.


Including the above preparation sources for your Maths exam, one more power booster can only boost up your scores, and also your confidence during your board exams. The power booster referred to here is the previous year's question paper for Mathematics for Class 12.


You can easily search for these previous year's question papers as per the specific year that you wish for on the Vedantu app or its official website. Here, the previous year's question papers are available with their exact solutions. If you refer to the previous year's questions, you can clearly know the pattern of your exam paper or atleast the type of pattern that you can expect in the current year as well. You also get the type of questions they may be asked in your upcoming exam, and also the way to go about solving those questions.

 

How solving Previous Year Mathematics Questions can be an Essential Part in Your Board Exam Preparations?

Most students are least confident and also fearful when it comes to mathematics and implementing its formula to get accurate answers in the prescribed amount of time. But, with practice, we can attain this efficiency by practicing more and more and bringing out those exact answers in very less time. To attain this, we need to be very regular without practice in this particular subject. 


Besides practicing from your textbooks, you can also solve previous year's question papers for the Class 12 ISC board, so that you will also get a glimpse of how your real exam is going to be and what it is actually going to look like! Choose any previous year's question paper of Class 12 for Mathematics and start solving it for analyzing yourself. Understand which chapters require more of your effort and time and invest and prioritize those chapters first until you yourself are satisfied, then again check start solving some other previous year's question paper, and start analyzing yourself.

FAQs on ISC Class 12 Mathematics Question Paper - 2019 Board Examinations

1. If f : A ⟶ A and A = R(8/5), show that the function f(x) = [(8x + 3)/(5x - 8)] is one-one onto. Hence, find f-1.

The given function is:

f : A ⟶ A and A = R - {8/5}

f(x) = [(8x + 3)/(5x - 8)]

Let x1, x2 be any two elements of set A, such that

f(x1) = f(x2)

[(8x1 + 3)/(5x1 - 8)] = [(8x2 + 3)/(5x2 - 8)]

= 40x1x2 - 64x1 + 15x2 - 24 = 40x1x2 + 15x1 - 64x2 - 24

15x2 + 64x2 = 15x1 + 64x1

79x1 = 79x2

x2 = x1

 Thus, f is one-one, for all x1, x2 ∈ A.

Let y be an arbitrary element of A, then

f(x) = y

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

8x + 3 = 5xy - 8y

8x - 5xy = -8y - 3

x(5y - 8) = 8y + 3

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

Clearly, x = [(8y + 3)/(5y - 8)] is a real number for all y ≠ 8/5.

⇒ Corresponding to each y ∈ A, there exists [(8y + 3)/(5y - 8)] ∈ A, such that f[(8y + 3)/(5y - 8)] = y

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

Now,

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

⇒f-1(y) = [(8y + 3)/(5y - 8)] for all y ∈ R - {8/5}

2. Where can I get previous years Mathematics question papers for the ISC Class 12 board?

You can get these previous years Mathematics question papers for ISC  by clicking here or just visit the official website of Vedantu or download its app from the Google Play Store and you can get a variety of study and learning materials for your exam preparations included with the previous question papers for mathematics for ISC board.

3. How can previous year question papers for Maths help me to attain good marks in the ISC board exams?

The subject Mathematics is all about practice and only by practicing it, you can attain good marks in this particular subject. Previous year's question papers contain not only the questions but the solutions as well so that you exactly know how to go about solving a particular question. You can understand where you are getting wrong and can correct yourself accordingly by referring to the answers which are already mentioned in the previous year's question paper.

3. Is it compulsory to solve the previous year questions before board exams?

It is always recommended to solve previous year's questions before the beginning of any exam to save yourself from the anxiety before the exams, and also to build confidence because now you are aware of the patterns or the type of questions that can be asked in your exam. The practice has made you so experienced that you know how to invest your time wisely during your exam.

5. Is the ISC exam tougher than ICSE?

ICSE exams are conducted for the students of the matriculation level and ISC exams are conducted for the ones of the intermediary levels. So, yes, the ICSE exam is tougher than ISC. It requires double effort and understanding of the subjects. Though the subjects asked in the ICSE board are lesser than ISC but here, you can score well mainly because of the academic stream that you go for because you are interested in them.

6. Can I score well in mathematics in my board exams of Class 12?

You can always score excellent marks in mathematics but it all depends on the amount of practice that you put in this particular subject and also, it depends on the understanding that you have acquired so far in your previous classes. So always practice before your Maths exam, resolve any doubts that you have regarding any topic then and there. Last but not the least, try to develop an interest in this interesting subject by investing in it all around.