These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.

(3x² -9x + 5)^{9}

Solution:

Let y = (3x² -9x + 5)^{9}

∴ \(\frac { dy }{ dx }\) = (3x² -9x + 5)^{8}. \(\frac { d }{ dx }\)(3x² -9x + 5)

= 9(3x² -9x + 5)^{8} (6x – 9)

= 27 (3x² -9x + 5)^{8} (2x – 3)

Question 2.

sin³x + cos^{6} x

SoL

Let y = sin³x + cos^{6} x

∴ \(\frac { dy }{ dx }\) = 3 sin²x . cosx + 6cos5x (- sinx)

= 3 sinx cosx (sinx – 2 cos^{4}x)

Question 3.

(5x)^{3cos2x}

Solution:

Let y = (5x)^{3cos2x}

Taking logarithmon both sides,

∴ log y = 3 cos 2x log 5x

Differentiating both sides, w.r.t. x,

Question 4.

sin^{-1} (x\(\sqrt{x}\)), 0 ≤ x ≤ 1

Solution:

Question 5.

\(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}},-2<x<2\)

Solution:

Question 6.

\(\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0<x<\frac{\pi}{2}\)

Solution:

Question 7. \((\log x)^{\log x}, x>1\)

Solution:

Let y = \((\log x)^{\log x}\)

Taking logarithmon both sides,

∴ log y = log x(log log x)

Differentiating both sides, w.r.t. x,

\(\frac { 1 }{ y }\) \(\frac { dy }{ dx }\) = logx.\(\frac { 1 }{ log x }\).\(\frac { 1 }{ x }\) + \(\frac { 1 }{ x }\).log log x

∴ \(\frac { dy }{ dx }\) = (log x)^{log x} [\(\frac { 1 }{ x }\) + \(\frac { log log.x }{ x }\)]

Question 8.

cos (a cos x + b sin x), for sorne constant a and b.

Solution:

Let y = cos (a cosx + b sinx)

\(\frac { dy }{ dx }\) = sin (a cosx + h sinx).

\(\frac { dy }{ dx }\)(a cosx + b sinx)

= – sin (a cosx + b sinx) [- a sinx + b cosx]

= (a sinx – b cosx).sin (a cosx + b sinx)

Question 9.

(sin x – cos x)^{sin x-cos x}, \(\frac { π }{ 4 }\)< x < \(\frac { 3π }{ 4 }\)

Solution:

When \(\frac { π }{ 4 }\)< x < \(\frac { 3π }{ 4 }\), then sin x > cos x

so that sin x – cos x is positive.

Let y = (sinx – cosx)^{sin x-cos x}

Taking logarithm on both sides,

∴ logy = (sinx – cosx) log (sinx – cosx)

Differentiating both sides w.r.t. x,

\(\frac { 1 }{ y }\) \(\frac { dy }{ dx }\)

=( sinx – cosx)\(\frac { 1 }{ sin x-cosx }\) .\(\frac { d }{ dx }\)(sinx-cosx) + log (sinx – cosx). (cosx + sinx)

= (cosx + sinx) + log (sinx – cosx) . (cosx + sinx)

= (cosx + sinx) [1 + log (sinx – cosx)]

∴ \(\frac { dy }{ dx }\) = (sin x – cos x)^{sin x-cos x}(cosx + sinx) [1 + log(sinx – cosx)]

Question 10.

x^{x} + x^{a} + a^{x} + a^{a} for some fixed a > 0 and x > 0.

Solution:

Question 11.

\(x^{x^{2}-3}+(x-3)^{x^{2}}\), for x > 3

Solution:

Let u = x\(x^{x^{2}-3}\) and v = (x – 3)^{x²}

Differentiating w.r.t. x,

∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)

u = x^{x²-3}

Taking logarithm on bath sides,

∴ log u = (x² – 3)log

Differentiating both sides w.r.t. x,

Taking logarithm on bath sides,

∴ log v = x²log(x – 3)

Differentiating both sides w.r.t. x,

Question 12.

Find \(\frac { dy }{ dx }\), if y = 12(1 – cos t),

x = 10(t – sin t), \(\frac { – π }{ 2 }\) < t < \(\frac { π }{ 2 }\)

Solution:

Question 13.

Find \(\frac { dy }{ dx }\), if

y = sin^{-1} x + sin^{-1} \(\sqrt{1-x^{2}},-1 \leq x \leq 1\).

Solution:

Question 14.

If \(x \sqrt{1+y}+y \sqrt{1+x}\) = 0, for – 1 < x < 1, prove that \(\frac { dy }{ dx }\) = – \(\frac{1}{(1+x)^{2}}\)

Solution:

\(x \sqrt{1+y}+y \sqrt{1+x}\) = 0

\(x \sqrt{1+y}\) = – y\(\sqrt{1+y}\)

Squaring both sides,

x²(1 + y) = y²(1 + x)

x² + x²y = y² + y²x

x² – y² = y²x – x²y

(x – y)(x + y) = xy(x – y)

(x + y) = – xy

y + xy = – x

y(1 + x) = – x

∴ y = \(\frac { – x }{ 1+x }\)

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = \(\frac{(1+x)(-1)-(-x) 1}{(1+x)^{2}}\)

= \(\frac{-1-x+x}{(1+x)^{2}}\) = \(\frac{-1}{(1+x)^{2}}\), x ≠ – 1

Question 15.

If (x – a)² + (y – b)² = c², for some c > 0, prove that

\(\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}\)

is a constant, independent of a and b.

Solution:

Question 16.

If cos y = x cos (a + y), with

cos a ≠ ± 1, prove that \(\frac { dy }{ dx }\) = \(\frac{\cos ^{2}(a+y)}{\sin a}\)

Solution:

Question 17.

If x = a(cos t + t sin t) and y = a(sin t – t cos t), find \(\frac{d^{2} y}{d x^{2}}\)

Solution:

Question 18.

If f(x) = |x|³, show that f”(x) exists for all real x and find it.

Solution:

f(x) can be redefined as

f(x) = \(\left\{\begin{aligned}

x^{3}, & x \geq 0 \\

-x^{3}, & x<0 \end{aligned}\right.\) For x > 0 and x < 0, fix) is a polynomial function. Hence f(x) is differentiable for x > 0 and x < 0. ∴ For x > 0, f'(x) = 3x² and f”(x) = 6x

For x < 0, f'(x) = – 3x² and f”(x) = – 6x

or f”(x) = 6|x| exists for all x ∈ R

Question 19.

Using mathematical induction, prove that \(\frac { d }{ dx }\)(xⁿ) = \(n x^{n^{-1}}\) for all positive integers n.

Solution:

Hence P(k + 1) is true.

i.e., P(k + 1) is true whenever P(k) is true.

Hence by the principle of mathematical induction, \(\frac { d }{ dx }\)(xⁿ) = nx^{n-1} is true for positive integer n.

Question 20.

Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.

Solution:

sin (A + B) = sin A cos B + cos A sin B

Differentiating both sides w.r.t. x,

Question 21.

Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Solution:

Yes.

i. Let f(x) = |x – 1| + |x – 2|

Let g(x) = |x|, h(x) = x – 1, k(x) = x – 2.

Then g(x), h(x) and k(x) are continuous functions since h(x) and k(x) are polynomial functions and g(x) ¡s the modulus function.

at x = 1, since Lf’(1) ≠ Rf’(1)

Similarly we can show that f(x) ¡s not differentiable at x = 2.

Thus f(x) = |x – 1| + |x – 2| is continuous everywhere and not differentiable at exactly two points, namely at x = 1 or x = 2.

Question 22.

If y = \(\left|\begin{array}{ccc}

f(x) & g(x) & h(x) \\

l & m & n \\

a & b & c

\end{array}\right|\), prove that \(\frac { dy }{ dx }\) = \(\left|\begin{array}{ccc}

f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\

l & m & n \\

a & b & c

\end{array}\right|\)

Solution:

Question 23.

If y = \(e^{a \cos ^{-1} x}\), – 1 ≤ x ≤ 1, show that (1 – x²)\(\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y\) = 0

Solution:

f(x) = x + \(\frac { 1 }{ x }\)

f(x) is a continuous function in [1, 3]

Ax) is differentiable in(1,3)

f’(x) = 1 + \(\\frac{-1}{x^{2}}\) exists for x ∈ (1, 3)