Ncert Solutions Maths class 11th

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(i) False: As per the definition of a chord, it should intersect the circle at two distinct points.

(ii) False: The centre of a circle only bisects the diameter, which is the chord of the circle.

(iii) True: The equation of an ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

If we put a = b = 1, then

x² + y² = 1, which is an equation of a circle

(iv) True: Given x > y. Multiplying by –1, –x < y.

(v) False: Since | | is a prime number, therefore √11$$ is irrational.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(ii) q: The equation x² – 1 = 0 does not have a root lying between 0 and 2.

(i) Let us consider an equilateral triangle ABC where $\angle A=\angle B=\angle C=60°,$ then triangle ABC is not obtuse angled since all its angles are equal

$\therefore$ The statement p is not true.

(ii) 1 lies between 0 and 2. Also, 1 satisfies the equation x² – 1 = 0

$\left[?1^2-1=1-1=0\right]$

$\therefore$ The statement q is not true.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

Let p: If x is an integer and x² is even, then x is also even.

Let q: x be an integer and x² be even

Let r: x is even

We have to prove, using the method of contra positive whether $q\Rightarrow r$ is true.

[i.e., its contra positive $\sim r\Rightarrow \sim q$ is true]

Let $\sim r$ be true i.e., r be false

Let us assume that x is not even (integer) i.e., x is odd.

$\therefore x=2m+1$ where m is an integer

$\therefore x^2{\left(2m+1\right)}^2=4m+1+4m$

$=4m^2+4m+1$

$=2\left(2m^2+2m+1\right)$ $=2t+1t=2m^2+2m$ is an integer.

$\Rightarrow$ x² is also odd $\Rightarrow$ x is not even

$\Rightarrow q\text{\hspace{0.17em}isfalse}\left[\text{\hspace{0.17em}Bydefof}q\right]\Rightarrow \sim q\text{\hspace{0.17em}istrue}$

$\therefore \sim r\Rightarrow \sim q$ is true

Thus, by the method of contrapositive $q\Rightarrow r$ is true

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

Let p: If x is an integer and x² is even, then x is also even.

Let q: x be an integer and x² be even

Let r: x is even

We have to prove, using the method of contra positive whether $q\Rightarrow r$ is true.

[i.e., its contra positive $\sim r\Rightarrow \sim q$ is true]

Let $\sim r$ be true i.e., r be false

Let us assume that x is not even (integer) i.e., x is odd.

$\therefore x=2m+1$ where m is an integer

$\therefore x^2{\left(2m+1\right)}^2=4m+1+4m$

$=4m^2+4m+1$

$=2\left(2m^2+2m+1\right)$ $=2t+1t=2m^2+2m$ is an integer.

$\Rightarrow$ x² is also odd $\Rightarrow$ x is not even

$\Rightarrow q\text{\hspace{0.17em}isfalse}\left[\text{\hspace{0.17em}Bydefof}q\right]\Rightarrow \sim q\text{\hspace{0.17em}istrue}$

$\therefore \sim r\Rightarrow \sim q$ is true

Thus, by the method of contrapositive $q\Rightarrow r$ is true

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

Let a = 2, b = –2 be two real numbers.

Clearly, a² = b² (=4) but $a\ne b.$

Thus, the given statement is not true.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(a) (i) Contra positive

(ii) Converse

(b) (i) Contra positive

(ii) Converse

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(i) If you get a job, then your credentials are good.

(ii) If the Banana trees stay warm for a month, then the trees will bloom.

(iii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

(iv) If you want to score an A+ in the class, then you do all the exercises in the book.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

Five different ways are–

(i) A natural number is odd implies that its square is odd.

(ii) A natural number is odd only if its square is odd.

(iii) For a natural number to be odd it is necessary that its square is odd.

(iv) For the square of a natural number to be odd it is sufficient that the number is odd.

(v) If the square of a natural number is not odd, then the natural number is not odd.