Author name: Prasanna

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-ex-5-3/

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.3

Ex 5.3 Class 12 NCERT Solutions Question 1.
2x + 3y = sinx
Solution:
2x + 3y = sinx
Differentiating w.r.t x,
\(2+3\frac { dy }{ dx } =cosx \)
⇒ \(\frac { dy }{ dx } =\frac { 1 }{ 3 } (cosx-2)\)

Exercise 5.3 Class 12 Maths Ncert Solutions In Hindi Question 2.
2x + 3y = siny
Solution:
2x + 3y = siny
Differentiating w.r.t x,
\(2+3.\frac { dy }{ dx } =cosy\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { 2 }{ cosy-3 } \)

Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate both sides w.r.t. x,

Question 4.
xy + y² = tan x + y
Solution:
xy + y² = tan x + y
Differentiate both sides w.r.t. x,

Question 5.
x² + xy + y² = 100
Solution:
x² + xy + y² = 100
Differentiate both sides w.r.t. x,

Question 6.
x³ + x²y + xy² + y³ = 81
Solution:
x³ + x²y + xy² + y³ = 81
Differentiate both sides w.r.t. x,

Question 7.
sin² y + cos xy = π
Solution:
sin² y + cos xy = π
Differentiate both sides w.r.t. x,

Question 8.
sin²x + cos²y = 1
Solution:
Given that
sin²x + cos²y = 1
Differentiating both sides, we get

Question 9.
y = \({ sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right)\)
Solution:

Question 10.
y = \({ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 3 } } <x<\frac { 1 }{ \sqrt { 3 } } \)
Solution:

Question 11.
y = \({ cos }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1\)
Solution:
y = \({ cos }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1\)
put x = tanθ
y = \({ cos }^{ -1 }\left( \frac { 1-tan^{ 2 }\quad \theta }{ 1+{ tan }^{ 2 }\quad \theta } \right) ={ cos }^{ -1 }(cos2\theta )=2\theta\)
= 2θ = 2 tan^{-1 x}
i.e., y = 2 tan^-1x
Differentiating w.r.t x, \(\frac{d y}{d x}=\frac{2}{1+x^{2}}\)

Question 12.
y = \({ sin }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1\)
Solution:

Question 13.
y = \({ cos }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right)\), – 1 < x < 1
Solution:

Question 14.
y = \(sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } } \)
Solution:

Question 15.
y = \(sin^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) ,0<x<\frac { 1 }{ \sqrt { 2 } } \)
Solution:
y = \(sin^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) ,0<x<\frac { 1 }{ \sqrt { 2 } } \)
put x = tanθ
we get
y = \(sec^{ -1 }\left( \frac { 1 }{ { 2cos }^{ 2 }\theta -1 } \right) ={ sec }^{ -1 }\left( \frac { 1 }{ cos2\theta } \right) \)
y = \(sec^{ -1 }(sec2\theta )=2\theta ,\quad 2{ cos }^{ -1 }x \)
i.e., y = 2 cos^-1 x
Differentiating w.r.t.x,
∴ \(\frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}}\)

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-ex-5-2/

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.2

Class 12 Maths Chapter 5 Exercise 5.2 Question 1.
sin(x² + 5)
Solution:
Let y = sin(x² + 5),
put x² + 5 = t
y = sint
t = x²+5
\(\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(\frac { dy }{ dx } =cost.\frac { dt }{ dx } =cos({ x }^{ 2 }+5)\frac { d }{ dx } ({ x }^{ 2 }+5)\)
= cos (x² + 5) × 2x
= 2x cos (x² + 5)

Ex 5.2 Class 12 NCERT Solutions Question 2.
cos (sin x)
Solution:
let y = cos (sin x)
put sinx = t
∴ y = cost,
t = sinx
∴\(\frac { dy }{ dx } =-sin\quad t,\frac { dt }{ dx } =cos\quad x \)
\(\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =(-sint)\times cosx\)
Putting the value of t, \(\frac { dy }{ dx } =-sin(sinx)\times cosx\)
\(\frac { dy }{ dx } =-[sin(sinx)]cosx\)

Ncert Solutions For Class 12 Maths Chapter 5 Exercise 5.2 Question 3.
sin(ax+b)
Solution:
let = sin(ax+b)
put ax+bx = t
∴ y = sint
t = ax+b
\(\frac { dy }{ dt } =cost,\frac { dt }{ dx } =\frac { d }{ dx } (ax+b)=a\)
\(Now\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =cost\times a=acos\quad t\)
\(\frac { dy }{ dx } =acos(ax+b)\)

Exercise 5.2 Class 12 NCERT Solutions Question 4.
sec(tan(\(\sqrt{x}\)))
Solution:
let y = sec(tan(\(\sqrt{x}\)))
by chain rule
\(\frac { dy }{ dx } =sec(tan\sqrt { x } )tan(tan\sqrt { x } )\frac { d }{ dx } (tan\sqrt { x } )\)
\(\frac { dy }{ dx } =sec(tan\sqrt { x } ).tan(tan\sqrt { x } ){ sec }^{ 2 }\sqrt { x } .\frac { 1 }{ 2\sqrt { x } } \)

Question 5.
\(\\ \frac { sin(ax+b) }{ cos(cx+d) } \)
Solution:

Question 6.
cos x³ . sin²(x⁵) = y(say)
Solution:
Let y = cos x³ . sin²(x⁵)

Question 7.
\(2\sqrt { cot({ x }^{ 2 }) }\)
Solution:

Question 8.
cos(\(\sqrt{x}\))
Solution:

Question 9.
Prove that the function f given by f (x) = |x – 1|, x ∈ R is not differential at x = 1.
Solution:

∴ f is not differentiable at x = 1

Question 10.
Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differential at x = 1 and x = 2.
Solution:

Hence f is not differentiable at x = 2

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-ex-5-5/

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5

Ex 5.5 Class 12 NCERT Solutions Question 1.
cos x. cos 2x. cos 3x
Solution:
Let y = cos x. cos 2x . cos 3x,
Taking log on both sides,
log y = log (cos x. cos 2x. cos 3x)
log y = log cos x + log cos 2x + log cos 3x,
Differentiating w.r.t. x, we get

Exercise 5.5 Class 12 Maths Solutions Question 2.
\(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)
Solution:

Ex 5.5 Class 12 Maths Ncert Question 3.
(log x)^cosx
Solution:
let y = (log x)^cosx
Taking log on both sides,
log y = log (log x)^cosx
log y = cos x log (log x),
Differentiating w.r.t. x,

Exercise 5.5 Class 12 NCERT Solutions  Question 4.
x – 2^sinx
Solution:
let y = x – 2^sinx,
∴ y = u – v
Differentiating w.r.t. x,

Ch 5 Maths Class 12 Ex 5.5 NCERT Solutions Question 5.
(x+3)².(x + 4)³.(x + 5)⁴
Solution:
let y = (x + 3)².(x + 4)³.(x + 5)⁴
Taking log on both side,
logy = log [(x + 3)² . (x + 4)³ . (x + 5)⁴]
= log (x + 3)² + log (x + 4)³ + log (x + 5)⁴
log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
Differentiating w.r.t. x, we get

On simplification,
\(\frac { dy }{ dx }\) = (x + 3)(x + 4)²(x + 5)³(9x² + 70x + 133)

Question 6.
\({ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }\)
Solution:
Let u = \(\left(x+\frac{1}{x}\right)^{x}\) = \(\left(\frac{x^{2}+1}{x}\right)^{x}\)
v = \(x^{\left(1+\frac{1}{x}\right)}\)
Let y = u + v
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)
u = \(\left(\frac{x^{2}+1}{x}\right)^{x}\)
Taking logaritham on both sides,
logu = x log \(\left(\frac{x^{2}+1}{x}\right)\)
∴ log u = x(log(x²+ 1) – logx)
Differentiating both sides w.r.t.x,

v = \(x^{\left(1+\frac{1}{x}\right)}\)
Taking logaritham on both sides,
logv = (1 + \(\frac { 1 }{ x }\)) log x
Differentiating both sides w.r.t. x,

Question 7.
(log x)^x + x^logx
Solution:
Let u = (log x)^x, v = x^logx and y = u + v
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)
u = (log x)^x
Taking logaritham on both sides,
logu = x log(log x)
Differentiating both sides w.r.t. x,

v = x^{log x}
Taking logarithm on both sides,
logv = logx. logx = (logx)²
Differentiating both sides w.r.t. x,

Question 8.
(sin x)^x+sin^-1 \(\sqrt{x}\)
Solution:
Let y = (sin x)^x + sin^-1 \(\sqrt{x}\)
Let y = u + v
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)
u = (sin x)^x
\(\frac { du }{ dx }\) = (sin x)^x[x cotx + log sinx] … (2)
Let y = (sinx)^x
Taking logarithm on bath sides,
logy = x.log sin x
Differentiating both sides w.r.t. x,

Question 9.
x^sinx + (sin x)^cosx
Solution:
Let y = x^sinx + (sin x)^cosx
Differentiating both sides w.r.t. x,

(i) Let y = (sin x)^x
Taking logarithm on both sides,
logy = sinx. logx
Differentiating both sides w.r.t. x,

(ii) Let y = (sinx)^cosx 
Taking logarithm on both sides,
logy = cosx. log sinx
Differentiating both sides w.r.t. x,

Question 10.
\({ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } \)
Solution:
Let u = x^{x cosx} and v = \(\frac{x^{2}+1}{x^{2}-1}\)
Let y = u + v
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)
u = x^{x cosx}
Taking logarithm on both sides,
logu = x cosx . logx
Differentiating both sides w.r.t. x,

Differentiating both sides w.r.t. x,

Question 11.
\((x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}\)
Solution:
Let u = (x cosx)^x and v = \((x \sin x)^{\frac{1}{x}}\)
u = (x cosx)^x
Taking logarithm on both sides,
log u = x log (x cosx)
log u = x[log x + log cosx]
Differentiating w.r.t. x,

Question 12.
x^y + y^x = 1
Solution:
x^y + y^x = 1
let u = x^y and v = y^x
∴ u + v = 1,
\(\frac { du }{ dx } +\frac { dv }{ dx }=0\)
Now u = x^y
Taking logarithm on both sides,
log u = y log x
Differentiating w.r.t. x

Question 13.
y^x = x^y
Solution:
x^y = y^x
Taking logarithm on both sides,
y log x = x log y
Differentiating w.r.t. x

Question 14.
(cos x)^y = (cos y)^x
Solution:
(cos x)^y = (cos y)^x
Taking logarithm on both sides,
y log cosx = x log cosy
Differentiating both sides w.r.t. x,

Question 15.
xy = e^(x-y)
Solution:
log(xy) = log e^(x-y)
⇒ log(xy) = x – y
⇒ logx + logy = x – y
\( ⇒\frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } ⇒\frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) } \)

Question 16.
Find the derivative of the function given by f (x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸) and hence find f'(1).
Solution:
Let f(x) = y = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)
Taking log both sides, we get
logy = log [(1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)]
logy = log(1 + x) + log (1 + x²) + log(1 + x⁴) + log(1 + x⁸)
Differentiating w.r.t. x,

Question 17.
Differentiate (x² – 5x + 8) (x³ + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?
Solution:
(i) By using product rule
f’ = (x² – 5x + 8) (3x² + 7) + (x³ + 7x + 9) (2x – 5)
f = 5x⁴ – 20x³ + 45x² – 52x + 11.

(ii) y = x⁵ – 5x⁴ + 15x³ – 26x² + 11x + 72
(by expanding the product)
Differentiating w.r.t. x, dy
\(\frac { dy }{ dx }\) = 5x⁴ – 20x³ + 45x² – 52x + 11

(iii) y = (x² – 5x + 8)(x³ + 7x + 9)
Taking logarithm on both sides,
log y = log(x² – 5x 4- 8)(x³ + 7x + 9)
log y = log(x² – 5x + 8) + log(x³ + 7x + 9)
Differentiating, w.r.t x,

= (2x – 5) (x³ + 7x + 9) + (3x² + 7) (x² – 5x + 8)
= 5x⁴ – 20x³ + 45x² – 52x + 11
Yes, the answers are the same.

Question 18.
If u, v and w are functions of w then show that
\(\frac { d }{ dx } (u.v.w)=\frac { du }{ dx } v.w+u.\frac { dv }{ dx } .w+u.v\frac { dw }{ dx } \)
in two ways-first by repeated application of product rule, second by logarithmic differentiation.
Solution:
Let y = u.v.w
(a) Product rule

(b) Logarithmic differentiation
y = uvw
Taking logarithm on both sides, we get
log y = log uvw
i.e., log y = log u + log v + log w
Differentiating both sides w.r.t. x, we get

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-ex-5-6/

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.6

Ex 5.6 Class 12 Maths Ncert Solutions Question 1.
x = 2at², y = at⁴
Solution:

Ex 5.6 Class 12 NCERT Solutions Question 2.
x = a cos θ, y = b cos θ
Solution:

Question 3.
x = sin t, y = cos 2t
Solution:

Question 4.
\(x=4t,y=\frac { 4 }{ t } \)
Solution:

Question 5.
x = cos θ – cos 2θ, y = sin θ – sin 2θ
Solution:

Question 6.
x = a(θ – sinθ), y = a(1 + cosθ)
Solution:

Question 7.
\(x=\frac { { sin }^{ 3 }t }{ \sqrt { cos2t } } \& y=\frac { { cos }^{ 3 }t }{ \sqrt { cos2t } } \)
Solution:

Question 8.
\(x=a\left( cost+log\quad tan\frac { t }{ 2 } \right) ,y=a\quad sint \)
Solution:

Question 9.
x = a sec θ, y = b tan θ
Solution:

Question 10.
x = a(cosθ + θsinθ), y = a(sinθ – θcosθ)
Solution:

Question 11.
If \(x=\sqrt { { a }^{ { sin }^{ -1 }t } } ,y=\sqrt { { a }^{ { cos }^{ -1 }t } } \) show that \(\frac { dy }{ dx } =-\frac { y }{ x } \)
Solution:
Given that

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-ex-5-1/

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.1

Ex 5.1 Class 12 NCERT Solutions Question 1.
Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
Solution:
(i) At x = 0. lim_x→0 f (x) = lim_x→0 (5x – 3) = – 3 and
f(0) = – 3
∴ f is continuous at x = 0

(ii) At x = – 3, lim_x→3 f(x)= lim_x→-3 (5x – 3) = – 18
and f( – 3) = – 18
∴ f is continuous at x = – 3

(iii) At x = 5, lim_x→5 f(x) = lim_x→5 (5x – 3) = 22 and
f(5) = 22
∴ f is continuous at x = 5

Class 12 Maths Chapter 5 Exercise 5.1 Question 2.
Examine the continuity of the function f(x) = 2x² – 1 at x = 3.
Solution:
lim_x→3 f(x) = lim_x→3 (2x² – 1) = 17 and f(3)= 17
∴ f is continuous at x = 3

Ex.5.1 Class 12 NCERT Solutions Question 3.
Examine the following functions for continuity.
(a) f(x) = x – 5
(b) f(x) = \(\\ \frac { 1 }{ x-5 } \), x ≠ 5
(c) f(x) = \(\frac { { x }^{ 2 }-25 }{ x+5 } \),x≠5
(d) f(x) = |x – 5|
Solution:
(a) f(x) = x – 5
f(x) is a polynomial.
Hence f(x) is continuous in R.

(b) f(x) = \(\\ \frac { 1 }{ x-5 } \)
f(x) is not defined at x = 5.
Hence f(x) is not continuous at x = 5.
At x ≠ 5, f(x) is a rational function.
Hence f(x) is continuous at x ≠ 5.

(c) f(x) = \(\frac { { x }^{ 2 }-25 }{ x+5 } \)
f(x) is not defined at x = – 5.
Hence f(x) is not continuous at x = – 5.
At x ≠ – 5, f(x) is a rational function.
Hence f(x) is continuous at x ≠ – 5.

(d) f(x) = |x – 5|
f(x) is redefined as
f(x) = \(\begin{cases}5-x ; & x<5 \\ x-5 ; & x \geq 5\end{cases}\)
For x < 5 and x > 5, f(x) is a polynomial function. Hence f(x) is continuous at x < 5 and x > 5.
At x = 5, f(x) is not defined uniquely, so take the left and right hand limits at x = 5

∴ f is continuous at x = 5
Thus f is continuous since it is continuous at all real values.
Another method
Let g(x) = |x| and h(x) = x – 5
(goh)(x) = g(h(x)) = g(x – 5)
= |x – 5| = f(x) .
Also g(x) and h(x) are continuous functions since the modulus function and the polynomial functions are continuous in R.
∴ f is a continuous function as it is the com-position of two continuous functions.

Exercise 5.1 Class 12 Maths Ncert Solutions Question 4.
Prove that the function f (x) = xⁿ is continuous at x = n, where n is a positive integer.
Solution:
f (x) = xⁿ is a polynomial which is continuous for all x ∈ R.
Hence f is continuous at x = n, n ∈ N.

Class 12 Maths Exercise 5.1 Question 5.
Is the function f defined by \(f(x)=\begin{cases} x,ifx\le 1 \\ 5,ifx>1 \end{cases}\) continuous at x = 0? At x = 1? At x = 2?
Solution:
(i) At x = 0
lim_x→0- f(x) = lim_→0- x = 0 and
lim_x→0+ f(x) = lim_→0+ x = 0 => f(0) = 0
∴ f is continuous at x = 0

(ii) At x = 1
lim_x→1- f(x) = lim_→1- (x) = 1 and
lim_x→1+ f(x) = lim_→1+(x) = 5
∴ lim_x→1- f(x) ≠ lim_→1+ f(x)
∴ f is discontinuous at x = 1

(iii) At x = 2
lim_x→2 f(x) = 5, f(2) = 5
∴ f is continuous at x = 2

Ex 5.1 Maths Class 12 NCERT Solutions Question 6.
\(f(x)=\begin{cases} 2x+3,if\quad x\le 2 \\ 2x-3,if\quad x>2 \end{cases}\)
Solution:
For x < 2 and x > 2. f(x) is a polynomial function. Hence f(x) is continuous at x < 2 and x > 2.
At x = 2
lim_x→2+ f(x) = lim_→2+ 2x – 3 = 2(2) – 3 = 1
lim_x→2- f(x) = lim_→2-(x) = 2(2) + 3 = 7
lim_x→2- f(x) ≠ lim_→2+ f(x), f is not continuous
at x = 2
Hence f is discontinuous at x = 2

Question 7.
\(f(x)=\begin{cases} |x|+3,if\quad x\le -3 \\ -2x,if\quad -3<x<3 \\ 6x+2,if\quad x\ge 3 \end{cases}\)
Solution:
For x ≤ – 3, f(x) = – x + 3
For – 3 < x < 3, f(x) = – 2x For x > 3, f(x) = 6x + 2
Thus for x < – 3, x ∈ (- 3, 3) and x > 3,
f(x) is a polynomial function. Hence f(x) is continuous.

∴ f is not continuous at x = 3

Question 8.
\(f(x)=\begin{cases} \frac { |x| }{ x } ;x\neq 0 \\ 0;x=0 \end{cases}\)
Solution:
The function f can be redefined as

Hence the point of discontinuity of f is at x = 0.

Question 9.
f(x) = \(\begin{cases} \frac { x }{ |x| } ;if\quad x<0 \\ -1,if\quad x\ge 0 \end{cases}\)
Solution:
Method I
For x < 0, f(x) is the quotient of two continuous functions. Hence f(x) is continuous for x < 0. For x > 0, f(x) is a constant function. Hence/x) is continuous for x > 0.
At x = 0

Hence f has no point of discontinuity.

Method II
The function f can be redefined as
f(x) = \(\begin{cases}-1 & \text { if } x<0 \\ -1 & \text { if } x \geq 0\end{cases}\)
i.e., f(x) = – 1 for x ∈ R
∴ f is a continuous function since f(x) is a constant function.
Hence f has no point of discontinuity.

Question 10.
f(x) = \(\begin{cases} x+1,if\quad x\ge 1 \\ { x }^{ 2 }+1,if\quad x<1 \end{cases}\)
Solution:
For x < 1 and x > 1, f(x) is a polynomial function. Hence f(x) is continuous at x < 1 and x > 1.

∴ f is a continuous at x = 1
Hence f has no point of discontinuity.

Question 11.
f(x) = \(\begin{cases} { x }^{ 3 }-3,if\quad x\le 2 \\ { x }^{ 2 }+1,if\quad x>2 \end{cases} \)
Solution:
For x < 2 and x > 2, f(x) is a polynomial function. Hence f(x) is continuous at x < 2 and x > 2.

∴ f is a continuous at x = 2
Hence f has no point of discontinuity.

Question 12.
f(x) = \(\begin{cases} { x }^{ 10 }-1,if\quad x\le 1 \\ { x }^{ 2 },if\quad x>1 \end{cases} \)
Solution:
For x < 1 and x > 1, f(x) is a polynomial function. Hence f(x) is continuous at x < 1 and x > 1.

∴ f is a continuous at x = 1
Hence f has no point of discontinuity of f is at x = 1

Question 13.
Is the function defined by f(x) = \(\begin{cases} x+5,if\quad x\le 1 \\ x-5,if\quad x>1 \end{cases} \) a continuous function?
Solution:
For x < 1 and x > 1, f(x) is a polynomial function. Hence f(x) is continuous at x < 1 and x > 1.
At x = 1, L.H.L.= lim_x→1- f(x) = lim_x→1- (x + 5) = 6,
R.H.L. = lim_x→1+ f(x) = lim_x→1+ (x – 5) = – 4
f(1) = 1 + 5 = 6,
f(1) = L.H.L. ≠ R.H.L.
f is not continuous at x = 1
At x = c < 1, lim_x→c (x + 5) = c + 5 = f(c)
At x = c > 1, lim_x→c (x – 5) = c – 5 = f(c)
∴ f is continuous at all points x ∈ R except x = 1.

Question 14.
f(x) = \(\begin{cases} 3,if\quad 0\le x\le 1 \\ 4,if\quad 1<x<3 \\ 5,if\quad 3\le x\le 10 \end{cases}\)
Solution:
For x ∈ (0, 1), x ∈ (1, 3) and x ∈ (3, 10), f(x) is a constant function. Hence f(x) is continuous at x ∈ (0, 1), x ∈ (1, 3) and x ∈ (3, 10)
At x = 1
L.H.L. = lim f(x) = 3,
R.H.L. = lim_x→1+ f(x) = 4 => f is discontinuous at
x = 1
At x = 3, L.H.L. = lim_x→3- f(x)=4,
R.H.L. = lim_x→3+ f(x) = 5 => f is discontinuous at
x = 3
∴ f is not continuous at x = 1 and x = 3.

Question 15.
f(x) = \(\begin{cases} 2x,if\quad x<0 \\ 0,if\quad 0\le x\le 1 \\ 4x,if\quad x>1 \end{cases}\)
Solution:
For x < 0 and x > 1, f(x) is a polynomial function.
Hence f(x) is continuous at x < 0, x > 1 and x ∈ (0, 1)

∴ f is a continuous at x = 1
Hence f is not continuous at x = 1

Question 16.
\(f(x)=\begin{cases} -2,if\quad x\le -1 \\ 2x,if\quad -1<x\le 1 \\ 2,if\quad x>1 \end{cases}\)
Solution:
For x < – 1 and x > 1, f(x) is a constant function.
For x ∈ (-1, 1), f(x) is constant polynomial function. Hence f(x) is continuous at x < – 1, x > 1 and x ∈ (- 1, 1)
At x = – 1, L.H.L. = lim_x→1- f(x) = – 2, f(-1) = – 2,
R.H.L. = lim_x→1+ f(x) = – 2
∴ f is continuous at x = – 1
At x= 1, L.H.L. = lim_x→1- f(x) = 2,f(1) = 2
∴ f is continuous at x = 1,
R.H.L. = lim_x→1+ f(x) = 2
Hence, f is continuous function.

Question 17.
Find the relationship between a and b so that the function f defined by
\(f(x)=\begin{cases} ax+1,if\quad x\le 3 \\ bx+3,if\quad x>3 \end{cases}\)
is continuous at x = 3
Solution:

Question 18.
For what value of λ is the function defined by
\(f(x)=\begin{cases} \lambda ({ x }^{ 2 }-2x),if\quad x\le 0 \\ 4x+1,if\quad x>0 \end{cases} \)
continuous at x = 0? What about continuity at x = 1?
Solution:
At x = 0, L.H.L. = lim_x→0- λ (x² – 2x) = 0 ,
R.H.L. = lim_x→0+ (4x+ 1) = 1, f(0)=0
f (0) = L.H.L. ≠ R.H.L.
∴ f is not continuous at x = 0,
whatever value of λ ∈ R may be
At x = 1, lim_x→1 f(x) = lim_x→1 (4x + l) = f(1)
∴ f is not continuous at x = 0 for any value of λ but f is continuous at x = 1 for all values of λ.

Question 19.
Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.
Solution:
Let c be an integer

∴ g is discontinuous at c.
Since c is an integer, g is discontinuous at all integral points.

Question 20.
Is the function defined by f (x) = x² – sin x + 5 continuous at x = π?
Solution:

∴ f is a continuous at x = π
Another method
x² + 5 and since are continuous functions.
∴ Hence x² + 5 – sinx = x² – sinx + 5 is continuous. Since sum of continuous func-tions is continuous.
∴ f(x) = 2 – since + 5 is continuous at x = π

Question 21.
Discuss the continuity of the following functions:
(a) f (x) = sin x + cos x
(b) f (x) = sin x – cos x
(c) f (x) = sin x · cos x
Solution:
(a) f(x) = sinx + cosx

Question 22.
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
(a) Let f(x) = cosx
At x = c, c ∈ R, lim_x→c cos x = cos c = f(c)
∴ f is continuous for all values of x ∈ R

(b) Let f(x) = cosecx
Domain of f is R- {nπ : n ∈ Z}
Let c ∈ Domain of f
lim_x→c f(x) = lim_x→c cosec x
= lim_x→c\(\frac { 1 }{ sinx }\) = \(\frac { 1 }{ sin c }\)
Since sinx is continuous
= cosec c = f(c)
∴ f is continuous at x = c
Since c is any real number in the domain, the cosecant function is continuous in R except for x = nπ, n ∈ Z

(c) Let f(x) = secx
Domain of f is R – \(\left\{(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}\right\}\)
Let c ∈ Domain of f
lim_x→c f(x) = lim_x→csec x = lim_x→c\(\frac { 1 }{ cosx }\)
= \(\frac { 1 }{ cos c }\) since cosx is continuous
= sec c = f(c)
∴ f is continuous at x = c
Since c is any real number in the domain, the secant function is continuous in R
except for x = (2n +1)\(\frac { π }{ 2 }\), n ∈ Z

(d) Let f(x) = cot x
Domain of f, R – {nπ, n Z}
Let c ∈ Domain of f

∴ f is continuous at x = c
Since c is any real number in the domain, the cotangent function is continuous in R except for x = nπ, n ∈ Z

Question 23.
Find all points of discontinuity of f, where
\(f(x)=\begin{cases} \frac { sinx }{ x } ,if\quad x<0 \\ x+1,if\quad x\ge 0 \end{cases}\)
Solution:
For x < 0, sinx and x are continuous functions.
∴ f(x) = \(\frac { sin x }{ x }\) is continuous when x < 0. For x > 0, f(x) = x + 1 is a polynomial function
∴ f(x) = x + 1 is continuous.
At x = 0

f is continuous at x = 0
Hence f is continuous in R and has no point of discontinuity.

Question 24.
Determine if f defined by f(x) = \(\begin{cases} { x }^{ 2 }sin\frac { 1 }{ x } ,if\quad x\neq 0 \\ 0,if\quad x=0 \end{cases}\) is a continuous function?
Solution:
At x = 0

Question 25.
Examine the continuity of f, where f is defined by f(x) = \(\begin{cases} sinx-cosx,if\quad x\neq 0 \\ -1,if\quad x=0 \end{cases}\)
Solution:
sinx and cosx are continuous functions. Hence sinx – cosx is a continuous function at x ≠ 0.
At x = 0

f is continuous at x = 0
Thus f is continuous function in R.

Question 26.
f(x) = \(\begin{cases} \frac { k\quad cosx }{ \pi -2x } ,\quad if\quad x\neq \frac { \pi }{ 2 } \quad at\quad x=\frac { \pi }{ 2 } \qquad \\ 3,if\quad x=\frac { \pi }{ 2 } \quad at\quad x=\frac { \pi }{ 2 } \end{cases} \)
Solution:

Question 27.
f(x) = \(\begin{cases} { kx }^{ 2 },if\quad x\le 2\quad at\quad x=2 \\ 3,if\quad x>2\quad at\quad x=2 \end{cases} \)
Solution:

Question 28.
f(x) = \(\begin{cases} kx+1,if\quad x\le \pi \quad at\quad x=\pi \\ cosx,if\quad x>\pi \quad at\quad x=\pi \end{cases} \)
Solution:

Question 29.
f(x) = \(\begin{cases} kx+1,if\quad x\le 5\quad at\quad x=5 \\ 3x-5,if\quad x>5\quad at\quad x=5 \end{cases} \)
Solution:

Question 30.
Find the values of a and b such that the function defined by
f(x) = \(\begin{cases} 5,if\quad x\le 2 \\ ax+b,if\quad 2<x<10 \\ 21,if\quad x\ge 10 \end{cases} \)
to is a continuous function.
Solution:
It is clear that, when x < 2, 2 < x < 10 and x > 10, the function f is continuous.

i.e., 10a + b = 21 …. (2)
Solving (1) and (2), we get a = 2 and b = 1

Question 31.
Show that the function defined by f(x)=cos (x²) is a continuous function.
Solution:
Now, f (x) = cosx², let g (x)=cosx and h (x) x²
∴ goh(x) = g (h (x)) = cos x²
Now g and h both are continuous ∀ x ∈ R.
f (x) = goh (x) = cos x² is also continuous at all x ∈ R.

Question 32.
Show that the function defined by f (x) = |cos x| is a continuous function.
Solution:
Let g(x) =|x|and h (x) = cos x, f(x) = goh(x) = g (h (x)) = g (cosx) = |cos x |
Now g (x) = |x| and h (x) = cos x both are continuous for all values of x ∈ R.
∴ (goh) (x) is also continuous.
Hence, f (x) = goh (x) = |cos x| is continuous for all values of x ∈ R.

Question 33.
Examine that sin |x| is a continuous function.
Solution:
Let g (x) = sin x, h (x) = |x|, goh (x) = g (h(x))
= g(|x|) = sin|x| = f(x)
Now g (x) = sin x and h (x) = |x| both are continuous for all x ∈ R.
∴ f (x) = goh (x) = sin |x| is continuous at all x ∈ R.

Question 34.
Find all the points of discontinuity of f defined by f(x) = |x|-|x+1|.
Solution:
f(x) = |x| – |x + 1|. The domain of f is R.
Let g(x) = |x| and h(x) = x + 1
Then g(x) and h(x) are continuous functions.
(goh)(x) = g(h(x)) = g(x + 1) = |x + 1|
Since g and h are continuous functions, goh is also continuous.
∴ |x +1| is a continuous function in R.
Hence |x| – |x + 1| is also continuous in R
since difference of continuous functions is also continuous.
That is, f is continuous in R.
∴ f has no point of discontinuity.

These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-4-ex-4-1/

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.1

Class 12 Maths Chapter 4 Exercise 4.1 Solutions Question 1.
Evaluate the following determinant:
\(\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}\)
Solution:
\(\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}\)
= 2 x (- 1) – (- 5) x (4)
= – 2 + 20
= 18

12th Maths Chapter 4 Exercise 4.1 Question 2.
(i) \(\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}\)
(ii) \(\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}\)
Solution:
(i) \(\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}\)
= cosθ cosθ – (sinθ)(-sinθ)
= cos²θ + sin²θ
= 1

(ii) \(\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}\)
= (x² – x + 1) (x + 1) – (x + 1) (x – 1)
= x³ – x² + x + x² – x + 1 – x² + 1
= x³ – x² + 2

Question 3.
If \(A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\) then show that |2A|=|4A|
Solution:
\(A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\)
⇒ \(2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
L.H.S = |2A|
= \(2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
= – 24
4|A| = 4|\(\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|\)| = 4(2 – 8) = 4 x – 6 = – 24
∴ |2A| = 4|A|

Question 4.
\(A=\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right] \) , then show that |3A| = 27|A|
Solution:

Question 5.
Evaluate the following determinant:
(i) \(\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)
(ii) \(\left| \begin{matrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{matrix} \right| \)
(iii) \(\left| \begin{matrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} \right| \)
(iv) \(\left| \begin{matrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)
Solution:
(i) \(\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)

Question 6.
If \(\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right] \), find |A|
Solution:
|A| = \(\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right] \)
= 1(-9+12)-1(-18+15)-2(8-5)
= 0

Question 7.
Find the values of x, if
(i) \(\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
(ii)\(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
Solution:
(i) \(\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
⇒ 2 – 20 = 2x² – 24
⇒ x² = 3
⇒ x = ±\(\sqrt{3}\)

(ii) \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
or
2 × 5 – 4 × 3 = 5 × x – 2x × 3
⇒ x = 2

Question 8.
If \(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\), then x is equal to
(a) 6
(b) +6
(c) -6
(d) 0
Solution:
(b) \(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\)
⇒ x² – 36 = 36 – 36
⇒ x² = 36
⇒ x = ± 6